Answer:
Is plausible that the successive throws are independent
Step-by-step explanation:
1) Table with info given
The observed values are given by the following table
__________________________________________________
First shot Made Second shot missed Total
__________________________________________________
Made 152 33 185
Missed 37 8 45
__________________________________________________
Total 189 41 230
2) Calculations and test
We are interested on check independence and for this we need to conduct a chi square test, the next step would be find the expected value:
Null hypothesis: Independence between two successive free throws
Alternative hypothesis: No Independence between two successive free throws
_____________________________________________________
First shot Made Second shot missed
_____________________________________________________
Made 189(185)/230=152.0217 41(185)/230=32.9783
Missed 189(45)/230=36.9783 41(45)/230=8.0217
_____________________________________________________
On this case all the expected values are higher than 5 and the sample size 230 is enough to apply the chi squared test.
3) Calculate the chi square statistic
The statistic for this case is given by:
[tex]\chi_{cal}^2 =\sum \frac{(O_i -E_i)}{E_i}[/tex]
Where O represent the observed values and E the expected values. Replacing the values that we got we have this
[tex]\chi_{cal}^2 =\frac{(152-152.0217)^2}{152.0217}+\frac{(33-32.9783)^2}{32.9783}+\frac{(37-36.9783)^2}{36.9783}+\frac{(8-8.0217)^2}{8.0217}=0.000003098+0.00001428+0.00001273+0.0.00005870=0.00008881[/tex]
Now with the calculated value we can find the degrees of freedom
[tex]df=(r-1)(c-1)=(2-1)(2-1)=1[/tex] on this case r means the number of rows and c the number of columns.
Now we can calculate the p value
[tex]p_v =P(\chi^2 >0.00008881)=0.9925[/tex]
On this case the pvalue is a very large value and that indicates that we can fail to reject the null hypothesis of independence. So is plausible that the successive throws are independent.
According to the Texas Department of State Health Services, there are approximately 17.7 million adults (age 18+) living in Texas, of which 1.45 million have diabetes. In 2010, 366,921 adult Texans were diagnosed with diabetes for the first time.
What is the incidence of diabetes in Texas? Express your answer in units of per 1000 persons per year
Answer:
20.73 ≈ 21 persons per year
Step-by-step explanation:
Data provided in the question:
Total population of Texas = 17.7 million
People having diabetes = 1.45 million
Number of new cases of diabetes diagnosed for the first time = 366,921
Now,
the incidence of diabetes in Texas
= [ Number of new cases diagnosed ] ÷ [ Total population ]
= 366,921 ÷ 17.7 million
= 366,921 ÷ 17,700,000
= 0.02073
thus,
the incidence of diabetes in Texas per 1000 persons
= 0.02073 × 1,000
= 20.73 ≈ 21 persons per year
A random sample of 49 text books purchased at a local bookstore showed an average price of $122 with a population standard deviation of $15. Let u (new) be the true mean cost of a text book sold by this store. Construct a confidence interval with a 90% degree of confidence. Clearly label the following:
a. Point estimate
b. Critical value,
c. Margin of error
d. Confidence interval
e. Interpretation (confidence statement).
Answer:
point estimate is $122critical value for the 90% confidence level (1.645)margin of error is $3.52590% confidence interval is $122±3.525there is 90% probability that true population average price of text books is in the range $122±$3.525Step-by-step explanation:
Confidence Interval can be calculated using P±ME where
P is the point estimate for the mean cost of a text book ( $122 )ME is the margin of error from the meanAnd margin of error (ME) can be calculated using the formula
ME=[tex]\frac{z*s}{\sqrt{N} }[/tex] where
z is the critical value for the 90% confidence level (1.645)s is the population standard deviation ($15) N is the sample size (49)Margin of error, ME=[tex]\frac{1.645*15}{\sqrt{49} }[/tex] = 3.525
Then 90% confidence interval is $122±3.525
To interpret this, there is 90% probability that true population average price of text books is in the range $122±$3.525
The accounting department analyzes the variance of the weekly unit costs reported by two production departments. A sample of 16 cost reports for each of the two departments shows cost variances of 2.3 and 5.6, respectively. Is this sample sufficient to conclude that the two production departments differ in terms of unit cost variance? Use a = .10.
Calculate the value of the test statistic (to 2 decimals).
Answer:
Test Statistic is 2.34
Step-by-step explanation:
File Attached. kindly view it.
A small town has 2100 inhabitants. At 8 AM, 80 people have heard a rumor. By noon half the town has heard it. At what time will 90% of the population have heard the rumor? (Do not round k in your calculation. Round the final answer to one decimal place.) hours after the beginning
Answer:
2.7 PM
Step-by-step explanation:
Since, the rate of spread is proportional to the product of fraction y of people who have heard the rumour and the fraction who have not heard,
[tex]\frac{dy}{dt}=ky(1-y)[/tex]
[tex]\frac{dy}{y(1-y)}=kdt[/tex]
[tex](\frac{1}{y}+\frac{1}{1-y})dy = kdt[/tex]
Integrating both sides,
[tex]\int (\frac{1}{y}+\frac{1}{1-y})dy = \int kdt[/tex]
[tex]\ln y - \ln (1-y) = kt + C[/tex]
[tex]\ln (\frac{y}{1-y}) = kt + C[/tex]
[tex]\frac{y}{1-y}=e^{kt + C}[/tex]
[tex]y = e^{kt+C} - y e^{kt+C}[/tex]
[tex]y(1+e^{kt+C}) = e^{kt+C}[/tex]
[tex]y = \frac{e^{kt+C}}{1+e^{kt+C}}[/tex]
If t = 0, ( at 8 AM ), y = [tex]\frac{80}{2100}[/tex]
[tex]\frac{80}{2100}= \frac{e^{0+C}}{1+e^{0+C}}[/tex]
[tex]\frac{4}{105}=\frac{e^C}{1+e^C}[/tex]
[tex]4 + 4e^C = 105e^C[/tex]
[tex]4 = 101e^C[/tex]
[tex]\implies e^C=\frac{4}{101}[/tex]
Now, at noon, i.e t = 4, y = [tex]\frac{1}{2}[/tex]
[tex]\frac{1}{2}=\frac{e^{4k}.\frac{4}{101}}{1+e^{4k}.\frac{4}{101}}[/tex]
[tex]\frac{1}{2}=\frac{4e^{4k}}{101+4e^{4k}}[/tex]
[tex]101 + 4e^{4k}=8 e^{4k}[/tex]
[tex]101 = 4e^{4k}[/tex]
[tex]\frac{101}{4}=e^{4k}[/tex]
[tex](\frac{101}{4})^\frac{1}{4} = e^k[/tex]
If [tex]y = \frac{90}{100}=\frac{9}{10}[/tex]
[tex]\frac{9}{10}= \frac{(\frac{101}{4})^\frac{t}{4}\times \frac{4}{101}}{1+(\frac{101}{4})^\frac{t}{4}\times \frac{4}{101}}[/tex]
Using graphing calculator,
t ≈ 6.722,
Hence, after 6.722 hours since 8 AM, i.e. on 2.7 PM ( approx ) the 90% of the population have heard the rumour.
Daniel is a front-desk manager at Refington Hotel, a mid-market hotel. During Daniel's shift last week, he received 29 customer complaints. A total of 428 guests had stayed in the hotel that week. Given this information, find out the errors per million opportunities.
Answer:
67,757 errors per million opportunities
Step-by-step explanation:
Assuming that each customer can only make a single complaint (1 error opportunity per customer), the number of errors per million opportunities (EPMO) is given by:
[tex]EPMO = \frac{complaints}{guests}*1,000,000 \\EPMO = \frac{29}{428}*1,000,000 \\EPMO = 67,757[/tex]
Refington hotel should expect 67,757 complaints per million guests.
Let R+ denote the set of positive real numbers. Let f : R × R+ → R be given by f(x, y) = x/y. (a) Is f an injective function? Prove your answer. (b) Is f a surjective function? Again, prove your answer. (c) Is f a bijection? Prove your answer.
Let [tex]f:\Bbb R\times\Bbb R^+\to\Bbb R[/tex].
a. [tex]f[/tex] is injective if [tex]f(x_1,y_1)=f(x_2,y_2)[/tex] for any two points [tex](x_1,y_1)[/tex] and [tex](x_2,y_2)[/tex], then the two points must be identical with [tex]x_1=x_2[/tex] and [tex]y_1=y_2[/tex].
[tex]f[/tex] is not injective because we can pick two points for which [tex]f[/tex] gives the same value:
[tex]f(2,1)=\dfrac21=2[/tex]
[tex]f(4,2)=\dfrac42=2[/tex]
b. [tex]f[/tex] is surjective if there exists [tex](x,y)[/tex] for which every point in the codomain [tex]\Bbb R[/tex] is obtained by [tex]f(x,y)[/tex].
This should be somewhat obvious if you consider 3 different cases:
[tex]f(x,y)<0[/tex] and can take on any negative real number for any choice of [tex]x<0[/tex][tex]f(x,y)=0[/tex] if and only if [tex]x=0[/tex][tex]f(x,y)>0[/tex] for any choice of [tex]x>0[/tex]So [tex]f[/tex] is surjective.
c. [tex]f[/tex] is bijective if it is both injective and surjective. The conclusions above show [tex]f[/tex] is not bijective.
Data on the pH of rain in Ingham County, Michigan, are as follows: 5.47 5.37 5.38 4.63 5.37 3.74 3.71 4.96 4.64 5.11 5.65 5.39 4.16 5.62 4.57 4.64 5.48 4.57 4.57 4.51 4.86 4.56 4.61 4.32 3.98 5.70 4.15 3.98 5.65 3.10 5.04 4.62 4.51 4.34 4.16 4.64 5.12 3.71 4.64 What proportion of the samples has pH below 5.0?
Answer:
[tex]\hat p =\frac{26}{39}=0.667[/tex]
Step-by-step explanation:
Some important concepts
A proportion refers to "the fraction of the total that possesses a certain attribute"
The data det ordered from the smallest to the largest value is:
3.1 ,3.71 ,3.71 ,3.74 ,3.98 ,3.98 ,4.15 ,4.16 ,4.16 ,4.32 ,4.34 ,4.51 ,4.51 ,4.56 ,4.57 ,4.57 ,4.57 ,4.61 ,4.62 ,4.63 ,4.64 ,4.64 ,4.64 ,4.64 ,4.86 ,4.96 ,5.04 ,5.11 ,5.12 ,5.37 ,5.37 ,5.38 ,5.39 ,5.47 ,5.48 ,5.62, 5.65 ,5.65 ,5.7
If we are interesed on the sample values below 5.0 we need to count how many values are below this number. If we do this we got that 26 numbers are below 5.0 and the total of numbers are 39.
so the proportion on this case would be"
[tex]\hat p =\frac{26}{39}=0.667[/tex]
To determine the proportion of rain samples with a pH below 5, count the number of data points less than 5.0 and divide it by the total number of data points. The pH value of the environment can vary greatly due to factors such as the presence of pollution.
Explanation:This question is about finding out the proportion of rain samples that have a pH below 5.0 in Ingham County, Michigan. To do this, you count the number of data points less than 5.0 and divide it by the total number of data points. The pH of the environment (rain, in this case) can vary greatly due to factors such as pollution, affecting the acidity or alkalinity of the rain. Normal rainwater has a pH between 5 and 6 due to the presence of dissolved CO₂ which forms carbonic acid. Anything below 7.0 is acidic and above 7.0 is alkaline.
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Mattel Corporation produces a remote-controlled car that requires three AA batteries. The mean life of these batteries in this product is 34 hours. The distribution of the battery lives closely follows the normal probability distribution with a standard deviation of 5.5 hours. As a part of their testing program Sony tests samples of 25 batteries.
What can you say about the shape of the distribution of sample mean?
What is the standard error of the distribution of the sample mean? (Round your answer to 4 decimal place.)
What proportion of the samples will have a mean useful life of more than 36 hours? (Round your answer to 4 decimal place.)
What proportion of the sample will have a mean useful life greater than 33.5 hours? (Round your answer to 4 decimal place.)
What proportion of the sample will have a mean useful life between 33.5 and 36 hours? (Round your answer to 4 decimal place.)
Answer:
0.0406, 0.8284,0.7887
Step-by-step explanation:
Given that Mattel Corporation produces a remote-controlled car that requires three AA batteries
X is N(34, 5.5)
Hence sample size of 25 would follow a t distribution with df = 24
This is because sample size <30
t distribution with df 24 would be bell shaped symmetrical about the mean and unimodal.
Std error of sample mean = std dev /sqrt n=[tex]\frac{5.5}{5} \\=1.1[/tex]
Prob (X>36) = [tex]P(t>\frac{36-34}{1.1} ) = P(t>1.82)\\= 0.04063[/tex]
i.e nearly 4.1% of the sample would have a mean useful life of more than 36 hours
X>33.5 implies [tex]t>-0.45[/tex]
=0.82837
=0.8284 proportion will have a mean useful life greater than 33.5 hours
Proportion between 33.5 and 36 hours
= [tex]0.3284+0.4593=0.7887[/tex]
The shape of the distribution of the sample mean is approximately normal due to the Central Limit Theorem. The standard error of the distribution of the sample mean can be calculated as 1.1 hours. Proportions of the sample mean falling above and between specific time intervals can be determined using z-scores and the Z-table.
Explanation:The shape of the distribution of the sample mean, in this case, is approximately normal. This is because the distribution of the battery lives closely follows the normal probability distribution. When taking a sample of 25 batteries, the Central Limit Theorem states that the distribution of the sample mean approaches a normal distribution regardless of the shape of the original distribution.
The standard error of the distribution of the sample mean can be calculated using the formula: standard deviation / square root of sample size. In this case, the standard deviation is 5.5 hours and the sample size is 25. Therefore, the standard error is 5.5 / √25 = 1.1 hours.
To determine the proportion of the samples that will have a mean useful life of more than 36 hours, we need to calculate the z-score first. The formula for z-score is: (sample mean - population mean) / standard error. Plugging in the given values, we get (36 - 34) / 1.1 = 1.82. By looking up the z-score in the Z-table, we find the corresponding proportion is approximately 0.0344.
Similarly, to find the proportion of the sample that will have a mean useful life greater than 33.5 hours, we calculate the z-score: (33.5 - 34) / 1.1 = -0.45. By looking up the z-score in the Z-table, we find the corresponding proportion is approximately 0.3264.
To determine the proportion of the sample that will have a mean useful life between 33.5 and 36 hours, we subtract the proportion of the sample that will have a mean useful life greater than 36 hours from the proportion of the sample that will have a mean useful life greater than 33.5 hours. Therefore, 0.3264 - 0.0344 = 0.2920.
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For population parameter μ = true average resonance frequency for all tennis rackets of a certain type, each of these is a confidence interval. (114.4, 115.6) and (114.1, 115.9) What is the value of the sample mean resonance frequency?
a. 114.5
b. 115.8
c. 114.1
d. 115.0
Answer:
X[bar]= 115
Step-by-step explanation:
Hello!
Every Confidence interval to estimate the population mean are constructed following the structure:
"Estimator" ± margin of error"
Wich means that the intervals are centered around the sample mean. To know the value of the sample mean you have to make the following calculation:
[tex]X[bar]= \frac{Upper bond + Low bond}{2}[/tex]
[tex]X[bar]= \frac{115.6+114.4}{2}[/tex] = 115
Since both intervals were calculated with the information of the same sample, you can choose either to calculate the sample mean.
I hope it helps!
The average US woman wears 515 chemicals on an average day from her makeup and toiletries. A random sample from California found that on average the California woman wears 325 chemicals per day with a standard deviation of 90.5.Which hypothesis test should be used to determine whether the sample contains less than the US average of wearing 515 chemicals per day?a. t-test for the population meanb. z-test for the population meanc. z-test for the population proportiond. t-test for the population proportion
Answer:
a. t-test for the population mean
Step-by-step explanation:
given that the average US woman wears 515 chemicals on an average day from her makeup and toiletries
A random sample from California found that on average the California woman wears 325 chemicals per day with a standard deviation of 90.5.
Note that here we have only sample std deviation and not population.
So better to use t test here.
Sample size is not given so whatever be the sample size here t test would be more appropriate.
Here we are testing for mean. So the correct answer is
a. t-test for the population mean
A farmer needs to enclose three sides of a garden with a fence (the fourth side is a cliff wall). The farmer has 43 feet of fence and wants the garden to have an area of 228 sq-feet. What should the dimensions of the garden be? (For the purpose of this problem, the width will be the smaller dimension(needing two sides); the length will be the longer dimension (needing one side). Additionally? The length should be as long as possible.)
Answer:
9.5 ft wide by 24 ft long
Step-by-step explanation:
A graphing calculator shows there are two solutions to the system of equations ...
2x + y = 43 . . . . . . fence length when x= width
xy = 228 . . . . . . . area
The solutions are ...
(width, length) = (9.5, 24) or (12, 19)
Since we want the length as long as possible, the choice of dimensions is ...
length = 24 feet; width = 9.5 feet.
The speed limit posted on a local highway is 75 mph. Is the average speed on that stretch of highway significantly more than 75mph? Forty vehicle’s speeds were recorded by speed detection devices. What would be the correct alternative hypothesis?
Answer: [tex]H_a: \mu >75[/tex]
Step-by-step explanation:
Alternative hypothesis [tex](H_a)[/tex]: It is a statement which always indicates that there is a significant difference between the groups being tested.
It always contains by < , > or ≠ sign .
Given claim : Is the average speed on that stretch of highway significantly more than 75mph?
Here objective is whether the average speed on that stretch of highway significantly more than 75mph.
Let [tex]\mu[/tex] be the population mean speed on that stretch of highway significantly more than 75mph.
Then, [tex]H_a: \mu >75[/tex]
Hence , the correct alternative hypothesis would be : [tex]H_a: \mu >75[/tex]
The correct alternative hypothesis for this question is that the average speed on the highway is significantly greater than 75mph.
Explanation:The correct alternative hypothesis for this question would be that the average speed on that stretch of highway is significantly greater than 75mph.
To determine if the average speed is significantly more than 75mph, a statistical test can be conducted. This test would compare the speeds of the 40 recorded vehicles to the 75mph speed limit and calculate the average speed. If the average speed is significantly higher than 75mph, it would support the alternative hypothesis.
For example, if the average speed of the 40 recorded vehicles is found to be 80mph with a small p-value, it would indicate that the average speed on that stretch of highway is significantly greater than 75mph.
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Computer output from a regression analysis is provided. Coefficients: Estimate Std. Error t value p-value (Intercept) 7.2960 14.5444 0.502 0.62200 X 1.6370 0.5453 3.002 0.00765 We want to do the hypothesis test to see if the slope in the population is different from zero? That is, do the hypothesis test to see if we have a statistically significant linear relationship. What is your decision on the hypothesis test and why? Use a level of significance of .05.
Answer:
Step-by-step explanation:
Hello!
You need to test the hypothesis that the slope of the regression is cero.
I've run in the statistic software the given data for Y and X and estimated the regression line:
Yi= 7.82 -1.60Xi
Where
a= 7.82
b=-1.60
Sb= 3.38
The hypothesis is:
H₀: β = 0
H₁: β ≠ 0
α: 0.05
This is a two-tailed test, the null hypothesis states that the slope of the regression is cero, this means that if the null hypothesis is true, there is no linear regression between Y and X.
The statistic for this test is a Student-t
t= b - β ~t[tex]_{n-2}[/tex]
Sb
The critical values are:
Left: [tex]t_{n-2; \alpha /2} = t_{2; 0.025} = -4.303[/tex]
Right: [tex]t_{n-2; \alpha /2} = t_{2; 0.975} = 4.303[/tex]
t= -1.60 - 0 = -0.47
3.38
the p-value is also two-tailed, you can calculate it by hand:
P(t ≤ -0.47) + (1 - P(t ≤ 0.47) = 0.3423 + (1 - 0.6603) =0.6820
With the level of significance of 5%, the decision is to not reject the null hypothesis. This means that the slope of the regression is equal to cero, i.e. there is no linear regression between the two variables.
I hope this helps!
The hypothesis test, utilizing a t-statistic of 3.002 and a p-value of 0.00765, rejects the null hypothesis, demonstrating a statistically significant linear relationship. The slope coefficient (β₁ = 1.637) suggests a moderately strong relationship.
Hypothesis Test for a Linear Relationship
In this hypothesis test, we are trying to determine whether there is a statistically significant linear relationship between the independent and dependent variables. The null hypothesis (H₀) states that there is no linear relationship (β₁ = 0), while the alternative hypothesis (H₁) states that there is a linear relationship (β₁ ≠ 0).
The test statistic used in this case is the t-statistic, which is calculated as the ratio of the estimated slope (β₁) to its standard error. The p-value is the probability of obtaining a test statistic as extreme or more extreme than the observed test statistic, assuming that the null hypothesis is true.
In this particular example, the t-statistic is 3.002 and the p-value is 0.00765. The level of significance, which is the threshold for rejecting the null hypothesis, is typically set at 0.05.
Since the p-value (0.00765) is less than the level of significance (0.05), we reject the null hypothesis. This means that we have statistically significant evidence to conclude that there is a linear relationship between the independent and dependent variables. In other words, the slope of the regression line is not equal to zero, indicating that there is a change in the dependent variable as the independent variable changes.
Interpretation of Results
The rejection of the null hypothesis in this case indicates that there is a statistically significant linear relationship between the independent and dependent variables. This means that we can be confident that the observed relationship is not due to chance. The magnitude of the t-statistic (3.002) suggests that the relationship is moderately strong.
The interpretation of the slope coefficient (β₁) is that for every one-unit increase in the independent variable, the dependent variable is expected to increase by 1.637 units, on average. The standard error (0.5453) indicates the variability of the slope estimate.
In conclusion, the hypothesis test provides strong evidence to support the existence of a statistically significant linear relationship between the independent and dependent variables. The magnitude of the slope coefficient indicates that the relationship is moderately strong.
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According to the United Nations, in the year 2002, the population of the world was 6.1 billion people and was growing at an annual rate of about 1.5%. If this pattern were to continue, then every year, the population would be 1.015 times the population of the previous year. Thus, if P(t) is the world population (in billions) t years after the base year 2002.
(a) What was the population in 2004?
(b) What will the population be in 2010?
Answer:
(a) 6.2843 billion
(b) 6.8716 billion
Step-by-step explanation:
Since there is a constant growth rate of 1.5% per year, and the population in 2002 was 6.1 billion people. The general equation for the total world population, in billions, after 2002 is:
[tex]P(t) = 6.1*(1+0.015)^t[/tex]
With t being the time, in years, after 2002.
a) What was the population in 2004?
[tex]t=2004-2002=2\\P(2) = 6.1*(1.015)^2\\P(2) = 6.2843 \ billion[/tex]
b) What was the population in 2010?
[tex]t=2010-2002=8\\P(8) = 6.1*(1.015)^8\\P(8) = 6.8716 \ billion[/tex]
Jason had $32. He spent all the money buying four CDs for x dollars each and two magazines for y dollars each. If Jason had bought five CDs and two magazines, he would have run short by $4. The following system of equations models this scenario:
4x + 2y = 32
5x + 2y = 36
Use the system of equations to solve for x and y.
(4, 8)
(5, 6)
(8, 4)
(6, 5)
Answer:
(4,8)
Step-by-step explanation:
We have the two equations:
[tex]$ 4x + 2y = 32 \hspace{15mm} .....(1) $[/tex] and
[tex]$ 5x + 2y = 36 \hspace{15mm} .....(2) $[/tex]
Subtracting (1) from (2):
⇒ 5x + 2y -4x - 2y = 36 - 32
⇒ x = 4
Substituting the value of x in (1), we get:
4(4) + 2y = 32
⇒ 2y = 16
⇒ y = 8
We write the solution in ordered pair as (x,y) = (4,8).
Answer:
The answer is the ordered pairs (4,8) because we used the system equations to solve for x and y.
Step-by-step explanation:
In a study of the accuracy of fast food drive-through orders, one restaurant had 32 orders that were not accurate among 367 orders observed. Use a 0.05 significance level to test the claim that the rate of inaccurate orders is equal to 10%. Does the accuracy rate appear to be acceptable?
Identify the rest statistic for this hypothesis test. Round to two decimal places.
Identify the P-value for this hypothesis test. Round to two decimal places.
Identify the conclusion for this hypothesis tes.
Does the accuracy rate appear to be acceptable?
Answer:
Null hypothesis:[tex]p=0.1[/tex]
Alternative hypothesis:[tex]p \neq 0.1[/tex]
[tex]z=\frac{0.087 -0.1}{\sqrt{\frac{0.1(1-0.1)}{367}}}=-0.83[/tex]
[tex]p_v =2*P(z<-0.83)=0.41[/tex]
The p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of orders that were not accurate is not significant different from 0.1 or 10% .
and the accuracy of the test yes is acceptable since the p value obtained is large enough to fail to reject the null hypothesis.
Step-by-step explanation:
Data given and notation n
n=367 represent the random sample taken
X=32 represent the orders that were not accurate
[tex]\hat p=\frac{32}{367}=0.087[/tex] estimated proportion of orders that were not accurate
[tex]p_o=0.1[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95
z would represent the statistic (variable of interest)
[tex]p_v{/tex} represent the p value (variable of interest)
Concepts and formulas to use
We need to conduct a hypothesis in order to test the claim that the rate of inaccurate orders is equal to 10%:
Null hypothesis:[tex]p=0.1[/tex]
Alternative hypothesis:[tex]p \neq 0.1[/tex]
When we conduct a proportion test we need to use the z statisitc, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.087 -0.1}{\sqrt{\frac{0.1(1-0.1)}{367}}}=-0.83[/tex]
Statistical decision
It's important to refresh the p value method or p value approach . "This methos is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.
The significance level provided [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a bilateral test the p value would be:
[tex]p_v =2*P(z<-0.83)=0.41[/tex]
So the p value obtained was a very high value and using the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of orders that were not accurate is not significant different from 0.1 or 10% .
A staff member at UF's Wellness Center is interested in seeing if a new stress reduction program will lower employees high blood pressure levels. Twenty people are selected and have their blood pressure measured. Each person then participates in the stress reduction program. One month after the stress reduction program, the blood pressure levels of the employees were measured again. Did the program reduce the average blood pressure level? The 95% confidence interval was (5.6, 10.2). What can we expect will be the p-value for a two sided test using this data?
Answer:
The average blood pressure were higher than after
Step-by-step explanation:
As twenty people are selected and their blood pressure measured. After one month as stress reduction program BP were measured .Confidence interval is 95% so we conduct that blood pressure were higher than after reduction program.
Answer:
The p-value should be smaller than 0.05.
Step-by-step explanation:
The only information we have is the 95% confidence interval for the difference of means. As the lower bound is positive, we are 95% confident that the reduction program had a effect in the blood pressure level.
Then, as the program had a effect in the blood pressure level, we know that the null hypothesis, that states that the blood pressure levels would no change significantly, is rejected.
If the significance level is 0.05, according to the confidence of the interval, to reject the null hypothesis, the p-value had to be lower than the significance level.
Then, the p-value should be lower than 0.05.
please help
1 through 5
Answer:
Step-by-step explanation:
A random sample of size n1 = 16 is selected from a normal population with a mean of 75 and standard deviation of 8. A second random sample of size n2 = 9 is taken independently from another normal population with mean 70 and standard deviation of 12. Let X1 and X2 be the two sample means. Find
(a) The probability that X1 − X2 exceeds 4.
(b) The probability that 3.5 < X1 − X2 < 5.5.
Answer:
the answer is in the attached image below
Step-by-step explanation:
The sum of independent normally distributed random variable is also a normally distributed random variable. The needed probabilities are given as:
(a) The probability that X1 − X2 exceeds 4 is 0.5279 approx(b) The probability that 3.5 < X1 − X2 < 5.5 is 0.0836 approxWhat is the distribution of random variable which is sum of normal distributions?Suppose that a random variable X is formed by n mutually independent and normally distributed random variables such that:
[tex]X_i = N(\mu_i , \sigma^2_i) ; \: i = 1,2, \cdots, n[/tex]
And if
[tex]X = X_1 + X_2 + \cdots + X_n[/tex]
Then, its distribution is given as:
[tex]X \sim N(\mu_1 + \mu_2 + \cdots + \mu_n, \: \: \sigma^2_1 + \sigma^2_2 + \cdots + \sigma^2_n)[/tex]
For the given case, let we take:
[tex]X_1[/tex] = Random variable assuming values of sample 1[tex]X_2[/tex] = Random variable assuming values of sample 2Then, we have:
[tex]X_1 \sim N(\mu = 75, \sigma^2 = (8)^2 = 64)\\\\X_2 \sim N(\mu = 70, \sigma^2 = (12)^2 = 144)\\[/tex]
Then, The random variable [tex]-X_2[/tex] has all negative values than of [tex]X_2[/tex], so its mean will also become negative, but standard deviation would be same(since its measure of spread which would be same for [tex]-X_2[/tex] .
Thus, [tex]-X_2 \sim N(-70, 12^2) = N(-70, 144)[/tex]
Thus, we get:
[tex]X = X_1 - X_2\\\\X \sim N(75 -70, \sigma^2 = 64 + 144)\\\\X \sim N(5, 208)[/tex]
Thus, mean of X is 5, and standard deviation is
[tex]\sqrt{208} \approx 14.42[/tex]
(positive root since standard deviation is non negative quantity)
Thus, calculating the needed probability with the use of z-scores:
Case 1: P( X1 − X2 exceeds 4)P(X > 4)
Converting to standard normal distribution, we get
[tex]Z = \dfrac{X - \mu}{\sigma} \approx \dfrac{X - 5}{14.42}\\\\P(X > 4) \approx P(Z > \dfrac{4-5}{14.42}) \approx P(Z > -0.07) = 1-P(Z \leq -0.07)\\[/tex]
Using the z-tables, the p value for z = -0.07 is 0.4721
p value for Z = z gives [tex]P(Z \leq z) = p[/tex]
And therefore,
[tex]P(Z \leq -0.07) = 0.4721\\\\\rm and\: thus\\\\P(X > 4) = P(Z > -0.07) = 1 - P(Z \leq -0.07) = 1 - 0.4721 = 0.5279[/tex]
b) The probability that 3.5 < X1 − X2 < 5.5.P( 3.5 < X1 − X2 < 5.5 ) = P( 3.5 < X < 5.5)
Using z scores, we get:
[tex]P( 3.5 < X < 5.5) = P(X < 5.5) - P(X < 3.5) \\\\P(3.5 < X < 5.5) \approx P(Z < \dfrac{5.5-5}{14.2}) - P(Z < \dfrac{3.5 - 5}{14.2})\\\\P(3.5 < X < 5.5) \approx P(Z < 0.113) - P(Z < -0.104)[/tex]
Using the z-tables, p value for z = 0.113 is 0.5438
and p value for z = -0.104 is 0.4602
Thus,
[tex]P(3.5 < X < 5.5) \approx P(Z < 0.113) - P(Z < -0.104)\\P(3.5 < X < 5.5) \approx 0.5438 - 0.4602 = 0.0836[/tex]
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A manager must select three coders from her group to write three different software projects. There are 7 junior and 3 senior coders in her group. The first project can be written by any of the coders. The second project must be written by a senior person and the third project must be written by a junior person. How many ways are there for her to assign the three coders to the projects if no person can be assigned to more than one project?
Answer:
The total number of ways are 168.
Step-by-step explanation:
Consider the provided information.
There are 7 junior and 3 senior coders in her group.
The first project can be written by any of the coders. The second project must be written by a senior person and the third project must be written by a junior person.
For second project we have 3 choices and for third project we have 7 choices.
Now there are 2 possible case:
Case I: If first and second coder is senior, then the total number of ways are:
[tex]3\times 2\times 7=42[/tex]
Case II: If first and third coder is junior, then the total number of ways are:
[tex]7\times 3\times 6=126[/tex]
Hence, the total number of ways are: 42+126=168
The manager has 180 different ways to assign the three coders to the three different projects based on the given conditions.
Explanation:The question is about the number of ways to assign three coders to three different projects, respecting certain conditions. This is a combinatorics problem. The first coder can be selected from any coder in the group (10 coders) so we have ten options. The second coder must be a senior so we have 1×3 options (as we have already selected one coder for the first project and we can't assign them to more than one project). Similarly, the third coder must be a junior from the remaining 9 coders (3 seniors and 6 juniors), hence we have 1×6 options for this choice. So, the total number of ways to assign the coders is the product of these options: 10×3×6 = 180 different ways.
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A Harris Interactive Survey for InterContinental Hotels and Resorts asked respondents, "When traveling internationally, do you generally venture out on your own to experience culture, or stick with your tour group and itineraries?" The survey found that 23% of the respondents stick with their tour group.
In a sample of 6 internationals travellers, what is the probability that 2 will stick with their tour group?
Answer:the probability that 2 will stick with their tour group is 0.21
Step-by-step explanation:
The survey found that 23% of the respondents stick with their tour group. This means that the probability of success,
p is 23/100 = 0.23
Probability of failure, q = 1 - p = 1 - 0.23
q = 0.77
Number of international travellers sampled, n is 6
Assuming a binomial distribution for the responses, the formula for binomial distribution is
P(x = r) = nCr × q^(n-1) × p^r
To determine the probability that 2 will stick with their tour group, we would determine
P( x= r = 2). It becomes
6C2 × 0.77^(6-1) × 0.23^2
= 15 × 0.77^5 × 0.0529
= 0.21
The probability that exactly 2 out of 6 international travelers will stick with their tour group, given that the probability of an individual sticking with the group is 23%, is approximately 27.81%.
The problem described involves calculating the probability of a specific number of successes in a series of independent trials and is an example of a binomial probability problem. In this case, with the probability that a person sticks with their tour group being 23%, or 0.23, and a sample of 6 international travelers, we want to find the probability that exactly 2 out of these 6 travelers stick with their group. This can be calculated using the binomial probability formula:
P(X = k) = [tex]C(n, k) \times p^k \times (1-p)^{n-k}[/tex]
Where:
P(X = k) is the probability of k successes in n trials
C(n, k) is the combination of n things taken k at a time
p is the probability of success on a single trial
n is the number of trials
k is the number of successes in n trials
For our problem:
P(X = 2) = C(6, 2) x [tex]0.23^2[/tex]x[tex](1-0.23)^{6-2}[/tex]
Calculating the combination, we have C(6, 2) = 6! / (2! x (6-2)!).
Thus:
P(X = 2) = 15 x [tex](0.23^2)[/tex] x [tex](0.77^4)[/tex]
P(X = 2) = 15 x 0.0529 x 0.3515
P(X = 2) = 0.2781
Therefore, the probability that exactly 2 of the 6 international travelers will stick with their tour group is approximately 0.2781, or 27.81%.
Suppose that a magnet high school includes grades 11 and 12, with half of the students in each grade. 60% of the senior class and 10% of the junior class are taking calculus. Suppose a calculus student is randomly selected to accompany the math teachers to a conference. What is the probability that the student is a junior? (Enter your answer as a fraction.)
Answer: Our required probability is [tex]\dfrac{1}{7}[/tex]
Step-by-step explanation:
Since we have given that
P(Junior ) = [tex]\dfrac{1}{2}[/tex]
P(Senior) = [tex]\dfrac{1}{2}[/tex]
Let the given event be 'C' taking calculus.
P(C|J) = 10% = 0.10
P(C|S) = 60% = 0.60
We need to find the probability that the student is a junior.
So, our required probability is given by
[tex]P(J|C)=\dfrac{P(J).P(C|J)}{P(S).P(C|S)+P(J).P(C|J)}\\\\P(J|C)=\dfrac{0.5\times 0.1}{0.5\times 0.1+0.5\times 0.6}\\\\P(J|C)=\dfrac{0.05}{0.05+0.3}\\\\P(J|C)=\dfrac{0.05}{0.35}\\\\P(J|C)=\dfrac{5}{35}\\\\P(J|C)=\dfrac{1}{7}[/tex]
Hence, our required probability is [tex]\dfrac{1}{7}[/tex]
The mean life span of a brand name tire is 50,000 miles. Assume that the life spans of the tires are normally distributed, and the population standard deviation is 800 miles.
a. If you select one tire, what is the probability that its life span is less than 48,500 miles?
b. If you select 100 tires, what is the probability that their mean life span is more than 50,200 miles?
Answer:
a) [tex]P(X <48500)=0.0304[/tex]
b) [tex]P(\bar X>50200)=1-0.994=0.0062[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Let X the random variable that represent the mean life span of a brand name tire, and for this case we know the distribution for X is given by:
[tex]X \sim N(\mu=50000,\sigma=800)[/tex]
Part a
We want this probability:
[tex]P(X<48500)[/tex]
The best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\sigma}[/tex]
If we apply this formula to our probability we got this:
[tex]P(X <48500)=P(Z<\frac{48500-50000}{800})=P(Z<-1.875)=0.0304[/tex]
Part b
Let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
On this case [tex]\bar X \sim N(50000,\frac{800}{\sqrt{100}})[/tex]
We want this probability:
[tex]P(\bar X>50200)=1-P(\bar X<50200)[/tex]
The best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
If we apply this formula to our probability we got this:
[tex]P(\bar X >50200)=1-P(Z<\frac{50200-50000}{\frac{800}{\sqrt{100}}})=1-P(Z<2.5)[/tex]
[tex]P(\bar X>50200)=1-0.994=0.0062[/tex]
Final answer:
To determine the probability that a tire's lifespan is less than a certain value, you calculate its Z-score and look up the probability in a standard normal distribution table. For sample means, use the standard error to calculate Z-scores. In hypothesis testing, compare the p-value to α to decide on the claim.
Explanation:
Calculating Probabilities for Normally Distributed Tires' Lifespan
To calculate the probability that a single tire has a lifespan of less than 48,500 miles, we can use the Z-score formula. The formula for the Z-score is (X - μ) / σ, where X is the value we are looking at (48,500 miles), μ is the mean (50,000 miles), and σ is the population standard deviation (800 miles). Calculating the Z-score gives us (48,500 - 50,000) / 800 = -1.875. We can then look up this Z-score on a standard normal distribution table or use a calculator to find the corresponding probability.
For part b, looking at the probability for the mean of 100 tires being more than 50,200 miles, we would use the standard error of the mean, σ/√ n, where n is the sample size (100 in this case). We calculate the Z-score with this new standard deviation and find the probability corresponding to that Z-score using the same method.
It is essential to remember that for larger samples, the standard deviation of the mean decreases, and the calculations have to reflect this change.
Hypothesis Testing for Tires' Lifespan
Using a hypothesis test, we can determine if there is enough evidence to support or reject a claim about the population mean. In a hypothesis test, we compare the p-value to the level of significance (α = 0.05). If the p-value is lower than α, we reject the null hypothesis. In the example given with an alpha of 0.05 and a p-value of 0.0103, we would reject the null hypothesis, concluding that the average lifespan of the tires is likely less than the claimed 50,000 miles.
In 2008 the Pew Research Center interviewed a random sample of 5,566 registered voters and found that 36% were Democrats, 27% Republicans, and 37% Independents.
Let’s assume that this was the actual distribution for political party affiliation for all registered voters in 2008.
We can use a chi-square goodness-of-fit test to answer which of one the following research questions?
a. Is the proportion of Democrats larger than the proportion of Republicans this year?
b. Is the percentage of registered voters that are Independent higher now than the 37% reported in 2008?
c. Is the distribution of political party affiliation for all registered voters in 2012 the same as stated for 2008?
Final answer:
The research question that can be answered using a chi-square goodness-of-fit test is option c: Is the distribution of political party affiliation for all registered voters in 2012 the same as stated for 2008?
Explanation:
The research question that can be answered using a chi-square goodness-of-fit test in this scenario is option c: Is the distribution of political party affiliation for all registered voters in 2012 the same as stated for 2008?
In a chi-square goodness-of-fit test, we compare the observed frequencies (in this case, the distribution of political party affiliation in 2008) with the expected frequencies (the distribution we are testing, which is the distribution of political party affiliation in 2012). If the observed frequencies differ significantly from the expected frequencies, we can conclude that there is a significant difference between the two distributions.
In this case, the observed distribution is 36% Democrats, 27% Republicans, and 37% Independents in 2008. We want to test if this distribution is the same as the distribution in 2012. By performing a chi-square goodness-of-fit test, we can determine if there is a significant difference between the two distributions.
Find the vertices and foci of the hyperbola with equation quantity x plus 4 squared divided by 9 minus the quantity of y plus 3 squared divided by 16 = 1.
Answer:
vertices: (-7, -3), (-1, -3)foci: (-9, -3), (1, -3)Step-by-step explanation:
For a hyperbola of the form ...
[tex]\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1[/tex]
The vertices are located at (h±a, k), and the foci are located at (h±c, k), where ...
[tex]c=\sqrt{a^2+b^2}[/tex]
Here, we have (h, k) = (-4, -3), a=3, b=4, and c=√(9+16) = 5.
So, the points of interest are ...
vertices: (-4±3, -3) . . . . shown red on the graphfoci: (-4±5, -3) . . . . . . . . shown green on the graphAnswer:
previous was correct
Step-by-step explanation:
If 1/3 of the school is 6th graders and 1/2 are girls what fraction are 6th grade girls
Answer: 1/6
Step-by-step explanation:
Suppose that the bacteria in a colony grow unchecked according to the Law of Exponential Change. The colony starts with 1 bacterium and triples in number every 20 minutes. How many bacteria will the colony contain at the end of 24 hours?
Answer: There are 2.25×10³⁴ bacteria at the end of 24 hours.
Step-by-step explanation:
Since we have given that
Number of bacteria initially = 1
It triples in number every 20 minutes.
So, [tex]\dfrac{20}{60}=\dfrac{1}{3}[/tex]
So, our equation becomes
[tex]y=y_0e^{\frac{1}{3}k}\\\\3=1e^{\frac{1}{3}k}\\\\\ln 3=\dfrac{1}{3}k\\\\k=\dfrac{1.099}{0.333}=3.3[/tex]
We need to find the number of bacteria that it will contain at the end of 24 hours.
So, it becomes,
[tex]y=1e^{24\times 3.3}\\\\y=e^{79.1}\\\\y=2.25\times 10^{34}[/tex]
Hence, there are 2.25×10³⁴ bacteria at the end of 24 hours.
The colony will contain approximately 3,486,784,401 bacteria at the end of 24 hours.
Explanation:The bacteria colony starts with 1 bacterium and triples in number every 20 minutes. To determine the number of bacteria at the end of 24 hours, we need to calculate the number of 20-minute intervals in 24 hours. There are 24 hours in a day, so there are 24 intervals of 20 minutes in a day. Therefore, the bacteria would triple in number 24 times. Starting with 1 bacterium, after 24 intervals, the number of bacteria would be:
1 * 3^24 = 1 * 3,486,784,401
So, at the end of 24 hours, the colony would contain approximately 3,486,784,401 bacteria.
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Identify the asymptotes of each function. 1/(x-3) - 6
[tex]\bf \cfrac{1}{x-3}-6\implies \stackrel{\textit{using the LCD of x-3}}{\cfrac{1-(x-3)6}{x-3}}\implies \cfrac{1-6x+18}{x-3}\implies \cfrac{-6x+19}{x-3}[/tex]
for the vertical asymptote, we simply zero out the denominator and solve for "x"
x - 3 = 0
x = 3 <---- that's the only vertical asymptote
for the horizontal asymptote, well, let's notice the degrees of the numerator and denominator, in this case, the degree of the numerator is 1, and the degree of the denominator is 1, thus when that occurs, the horizontal asymptote occurs at the fraction from the leading terms' coefficients.
[tex]\bf \cfrac{-6x+19}{1x-3}\implies \stackrel{\textit{horizontal asymptote}}{\cfrac{-6}{1}\implies -6 = y}[/tex]
In a sample of 83 walking canes, the average length was found to be 34.9in. with a standard deviation of 1.5. Give a point estimate for the population standard deviation of the length of the walking canes. Round your answer to two decimal places, if necessary.
Answer:
The point estimate for the population standard deviation of the length of the walking canes is 0.16.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].
In this problem
The point estimate of the standard deviation is the standard deviation of the sample.
In a sample of 83 walking canes, the average length was found to be 34.9in. with a standard deviation of 1.5. By the Central Limit Theorem, we have that:
[tex]s = \frac{1.5}{\sqrt{83}} = 0.16[/tex]
The point estimate for the population standard deviation of the length of the walking canes is 0.16.
According to a study conducted by an organization, the proportion of Americans who were afraid to fly in 2006 was 0.10. A random sample of 1 comma 400 Americans results in 154 indicating that they are afraid to fly. Explain why this is not necessarily evidence that the proportion of Americans who are afraid to fly has increased.
Final answer:
The slightly higher proportion of Americans who indicated they are afraid to fly in a random sample is not necessarily evidence of an increased fear of flying overall. This variation could be due to sample variability, and without statistical analysis like hypothesis testing or confidence interval construction, one cannot conclude a genuine increase in fear of flying among Americans.
Explanation:
A study conducted by an organization found that the proportion of Americans who were afraid to fly in 2006 was 0.10. When a random sample of 1,400 Americans results in 154 individuals indicating that they are afraid to fly, it might initially seem like evidence that the proportion of Americans afraid to fly has increased. However, this is not necessarily indicative of an overall trend. To understand why we need to consider statistical variability and the concept of a confidence interval.
Statistical fluctuations in samples can lead to results that differ from the actual population proportion. In this case, the sample proportion of individuals afraid to fly is roughly 0.11 (154/1400), only slightly higher than the reported 0.10 from 2006. Without conducting a hypothesis test or constructing a confidence interval around the sample proportion, it's not possible to definitively say whether this difference is statistically significant or just due to random chance.
Furthermore, sample size plays a crucial role in the reliability of estimates. Although a sample size of 1,400 may seem large, the inherent randomness in sample selection could still lead to results differing from the true population proportion. In summary, a slight increase in the sample proportion of individuals afraid to fly, compared to a previous study, is not conclusive evidence of a trend without further statistical analysis to support such a claim.
The data from a sample of 1,400 Americans revealing 154 individuals who are afraid to fly does not necessarily indicate an increase in the proportion of Americans with this fear compared to 2006. Statistically significant evidence through a hypothesis test is required to make such a conclusion since sample data can fluctuate due to random chance.
Explanation:A random sample of 1,400 Americans resulted in 154 indicating that they are afraid to fly, which might suggest an increase from the 2006 study that showed a 0.10 proportion of Americans had this fear. However, this is not necessarily evidence that the proportion of Americans who are afraid to fly has increased due to potential sampling error or the natural fluctuation inherent in sample data. To determine if there is a statistically significant increase, one would need to perform a hypothesis test comparing the sample proportion to the known proportion of 0.10.
Variability can occur in survey results, and a single sample might not represent the entire population accurately. Also, the difference observed might be a result of random chance. Therefore, it's essential to conduct statistical tests to infer whether the observed change is meaningful and not due to sampling variability.