Median: 133.5
Lower/First quartile: 128
Upper/Third quartile: 139
Interquartile range: 139 - 128 = 11
so your interquartile range is 11
SurveyMonkey is a popular website for hosting online surveys and from time to time they will also create surveys based on a subject they are interested in answering questions about. Given the widespread adoption of dating sites and apps, they wanted to learn how people feel about them. It was found that out of 400 responses, 100 have used or are currently using online dating services. The average age of those that were using online dating services was 35 years old with a sample standard deviation of 12 years. What distribution should you use to compute a confidence interval for the average age of people using online dating services
Answer:
So we use the t-distribution to compute a confidence interval for the average age of people using online dating services
Step-by-step explanation:
Confidence interval for a mean.
We have to decide between the t-distribution and the z-distribution.
T-distribution: We use the sample standard deviation.
Z-distribution: We use the population standard deviation.
In this problem:
The average age of those that were using online dating services was 35 years old with a sample standard deviation of 12 years.
So we use the t-distribution to compute a confidence interval for the average age of people using online dating services
Basic Computation: Setting Hypotheses Suppose you want to test the claim that a population mean equals 30 .
(a) State the null hypothesis. ___________________________
(b) State the alternate hypothesis if you have no information regarding how the population mean might differ from 30. _________________
(c) State the alternate hypothesis if you believe (based on experience or past studies) that the population mean may be greater than 30. _______________
(d) State the alternate hypothesis if you believe (based on experience or past studies) that the population mean may not be as large as 30. ______________
Answer:
Part a
Null hypothesis: [tex]\mu =30[/tex]
Part b
Alternative hypothesis: [tex]\mu \neq 30[/tex]
Part c
Alternative hypothesis: [tex]\mu > 30[/tex]
Part d
Alternative hypothesis: [tex]\mu < 30[/tex]
Step-by-step explanation:
Previous concepts
A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".
The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".
The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".
Solution to the problem
For this case we want to test if the true mean is equal to 30 or no.
Part a
Null hypothesis: [tex]\mu =30[/tex]
Part b
Alternative hypothesis: [tex]\mu \neq 30[/tex]
Part c
Alternative hypothesis: [tex]\mu > 30[/tex]
Part d
Alternative hypothesis: [tex]\mu < 30[/tex]
The domain of the function is
The range of the function is
Answer:
Domain is where the function is defined. In other words, all valid values of x that allows us to find y.
Range is values of y for the function.
Step-by-step explanation:
Here is the example of function [tex]y = \sqrt{x}[/tex], so domin is between 0 and positive infinity. x cannot be negative.
The range is from 0 (because that's the lowest value of y) to positive infinity
Answer:
all nonzero real numbers for both of them
Step-by-step explanation:
e2021
A process is producing a particular part where the thickness of the part is following a normal distribution with a µ = 50 mm and σ = 5 mm. If a random sample of 25 parts were taken, what is the probability that this selected sample has an average thickness greater than 53?
Answer:
0.13% probability that this selected sample has an average thickness greater than 53
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
In this problem, we have that:
[tex]\mu = 50, \sigma = 5, n = 25, s = \frac{5}{\sqrt{25}} = 1[/tex]
What is the probability that this selected sample has an average thickness greater than 53?
This is 1 subtracted by the pvalue of Z when X = 53. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{53 - 50}{1}[/tex]
[tex]Z = 3[/tex]
[tex]Z = 3[/tex] has a pvalue of 0.9987
1 - 0.9987 = 0.0013
0.13% probability that this selected sample has an average thickness greater than 53
A recent survey in the N.Y. Times Almanac indicated that 48.8% of families own stock. A broker wanted to determine if this survey could be valid. He surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate?
Answer:
There is enough statistical evidence to support the claim that the survey is not accurate.
Step-by-step explanation:
We have to perform a test of hypothesis on the proportion.
The claim is that the proportion of families that own stock differs from 48.8%.
Then, the null and alternative hypothesis are:
[tex]H_0: \pi=0.488\\\\H_a:\pi\neq0.488[/tex]
The significance level is 0.05.
The sample. of size n=250, has a proportion of p=0.568.
[tex]p=X/n=142/250=0.568[/tex]
The standard error of the proportion is
[tex]\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.488*0.512}{250}}=\sqrt{0.00099}=0.032[/tex]
The z-statistic can now be calculated:
[tex]z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.568-0.488-0.5/250}{0.032}=\dfrac{0.078}{0.032}=2.4375[/tex]
The P-value for this two-tailed test is then:
[tex]P-value=2*P(z>2.4375)=0.015[/tex]
As the P-value is smaller than the significance level, the effect is significant. The null hypothesis is rejected.
There is enough evidence to support the claim that the survey is not accurate.
The suffecient eveidence to insure that the percentage of families who won stock is different from 48.8%.
Null Hypothesis:The hypothesis that there is no significant difference between specified populations, any observed difference being due to sampling or experimental error.
The null hypothesis can be stated as shown below:
[tex]H_{0} :P=48.8[/tex]%
That is, the percentage of families who own stock is [tex]48.8[/tex]%.
The alternative hypothesis can be stated as shown below:
[tex]H_{a} :p\neq 48.8[/tex]%
That is, the percentage of families who own stock is different from [tex]48.8[/tex]%
The [tex]z[/tex] test statistic is given below:
[tex]z=\frac{\hat{p-p}}{\sqrt{\frac{p\left ( 1-p \right )}{n}}}[/tex]
Here, [tex]\hat{p}[/tex] is the observed proportion then, [tex]\hat{p}[/tex] is calculated by,
[tex]\hat{p}=\frac{x}{n} \\=\frac{142}{250} \\=0.568[/tex]
Therefore,
[tex]z=\frac{\hat{p}-p}{\sqrt{\frac{p\left ( 1-p \right )}{n}}} \\=\frac{0.08}{0.0316}\\ =2.532[/tex]
Calculating the p-value be excel as attached below then we get,
The p-value is 0.01.
This p-value is called the actual level of significance.
The value of alpha is given as [tex]0.05[/tex]
Decision: The rejection of the null hypothesis [tex]H_0[/tex] because [tex]p-value < \alpha[/tex]
So, reject the null hypothesis.
Conclusion: The p-value is less than considered level of significance 5%.
Therefore the null hypothesis get rejected while the alternative hypotheisis is accepted.
Hence, there is suffecient eveidence to insure that the percentage of families who won stock is different from 48.8%.
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Suppose that prior to conducting a coin-flipping experiment, we suspect that the coin is fair. How many times would we have to flip the coin in order to obtain a 99% confidence interval of width of at most .18 for the probability of flipping a head? (note that the z-score was rounded to three decimal places in the calculation)
To determine the sample size needed for a 99% confidence interval with a maximum width of .18 for a coin flip experiment, we must rely upon statistical principles such as the law of large numbers and the relationship between sample size and confidence interval width. Although we cannot give a specific number without more details, it is generally true that a larger sample size (i.e., more coin flips) results in a narrower confidence interval.
Explanation:Your question pertains to finding the necessary sample size for a particular width of a confidence interval in a coin-flip experiment. A confidence interval represents the range where we are certain that the parameter, in this case, the probability of getting a head, lies to a certain degree, say 99%. Narrowing down this interval or making it smaller would require a larger number of coin flips or trials. The probable outcome of our experiment does indeed align with the theoretical probability when a large number of trials is conducted, which is also known as the law of large numbers.
The calculation of the size of a confidence interval involves standard deviation, confidence level (z-score) and sample size. By adjusting these parameters, we can alter the size of a confidence interval. For example, a 90% confidence interval would be smaller compared to a 95% interval due to the decreased degree of certainty. However, without exact numbers we cannot directly calculate how many flips are required for a 99% confidence interval of a certain width.
As a general rule of thumb, when conducting confidence interval tests, it would be safe to suggest that a larger sample size (which for a coin flip experiment would mean more coin flips) will result in a narrower confidence interval. Therefore, to achieve a 99% confidence interval of width at most .18, one would need net a large number of trials, or flips.
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8x-y+3x what is the answer for that
Answer:
11x-y
Step-by-step explanation:
You add 8x+3x and get 11x then you put the equation together
11x-y
HOPE THIS HELPS!!!
I think c is the right answer am I correct ?
Answer: yes
-5 is not greater than 4
-5
Step-by-step explanation:
Decompose 0.42 into tenths and hundredths
Answer:
0.40 + 0.02
Step-by-step explanation:
4 tenths plus 2 hundredths
There are 157 college students interviewed about their work schedules. 85 of them work during the day. 43 of them work nightshift. 28 students work both. How many work old dayshift?
Step-by-step explanation:
No of students work during day shift = 85
No of students work during night shift = 43
No of students working both shift = 28
So first we will divide 28 by 2 = 14
Then it means there were 14 students working in day shift and 14 students working in night shift
Total number of students work during day shift = 85 + 14 = 99
Given: ∠8 ≅ ∠16
Which lines must be parallel?
A) r and s
B) p and q
C) p and r
D) q and s
Hurry I need this one quick I don’t have a lot of time
Based on the converse of the Corresponding Angles Theorem, if ∠8 ≅ ∠16, then the lines that must be parallel are: A) r and s.
What is the converse of corresponding angles theorem?
If two lines are cut by a transversal, and the corresponding angles are congruent, then the lines are parallel, which is the converse of the Corresponding Angles Theorem.
In the image below, angles 8 and 16 are said to be congruent to each other, and both lie on lines r and s.
Therefore, based on the converse of the Corresponding Angles Theorem, lines r and s must be parallel (option A).
my question is on the picture ^
Answer:
C
Step-by-step explanation:
If the line is parallel that means 1.2 has to be the same. And passing through (5,0) means a plus five.
inDJKL side JK measures 10.6 inches side KL measures 7 inches inside JL measures 5 inches based on the information that is provided which could be a correct set of angles measured for these sides?
angle J= 23.2 angle K equals 33.5 angle owl equals 23.2
angle J equals 23.2 angle K equals 33.5 angle L equals 123.2
angle J equals 123.2 angle K equals 23.2 angle L equals 33.5
angle J equals 33.5 angle K equals 23.2 angle L equals 123.2
Answer: Angle J = 33.5°, angle K = 23.2°, angle L = 123.2°
Step-by-step explanation: i did the quiz and got it correct :D
YAHOO!!! YOU ARE CORRECT!!! LETS GIVE HIM AROUND OF APPLAUSE!!!
So thats why its "angle J equals 33.5 angle K equals 23.2 angle L equals 123.2" is the last option! (D.) U R smart, Einstein! ;)
Complete the statements to apply the triangle inequality rule to the given triangle. QS + QR > QR + RS > RS + QS >
Answer:
QS + QR > RS
QR + RS > QS
RS + QS > QR
Step-by-step explanation:
JUST TOOK THE TEST
The complete Triangle Inequality is
QS + QR > RSQR + RS > QSRS + QS > QRWhat is Triangle Inequality?The triangle inequality theorem defines the relationship between a triangle's three sides. The total of the lengths of the two sides of any triangle is always greater than the length of the third side, according to this theorem. In other words, the shortest distance between two unique points is always a straight line, according to this theorem.
According to Triangle Inequality " the sum of two side of the triangle is greater than the third side of Triangle."
Then, applying Triangle Inequality
QS + QR > RS
QR + RS > QS
RS + QS > QR
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A zoologist is studying four very closely related feline species. She wishes to compare their gestation periods. An observational study is conducted where the zoologist observes the gestation period of 50 randomly selected felines in each species. A one-way ANOVA test is run to test the hypothesis that the average gestation period is the same for each of the four species. A level of significance of 0.05 is used, and a P-value of 0.01 is calculated. Based on this result, the conclusion that can be drawn is that:
a. all of the average gestation periods are the same across all four species
b. at least some, but not all, of the gestation periods across all four species are the same
c. all of the average gestation periods across all four species are different
d. not all of the average gestation periods are the same across all four species
Answer:
[tex] p_v= 0.01[/tex]
Since the significance level is 0.05 we see that [tex]pv<\alpha[/tex] so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:
b. at least some, but not all, of the gestation periods across all four species are the same
Because is only to identify if AT LEAST one mean is different, NOT to conclude that the all the means are different.
Step-by-step explanation:
Previous concepts
Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".
The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"
Solution to the problem
The hypothesis for this case are:
Null hypothesis: [tex]\mu_{A}=\mu_{B}=\mu_{C}= \mu_D[/tex]
Alternative hypothesis: Not all the means are equal [tex]\mu_{i}\neq \mu_{j}, i,j=A,B,C,D[/tex]
In order to find the mean square between treatments (MSTR), we need to find first the sum of squares and the degrees of freedom.
If we assume that we have [tex]p=4[/tex] groups and on each group from [tex]j=1,\dots,p[/tex] we have [tex]n_j[/tex] individuals on each group we can define the following formulas of variation:
[tex]SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2 [/tex]
[tex]SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2 [/tex]
[tex]SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2 [/tex]
And we have this property
[tex]SST=SS_{between}+SS_{within}[/tex]
And in order to test this hypothesis we need to ue an F statistic and for this case the p value calculated is
[tex] p_v= 0.01[/tex]
Since the significance level is 0.05 we see that [tex]pv<\alpha[/tex] so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:
b. at least some, but not all, of the gestation periods across all four species are the same
Because is only to identify if AT LEAST one mean is different NOT to conclude that the all the means are different.
The correct option is (d)not all of the average gestation periods are the same across all four species.
Test Statistic:In a test of hypothesis, the test statistic is a function of the sample data used to decide whether or not to reject the null hypothesis.
Given that,
[tex]\alpha=0.05\\p-value=0.009[/tex]
[tex]H_0:[/tex]The average gestation period is the same for each of the four spices.
Since the p-value is less than the value of the [tex]\alpha[/tex].
So, we reject the [tex]H_0[/tex]
Not all of the average gestation periods are the same across all four species.
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Please show step by step solution
It is known that the population variance equals 484. With a 0.95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is
25
74
189
75
Answer:
b) 74
The sample size that needs to be taken if the desired margin of error is 5 or less is 74
Step-by-step explanation:
Explanation:-
Given population variance σ² = 484
σ = √484
σ = 22
The level of significance ∝=0.95
The z-score of 0.95 level of significance = 1.96
Given Margin of error = 5
we know that the margin of error is determined by
[tex]M.E = \frac{Z_{\alpha }S.D }{\sqrt{n} }[/tex]
cross multiplication, we get
[tex]\sqrt{n} = \frac{Z_{\alpha }S.D }{M.E}[/tex]
[tex]\sqrt{n} = \frac{1.96 X22 }{5} = 8.624[/tex]
Squaring on both sides, we get
n = (8.624) ²
n = 74.37 ≅74
Conclusion:-
The sample size that needs to be taken if the desired margin of error is 5 or less is 74
PLEASE HURRY NOW: Find the area of the rectangle below.
12 square units
24 square units
10 square units
20 square units
-1/7/-5/9
Step1:
Step2:
Step3:
Answer:
9/35
Step-by-step explanation:
-1/7 ÷ -5/9
Copy dot flip
-1/7 * -9/5
Multiply the numerators
-1*-9 = 9
Multiply the denominators
7*5 = 35
9/35
Step-by-step explanation:
[tex]\frac{-\frac{1}{7} }{-\frac{5}{9} }[/tex]
Invert the fraction in the denominator and multiply it by the fraction in the numerator.
[tex](-\frac{1}{7} )(-\frac{9}{5} )\\\\\frac{9}{35}[/tex]
What is/are the factor(s) in the "Sub-U-Like" study?
A. 30-second, 60-second and 90-second commercials
B. The number of students
C. Craving for Sub-U-Like
D. The length of the television program
E. Length and frequency of the commercial.
F. One, three, or five commercials during the 50-minute television program
Answer:
E. Length and frequency of the commercial.
Remaining Details of the Question:
A study investigated the effect of the length and the repetition of TV advertisements on students' desire to eat at a Sub-U-Like sandwich franchise. Sixty students watched a 50- minute television program that showed at least one commercial for Sub-U-Like during advertisement breaks. Some students saw a 30-second commerical, some saw a 60-second commercial, others a 90-second commerical. The same commerical was shown one, three, or five times during the program. After the viewing, each student was asked to rate their craving for a Sub-U-Like sandwich from the set ("Don't want to eat", "Neutral", "Want to eat"}
Explanation:
During the course of an experimental study, it is important to identify the variables of interest. These variables are referred to as the factors. Factors are controlled independent variables in an experimental study that can be manipulated by the experimenter.
In this case, it was indicated that the study investigated the effect of the length and the repetition of TV advertisements on students' desire to eat at a Sub-U-Like sandwich franchise. Therefore, the factors in the "Sub-U-Like" study are length and frequency of the commercial.
Final answer:
The factors in the "Sub-U-Like" study include the length and frequency of the commercial, the number of students, and the number of commercials during the TV program. Factors in the "Sub-U-Like" study, while not explicitly defined in the examples provided, can generally include variables such as length and frequency of commercials and the number of commercials shown, all of which can influence viewers' behaviors and perceptions in psychological or marketing research.
Explanation:
Factors in the "Sub-U-Like" study:
Factor A: Length and frequency of the commercial
Factor B: The number of students
Factor F: One, three, or five commercials during the 50-minute television program
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? Group of answer choices The p-value should be smaller than 0.95 The p-value should be larger than 0.95 The p-value should be smaller than 0.05. The p-value should be higher than 0.05.
Answer:
The p-value should be smaller than 0.05.
Step-by-step explanation:
We know that the 95% confidence interval for the difference of means is (5.6, 10.2). As the lower bound is larger than 0, we are 95% confident that the reduction program had had a effect in the blood pressure level.
Then, as the program had a significant effect in the blood pressure level, we know that the null hypothesis (that states that the blood pressure levels would no change significantly) is then 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. In conclusion, the p-value should be lower than 0.05.
Christian and Tanae both leave Disneyland at the same time. Christian travels north at 65 mph. Tanae travels south at 55 mph
apart? Which of the following equations would you use to solve this word problem?
Answer:
65t+55t=540
Step-by-step explanation:
65 multiplied by four and a half equals 292.5
55 multiplied by four and a half equals 247.5
add together to get 540
If Christian travels north at 65 mph. The equations that you would use to solve this word problem is: 65t+55t=540.
Equation:The equation will be: 65t+55t=540
Where:
t=time
65=Distance to north per hour
55=Distance to south per hour
540=Miles apart
Therefore the equations that you would use to solve this word problem is: 65t+55t=540.
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Which statements are true about a decagonal (10-sided) pyramid? Check all that apply.
It has one face that is a decagon.
It has two faces that are decagons.
It has 11 faces.
It has 20 edges.
It has 20 vertices.
Answer:
B. It has two faces that are decagons.
D. It has 20 edges.
E. It has 20 vertices.
Answer:b,d and e
Step-by-step explanation:
can someone help me with this
if a line with a slope of 6 crosses the y-axis at (0,-4), what is the equation of the line?
Answer:
y = 6x-4
Step-by-step explanation:
The slope intercept form of a line is
y = mx+b where m is the slope and b is the y intercept
We know the slope is 6 and the y intercept is -4
y = 6x + -4
y = 6x-4
Slope-intercept form: y = mx + b (m = slope and b = y-intercept)
Slope (m) = 6
y-intercept (b) = -4
Put into its final form.
y = 6x - 4
Best of Luck!
what is factored form of x2-12x-4?
Answer:
The expression is not factorable with rational numbers...
Step-by-step explanation:
|-1 1/5| and |-3/5| from least to greatest
Answer:
3/5 then 11/5
Step-by-step explanation:
These symbols "I I" represent absolute values.
11/5 equals to 2 1/5
3/5 is less than 2 1/5
Which method would determine the volume of the prism with dimensions 2 times 2 and one-fourth times 4 shown below?
A prism has a length of 2 and one-fourth, height of 4, and width of 2 inches.
There are 8 one-quarter cubes and 16 unit cubes so (8 times one-fourth) + 16 will give the volume.
There are 8 one-quarter cubes and 8 unit cubes so (8 times one-fourth) + 8 will give the volume.
There are 8 one-quarter cubes and 16 unit cubes so (8 times 4) + 16 will give the volume.
There are 8 one-quarter cubes and 8 unit cubes so (8 times 4) + 8 will give the volume.
Answer:
THE CORRECT ANSWERS (B.)Step-by-step explanation:
Answer:
The Answer is B.
Step-by-step explanation:
I took the test on Edg 2021 and got it right the first time.
We are interested in determining whether the variances of the sales at two music stores (A and B) are equal. A sample of 25 days of sales at store A has a sample standard deviation of 30, while a sample of 16 days of sales from store B has a sample standard deviation of 20. At 95% confidence, the null hypothesis _____.
What is 5.9 as a fraction.
THIS WILL HELP A LOT
Answer:
59/10
Step-by-step explanation:
5+9/10
50/10 + 9/10
=59/10
Answer:
it 59/10
Step-by-step explanation:
An automobile parts supplier owns a machine that produces a cylindrical engine part. This part is supposed to have an outside diameter of three inches. Parts with diameters that are too small or too large do not meet customer requirements and must be rejected. Lately, the company has experienced problems meeting customer requirements. The technical staff feels that the mean diameter produced by the machine is off target. In order to verify this, a special study will randomly sample 100 parts produced by the machine. The 100 sampled parts will be measured, and if the results obtained cast a substantial amount of doubt on the hypothesis that the mean diameter equals the target value of three inches, the company will assign a problem-solving team to intensively search for the causes of the problem.
a. The parts supplier wishes to set up a hypothesis test so that the problem-solving team will be assigned when the null hypothesis is rejected. Set up the null and alternative hypotheses for this situation.
Answer:
a) The null hypothesis is
H₀: μ₀ = 3 inches
And the alternative hypothesis is
Hₐ: μ₀ ≠ 3 inches
b) Check Explanation.
c) The $3000 is obviously the cost of a type I error.
Step-by-step explanation:
For hypothesis testing, the first thing to define is the null and alternative hypothesis.
The null hypothesis plays the devil's advocate and is always about the absence of significant difference between two proportions being compared. It usually maintains that random chance is responsible for the outcome or results of any experimental study/hypothesis testing. It usually contains the signs =, ≤ and ≥ depending on the directions of the test.
While, the alternative hypothesis takes the other side of the hypothesis; that there is indeed a significant difference between two proportions being compared. It usually confirms the the theory being tested by the experimental setup. It usually maintains that other than random chance, there are significant factors affecting the outcome or results of the experimental study/hypothesis testing. It usually contains the signs ≠, < and > depending on the directions of the test.
For this question, we want to verify that the mean diameter produced by the machine is really off-target.
Note that, that target is 3 inches.
Hence, the null hypothesis will be that that there is no difference between the mean diameter of the 100 cylinders sampled and the target of 3 inches.
And the alternative hypothesis will confirm the concerns of the technical staff that there is a significant difference between the mean diameter of the 100 cylinders sampled and the target diameter of 3 inches. That is, the mean diameter is off-target.
Mathematically,
The null hypothesis is
H₀: μ₀ = 3 inches
And the alternative hypothesis is
Hₐ: μ₀ ≠ 3 inches
b) A type I error involves rejecting the null hypothesis and accepting the alternative hypothesis when in reality, the null hypothesis is true. It involves saying there is significant evidence to show that the mean diameter of the sampled cylinders is indeed different from the targeted 3 in. (that is, the mean diameter of sampled cylinders is off-target), so they try to fix the issue when in reality, there is actually no significant difference between the mean diameter of sampled cylinders and the targeted 3 inches.
While a type II error involves failing to reject the null hypothesis when in reality it should have been rejected.
It entails not rejecting the null hypothesis and making conclusions based on the null hypothesis, when in reality, the alternative hypothesis should have been accepted together with its conclusion.
In this one the firm would conclude that they do not need to change anything as there is no significant difference between the mean diameter of the sampled cylinders and the targeted 3 inches, when in reality, there is a significant difference between the mean diameter of the sampled cylinders and the targeted 3 inches diameter.
c) The $3000 is obviously the cost of a type I error.
This is because the type I error is the one that involves tryin to fix the problem that did not even exist in reality.
Hope this Helps!!!
Answer:
Step-by-step explanation:
An Automobile Parts supplier owns a machine that produces a cylindrical engine part. The part is supposed to have an outside diameter of 3 inches in order for customers requirements to be met.
Lately, the company has been producing cylindrical engine parts with diameters that are either < 3inches or > 3inches. Technical staff feel that the mean diameter produced by the machine, from a random sample of 100 engine parts is not equal to 3 inches. The parts supplier wishes to set up a hypothesis test so that the problem-solving team can swing into action if or when the Null Hypothesis is rejected.
The null hypothesis is:
100 Cylindrical Engine Parts have a mean diameter = 3 inches
The alternative hypothesis is:
100 Cylindrical Engine Parts have a mean diameter that is NOT equal to 3 inches
[You can input the "not equal to" sign]
If the null hypothesis is rejected after statistical procedures, the supplier will proceed to employ the services of the problem-solving team!