Answer:
(0.6728, 0.7672)
(0.1330, 0.1870)
Step-by-step explanation:
Given that in a recent poll, 600 people were asked if they liked soccer, and 72% said they did.
Std error = [tex]\sqrt{\frac{pq}{n} } \\=\sqrt{\frac{0.72(0.28)}{600} } \\=0.0183[/tex]
Margin of error for 99% we would use the value [tex]z = 2.576[/tex]
Margin of error = [tex]2.576*SE\\=2.576*0.0183\\=0.0472[/tex]
Confidence interval lower bound = [tex]0.72-0.0472=0.6728[/tex]
Upper bound = [tex]0.72+0.0472=0.7672[/tex]
99% confidence interval for the true population proportion of people who like soccer.=(0.6728, 0.7672)
b) n =500
Sample proportion p=[tex]\frac{80}{500} =0.16[/tex]
Margin of error for 90% = [tex]1.645*\sqrt{\frac{0.16*0.84}{500} } \\\\=0.0270[/tex]
90% confidence interval for the true population proportion of people with kids. =[tex](0.16-0.0270, 0.16+0.0270)\\=(0.1330, 0.1870)[/tex]
In fall 2014, 36% of applicants with a Math SAT of 700 or more were admitted by a certain university, while 18% with a Math SAT of less than 700 were admitted. Further, 38% of all applicants had a Math SAT score of 700 or more. What percentage of admitted applicants had a Math SAT of 700 or more? (Round your answer to the nearest percentage point.)
To answer this question, we must calculate the percentage of the total population in each SAT score category that was admitted to the university. We find that 55% of admitted applicants had a Math SAT scores of 700 or more.
Explanation:First, multiply the overall percentage of applicants by the percentage admitted for each SAT score category. For those with scores of 700 or more, you get 36% * 38% = 13.68%. For those with scores less than 700, it's 18% * 62% = 11.16%. Then, add these results together to get the total percentage of all applicants who were admitted, which is 24.84%.
The next step is to calculate what fraction of this combined admitted students group got a Math SAT of 700 or more. The percentage of admitted students who had a SAT score of 700 or more is the percentage of admitted students in that category divided by the total percentage of all admitted students. So, you get 13.68% ÷ 24.84% = 55.05%. This rounds to 55% when expressed as a percentage, so we can say that 55% of admitted applicants had a Math SAT of 700 or more.
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Final answer:
To find the percentage of admitted applicants with a Math SAT of 700 or more, calculate the overall admission rate (AR), and then determine Group A's contribution to this rate. The percentage of admitted applicants with an SAT score of 700 or more is approximately 55% after rounding to the nearest percentage point.
Explanation:
To find the percentage of admitted applicants who had a Math SAT score of 700 or more, we can use the information provided to set up a weighted average problem. Let's denote the applicants with a Math SAT of 700 or more as Group A and those with a Math SAT of less than 700 as Group B.
From the information given:
36% of Group A were admitted.
18% of Group B were admitted.
38% of all applicants are in Group A; hence, 62% are in Group B (100% - 38%).
We want to find the percentage of all admitted students that had a Math SAT of 700 or more. The overall admission rate (AR) can be calculated as follows:
AR = (Percentage of Group A × Admission rate of Group A) + (Percentage of Group B × Admission rate of Group B)
AR = (38% × 36%) + (62% × 18%)
AR = 13.68% + 11.16%
AR = 24.84%
Next, we calculate the contribution of Group A to the overall admission rate:
Contribution from Group A = Percentage of Group A × Admission rate of Group A
Contribution from Group A = 38% × 36%
Contribution from Group A = 13.68%
Now, we find the percentage of the total admissions that were applicants with a Math SAT of 700 or more:
Percentage of admitted applicants with SAT ≥ 700 = (Contribution from Group A ÷ Overall admission rate) × 100
Percentage of admitted applicants with SAT ≥ 700 = (13.68% ÷ 24.84%) × 100
Percentage of admitted applicants with SAT ≥ 700 ≈ 55% (rounded to nearest percentage point)
You wish to examine whether hot air hand blowers or paper in a medical facility have an effect on the rate of respiratory incidence. You follow 98 health care workers in primary care clinic during flu season. You observe the following incidence of respiratory disease and preferred method of hand drying. 46 workers preferred paper for hand drying, 52 preferred hot air blowers. During the study period 29 of the 98 workers experienced a respiratory illness.
OBSERVED VALUES Sick Not Sick
Paper 19 27 49
HotAir 10 42 52
29 68 98
Your null and alternative hypothesis for the Chi Square test comparing the exposure (air vs paper) and outcome (ill or not ill is Hnull There is no association between the drying choice and ilness Halt There is an association between drying choice and illness
a.True
b. False
Answer:
There is an association between drying choice and illness
a.True
Step-by-step explanation:
H_0: There is no association between the drying choice and ilness
H_a: There is an association between drying choice and illness
(Two tailed chi square test)
Observations are given as per the following table
Observed
Sick Not sick
Paper 19 27 46
Hot air 10 42 52
29 69 98
Expected row total*col total/grand total
Paper 13.6122449 32.3877551 46
Hot air 15.3877551 36.6122449 52
29 69 98
Chi square =(observed-expected)^2/Expected
Paper 2.132484778 0.896261718 3.028746496
Hot air 1.886428842 0.792846905 2.679275747
4.01891362 1.689108623 5.708022243
Thus chi square = 5.708
degrees of freedom = (r-1)(c-1) = 1
p value = 0.0167
Since p <0.05 our alpha we reject null hypothesis
At 5% significance level, we can say
There is an association between drying choice and illness
a.True
please please PLEASE help! no one's been able to answer this yet and I have so much more homework left to do.
Write the slope-intercept form of the equation of the line.
1. through: (4,-4) slope=0
2. through: (5,-4) slope= -1/3
For the first one, since the slope is 0 the line must be just a horizontal line that passes through the given point. The equation would be y=-4 because the y coordinate for the point is -4.
For the second problem, utilize y=mx+b, the parent equation for slope intercept form (m-slope, b-y intercept).
y=mx+b
-4=-1/3(5)+b
b=-2 1/3
equation: y=-1/3x - 2 1/3
A group of GSU students in the Young Democrats Club wish to determine the likeability of their favorite candidate. They survey 2322 randomly selected registered voters and ask them to rate their candidate (on a "thermometer" from 0 to 100, where 0 means "very cold"and 100 means "very warm"feelings).Suppose the sample of n=2322 responses have a mean warmth of 64.16 with a sample standard deviation of s=26.34.Find the 95% confidence interval to estimate the population/national mean warmth rating for their candidate.A) Note: The critical value, tc, that we use to calculate the margin of error for a 95% Confidence Interval is tc = ____ (round to 4 decimal places)B) The margin of error for this confidence interval is: m = ____ (round to 4 decimal places)C) The 95% Confidence Interval for the likeability of their candidate is: (____ , ____) round each to 2 decimal places
Answer:
a) [tex]tc=\pm 1.9601[/tex]
b) [tex]m=1.9601 \frac{26.34}{\sqrt{2322}}=1.0714[/tex]
c) The 95% confidence interval is given by (63.09;65.23)
Step-by-step explanation:
1) Notation and definitions
n=2322 represent the sample size
[tex]\bar X= 64.16[/tex] represent the sample mean
[tex]s=26.34[/tex] represent the sample standard deviation
m represent the margin of error
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
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".
2) Calculate the critical value tc
In order to find the critical value is impornta to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. The degrees of freedom are given by:
[tex]df=n-1=2322-1=2321[/tex]
We can find the critical values in excel using the following formulas:
"=T.INV(0.025,2321)" for [tex]t_{\alpha/2}=-1.9601[/tex]
"=T.INV(1-0.025,2321)" for [tex]t_{1-\alpha/2}=1.9601[/tex]
The critical value [tex]tc=\pm 1.9601[/tex]
3) Calculate the margin of error (m)
The margin of error for the sample mean is given by this formula:
[tex]m=t_c \frac{s}{\sqrt{n}}[/tex]
[tex]m=1.9601 \frac{26.34}{\sqrt{2322}}=1.0714[/tex]
4) Calculate the confidence interval
The interval for the mean is given by this formula:
[tex]\bar X \pm t_{c} \frac{s}{\sqrt{n}}[/tex]
And calculating the limits we got:
[tex]64.16 - 1.9601 \frac{26.34}{\sqrt{2322}}=63.09[/tex]
[tex]64.16 + 1.9601 \frac{26.34}{\sqrt{2322}}=65.23[/tex]
The 95% confidence interval is given by (63.09;65.23)
A line is drawn on the coordinate plane that passes through the point (10,1) and has a slope of -0.5. what is the y intercept
Answer:
y int = 6
Step-by-step explanation:
simplistic plotting in my head
A line is drawn on the coordinate plane that passes through the point (10,1) and has a slope given as -0.5. The y intercept will be 6.
What is y-intercept of a function?The intersection of the graph of the function with the y-axis gives y-intercept of that function.
The y-intercept is the value of y on the y-axis at which the considered function intersects it.
Assume that we've got: y = f(x)
At y-axis, we've got x = 0, so putting it will give us the y-intercept.
Thus, y-intercept of y = f(x) is y = f(0)
A line is drawn on the coordinate plane that passes through the point (10,1) and has a slope given as -0.5.
So the equation of line
y = mx + c
1 = -0.5(10) + c
1 = -5 + c
c = 6
Hence, The cut-off point with the "y" axis is 6.
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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:
dominic bought a new alram clock that was on sale for 18,75 if this price represents a 30% discount from the original price what is the original price from the nearest cent?
Answer:
Step-by-step explanation:
Which statement about the siege of Savannah and its outcome is correct?
A
The siege was a failed attempt by American forces to retake Savannah from the British.
B
The siege was a failed attempt by British forces to retake Savannah from the Americans.
C
The siege was a successful attempt by American forces to retake Savannah from the British.
D
The siege was a successful attempt by British forces to retake Savannah from the Americans.
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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Construct a confidence interval for p 1 minus p 2 at the given level of confidence. x 1 equals395, n 1 equals 538, x 2 equals 444, n 2 equals 572, 90% confidence The researchers are 90 90% confident the difference between the two population proportions, p 1 minus p 2, is between .733 . 733 and .776 . 776.
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%.
Given the list of terms 1/3, 1, 5/3, 7/3,.... Find the 15th and −16th term.
Answer:
15th term =29/3
16th term = 31/3
Step-by-step explanation:
Given an arithmetic sequence with the first term a1 and the common difference d , the nth (or general) term is given by an=a1+(n−1)d .
First we find the 15th term
n=15
a1=1/3
d=1 - 1/3 = 2/3
Solution
1/3+(15-1)2/3
1/3+28/3
(1+28)/3
29/3
Lets find the 16th term
1/3+(16-1)2/3
1/3+30/3
(1+30)/3
31/3
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
An article reports that in a sample of 413 male college students, the average number of energy drinks consumed per month was 2.48 with a standard deviation of 4.87, and in a sample of 382 female college students, the average was 1.22 with a standard deviation of 3.23. Can you conclude that the mean number of energy drinks is greater for male students than for female students? Find the P-value and state a conclusion.
The P-value is 0.0000; we reject the null hypothesis and conclude males consume more energy drinks than females.
We will conduct a hypothesis test for the difference between two means. Here are the steps we'll follow:
1. **State the Hypotheses**:
Null hypothesis (H0): μ1 - μ2 ≤ 0 (The mean number of energy drinks consumed by male students is less than or equal to that consumed by female students.)Alternative hypothesis (H1): μ1 - μ2 > 0 (The mean number of energy drinks consumed by male students is greater than that consumed by female students.)2. **Check the Conditions**:
The samples are independent.The sample sizes are large enough (both greater than 30) for the Central Limit Theorem to apply, meaning the sampling distribution of the difference between means will be approximately normally distributed.3. **Calculate the Test Statistic**:
We'll use the formula for the test statistic for two independent samples (assuming equal variances are not assumed):
[tex]\[ t = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \][/tex]
where:
[tex]\(\bar{x}_1\) and \(\bar{x}_2\)[/tex] are the sample means,[tex]\(s_1^2\) and \(s_2^2\)[/tex] are the sample variances,[tex]\(n_1\) and \(n_2\)[/tex] are the sample sizes.4. **Calculate the Degrees of Freedom**:
We'll use the Welch-Satterthwaite equation to approximate the degrees of freedom for the t-distribution:
[tex]\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \][/tex]
5. **Find the P-value**:
Since we are conducting a one-tailed test, we will find the area to the right of our calculated t-value in the t-distribution with the calculated degrees of freedom.
6. **State the Conclusion**:
We will compare the P-value to the significance level (usually 0.05) and decide whether to reject the null hypothesis.
Let's perform the calculations.
The test statistic (t-value) for the difference in the mean number of energy drinks consumed per month between male and female college students is approximately 4.3285. The degrees of freedom for this test is approximately 720.78.
The P-value for this one-tailed test is approximately 0.00000857 (rounded to four decimal places, this is 0.0000). Since the P-value is significantly less than the common alpha level of 0.05, we reject the null hypothesis.
**Conclusion**: We can conclude that there is significant evidence at the 0.05 level to suggest that the mean number of energy drinks consumed by male college students is greater than that consumed by female college students.
every 3 hours, a machine produces 60 baskets. What is the unit rate?
a. 180 baskets per hour
b. 3 baskets per hour
c. 20 baskets per hour
d. 57 baskets per hour
Answer:
c. 20 baskets per hour
Step-by-step explanation:
In this context, "per" can be considered to mean "divided by." Then to find the unit rate in terms of baskets per hour, we compute ...
rate = (baskets)/(hours) = (60 baskets)/(3 hours) = (60/3) baskets/hour
rate = 20 baskets/hour . . . . matches choice C
The heights of adult men in America are normally distributed, with a mean of 69.3 inches and a standard deviation of 2.64 inches. The heights of adult women in America are also normally distributed, but with a mean of 64.5 inches and a standard deviation of 2.53 inches. a) If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)? z = 2.16 b) What percentage of men are SHORTER than 6 feet 3 inches? Round to nearest tenth of a percent.
Answer:
z-score is 2.16
98.46% of men are SHORTER than 6 feet 3 inches
Step-by-step explanation:
The heights of adult men in America are normally distributed, with a mean of 69.3 inches and a standard deviation of 2.64 inches.
Mean = [tex]\mu = 69.3 inches[/tex]
Standard deviation = [tex]\sigma = 2.64 inches[/tex]
a) If a man is 6 feet 3 inches tall, what is his z-score (to two decimal places)
x = 6 feet 3 inches
1 feet =12 inches
6 feet = 12*6 = 72 inches
So, x = 6 feet 3 inches = 72+3=75 inches
Formula : [tex]Z=\frac{x-\mu}{\sigma}[/tex]
[tex]Z=\frac{75-69.3}{2.64}[/tex]
[tex]Z=2.159[/tex]
So, his z-score is 2.16
No to find percentage of men are SHORTER than 6 feet 3 inches
We are supposed to find P(x< 6 feet 3 inches)
z-score is 2.16
Refer the z table
P(x< 6 feet 3 inches) =0.9846
So, 98.46% of men are SHORTER than 6 feet 3 inches
Write the equation of the circle with center(-3,-2) and (4,5) a point on the circle
Answer:
(x + 3)^2 + (y + 2)^2 = 98.
Step-by-step explanation:
(x - h)^2 + (y - k) ^2 = r^2 ( center is (h, k) and r = radius.)
So we have:
(x + 3)^2 + (y + 2)^2 = r^2
When x = 4 y = 5, so
(4 + 3)^2 + (5 + 2)^2 = r^2
r^2 = 7^2 + 7^2 = 98
.
Answer:
Step-by-step explanation:
(x+2)^2+(y+3)^2=100
A tank initially contains 200 gallons of brine, with 50 pounds of salt in solution. Brine containing 2 pounds of salt per gallon is entering the tank at the rate of 4 gallons per minute and is is flowing out at the same rate. If the mixture in the tank is kept uniform by constant stirring, find the amount of salt in the tank at the end of 40 minutes.
Answer:
y = 50
t=0
we find at end of 40 minutes value of salt is y = 242.735
Step-by-step explanation:
explanation has been defined in attachment
Final answer:
The amount of salt in the tank at the end of 40 minutes is 370 pounds.
Explanation:
To find the amount of salt in the tank after 40 minutes, we need to calculate the net change in the amount of salt in the tank during that time. The tank initially contains 200 gallons of brine with 50 pounds of salt, and brine is flowing in and out of the tank at a rate of 4 gallons per minute. Since the salt concentration in the brine entering the tank is 2 pounds per gallon, the amount of salt entering the tank each minute is 4 gallons x 2 pounds = 8 pounds.
Therefore, the net change in the amount of salt in the tank after 40 minutes is 40 minutes x 8 pounds/minute = 320 pounds. Adding this to the initial amount of salt in the tank, we get 50 pounds + 320 pounds = 370 pounds of salt in the tank at the end of 40 minutes.
Let us consider an arithmetic expression described by a tree constructed from tuples as follows:
1. An integer is described by a tuple int(N), where N is an integer.
2. An addition is described by a tuple add(X Y), where both X and Y are arithmetic expressions. 3. A multiplication is described by a tuple mul(X Y), where both X and Y are arithmetic
expressions.
4. A variable is described by a tuple var(A), where A is an atom giving the variable name.
An environment is a record with a label env and for each variable name there is a feature which corresponds to an integer value. For example, the record env(a:2 b:4) says that the variable "a" has value 2, whereas the variable "b" has value 4 (note that the Oz features should start with a lower- case letter). For example, add(var(a) mul(int(3) var(b))) is an arithmetic expression containing two variables (namely a and b) and its evaluation returns 14.
Give a specification and an Oz implementation of a function Eval that takes as arguments an arithmetic expression and an environment, and returns its value. For example, the call
{Eval add(var(a) mul(int(3) var(b))) env(a:2 b:4)} should return 14.
The question asks for an Oz function to parse and evaluate arithmetic expressions represented by tuples. The Eval function recursively breaks down the expressions into smaller parts until it‘s a single integer. Pattern matching is used to simplify the process of breaking down the expressions.
Explanation:For this task, we are required to implement an Oz function named Eval that evaluates arithmetic expressions built using tuples. Each tuple represents an arithmetic operation, integer or variable. The function uses an environment record to map each variable to an integer.
Here is a specification and possible Oz implementation of the Eval function:
fun {Eval Exp Env}
case Exp of int(N) then N
mul(X Y) then {Mul {Eval X Env} {Eval Y Env}}
add(X Y) then {Add {Eval X Env} {Eval Y Env}}
var(A) then {Find A Env}
else raise system.error end
Here, the Eval function breaks down the expression based on its pattern and recursively reduces the expression until it's a single integer. The Find function is a helper function that looks up the value of a variable in the environment. We can use pattern matching to simplify the process of breaking down the expressions, which is a powerful feature in Oz.
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The purpose of statistical inference is to provide information about the.a. sample based upon information contained in the population.b. population based upon information contained in the sample.c. population based upon information contained in the population.d. mean of the sample based upon the mean of the population.10. Random samples of size 81 are taken from an infinite population whose mean and standard deviation are 200 and 18, respectively. The distribution of the population is unknown. The mean and the standard error of the mean are.a. 200 and 18.b. 81 and 18.c. 9 and 2.d. 200 and 2.
Answer: The purpose of statistical inference is to provide information about the population based upon information contained in the sample.
The mean and standard error of the mean are 200 and 2 respectively.
Step-by-step explanation:
Statistical inference is used to draw conclusions about the some unknown parameter based on samples that are drawn from the original population.
The formula for central limit theorem can be stated as follows:
[tex]\begin{array}{l}\\{\mu _{\bar x}} = \mu \\\\{\sigma _{\bar x}} = \frac{\sigma }{{\sqrt n }}\\\end{array}[/tex]
Here, [tex]\mu[/tex] is the population mean, [tex]\sigma[/tex] is the population standard deviation, [tex]{\mu _{\overline x }}[/tex] is the sample mean, [tex]{\sigma _{\overline x }}[/tex] is the sample variance and [tex]n[/tex] is the sample size.
The option (a), “sample based upon information contained in the population” is incorrect because the estimate is calculated based on sample.
The option (c), “population based upon information contained in the population” is incorrect because statistical inference helps to draw conclusions about some unknown parameter based on sample data.
The option (d), “mean of the sample based upon the mean of the population” is incorrect because statistical inference draws some conclusions about some unknown parameter of population based on sample and not based on mean of population.
As the statistical inference is used to draw some conclusions based on sample data, options (a), (c), and (d) are the incorrect answer choices.
Statistical inference draws some conclusions or estimate the unknown parameter based on sample drawn from population. That is, the conclusion about population is made based on the sample data. Hence, option (b), “population based upon information contained in the sample” is the correct answer in the provided scenario.
As the statistical inference draw some conclusions about the population based upon information contained in sample, option (b) can be considered as correct answer choice.
According to provided data, the population mean is equal to 200, population variance is equal to 18 and sample size is equal to 81.
The mean of sampling means can be calculated as:
[tex]\begin{array}{c}\\{\mu _{\bar x}} = \mu \\\\ = 200\\\end{array}[/tex]
The standard error of the mean can be calculated as:
[tex]\begin{array}{c}\\{\sigma _{\bar x}} = \frac{\sigma }{{\sqrt n }}\\\\ = \frac{{18}}{{\sqrt {81} }}\\\\ = \frac{{18}}{9}\\\\ = 2\\\end{array}[/tex]
So, the options (a), (b) and (c) are considered as incorrect options because the values do not match with the obtained values of mean and the standard error of the mean..
The options (a), (b) and (c) are considered as incorrect answer choice because the obtained value of mean and standard error of the mean is 200 and 2 respectively.
The obtained value of mean and standard error of the mean is 200 and 2 respectively. So, it can be said that option (d) is the correct answer in the provided scenario.
Statistical inference aims to draw conclusions about a population based on sample data, and for random samples of size 81 from a population with a mean of 200 and standard deviation of 18, the mean and standard error of the sample mean are 200 and 2, respectively.
Explanation:The purpose of statistical inference is to provide information about the population based upon information contained in the sample. This includes estimating a population parameter like the mean or proportion by analyzing a subset of collected data. When dealing with samples, specifically random samples, we use inferential statistics to make assumptions about the larger population from which the sample is drawn.
When you have random samples of size 81 from an infinite population with a known mean (μ = 200) and standard deviation (σ = 18), and the distribution of the population is unknown, the distribution of the sample means will approach a normal distribution according to the Central Limit Theorem. The mean of the sampling distribution of the sample mean will be the same as the population mean, so the correct answer is 200.
The standard error of the mean (SEM) is calculated as the population standard deviation divided by the square root of the sample size (n). Thus, SEM = σ/√n = 18/√81 = 18/9 = 2. Therefore, the answer is 200 for the mean and 2 for the standard error of the mean.
What is the answer to 2x9+(2x14
Answer:
46
Step-by-step explanation:
The order of operations is to multiply before you divide.
2x9= 18
2x14=28
18+28=46
Answer:
46
Step-by-step explanation:
2*9=18
2*14=28
18+28=46
Consider a triangle with angles 30o-60o-90o, with hypotenuse of length 2. Calculate the lengths of the other 2 sides and store the two calculated lengths as well as the given length in a list called sides. Use indexing to display the lengths of each of the sides in a print statement. Make sure the lengths have 2 decimal places!
Answer:
1.00 and 1.73
Step-by-step explanation:
1) Calculating the 1st and the 2nd leg using trigonometric function sine:
[tex]sin (30^{0})=\frac{c}{2}\Rightarrow \frac{1}{2}=\frac{c}{2}\Rightarrow 2*\frac{1}{2}=2*\frac{c}{2}\Rightarrow c=1\\sin(60^0)=\frac{b}{2}\Rightarrow \frac{\sqrt{3}}{2}=\frac{b}{2}\Rightarrow b=\sqrt{3}\Rightarrow\: b\approx \:1.73[/tex]
Once one leg is found there is another way to find the other leg, using Pythagorean Theorem too.
[tex]a^{2}=b^{2}+c^{2}\Rightarrow 2^{2}=b^{2}+1^{2}\Rightarrow 4-1=b^{2}\Rightarrow b=\sqrt{3} b\approx 1.73[/tex]
Sides
a hypotenuse: 2.00
-------------------------------
b leg: 1.73
--------------------------------
c leg: 1.00
--------------------------------
Geoff has been running a restaurant in the city for many years. He knows from experience that the proportion of customers that leave tips at this restaurant is 0.56. Geoff has recently opened a restaurant in a suburban area. It seems that tipping habits are different in the suburbs: Geoff believes that the proportion of customers that tip at his new restaurant (p) may be different to 0.56. He conducts a hypothesis test to find out. The null and alternative hypotheses are: H0: p = 0.56 Ha: p ≠ 0.56 Geoff takes a random sample of 113 customers and notes whether or not they leave a tip. In this sample, the proportion of people that leave a tip is 0.60. a)Calculate the test statistic (z). Give your answer to 2 decimal places. z = 0.857 b)Calculate the P-value. Give your answer as a decimal to 4 decimal places. P-value = 0.3898
Answer:
(a) 0.87 (b) 0.3843
Step-by-step explanation:
We have the null and alternative hypotheses [tex]H_{0}: p = 0.56[/tex] and [tex]H_{a}: p\neq 0.56[/tex] (two-tailed alternative). There is a large sample size of n = 113 customers and a point estimate of p that is [tex]\hat{p} = 0.60[/tex] with an estimate standard deviation given by [tex]\sqrt{\hat{p}(1-\hat{p})/n} = \sqrt{0.60(1-0.60)/113} = 0.0461[/tex]. Then, the test statistic z which comes from a standard normal distribution is given by
(a) [tex]z = \frac{\hat{p}-0.56}{\sqrt{\hat{p}(1-\hat{p})/n}}=(0.60-0.56)/0.0461 = 0.8677[/tex]
(b) The p-value is given by 2P(Z > 0.87) = 0.3843, this because we are dealing with a two-tailed alternative.
A confidence interval is constructed for an unknown population proportion, p. A sample is collected, and the 95% confidence interval is calculated to be 0.40 ± 0.06. Based on this information, it is most accurate to say that there is approximately 95% confidence in the assertion that:
Answer: The population proportion is between 0.34 and 0.46 .
Step-by-step explanation:
Interpretation of 95% confidence interval : A person can be about 95% confident that the true population parameter lies in the interval.
A confidence interval is constructed for an unknown population proportion, p. A sample is collected, and the 95% confidence interval is calculated to be 0.40 ± 0.06.
i.e. Lower limit = [tex]0.40-0.06=0.34[/tex]
Upper limit = [tex]0.40+0.06=0.46[/tex]
i.e. 95% confidence interval = (0.34, 0.46)
i.e. A person can be about 95% confident that the true population proportion (p) lies in the interval (0.34, 0.46).
⇒ It is most accurate to say that there is approximately 95% confidence in the assertion that:
The population proportion is between 0.34 and 0.46 .
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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In the casino game of Roulette, there are 38 slots on the wheel. Of the 38 slots, 18 are colored red, 18 are colored black, and 2 are colored green. For a single game, the wheel is spun and a ball is rolled. Eventually, the ball will randomly fall into one of the slots. For the next game, the wheel is respun and the ball rerolled, making the outcome independent of the previous outcome.(a) On a single game, what is the probability that the outcome is red? P(red) = 18/38 = 0.474.(b) Suppose a casino patron plays the game 20 consecutive times and bets that the outcome will be red each time. What is the probability that the patron is correct (and wins) on exactly 10 of the games?(c) Suppose a casino patron plays the game 20 consecutive times and bets that the outcome will be red each time. What is the probability that the patron is correct (and wins) on at most 10 of the games?(d) Suppose a casino patron plays the game 20 consecutive times and bets that the outcome will be red each time. What is the probability that the patron is correct (and wins) on at least 15 of the games?
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 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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Angles α and β are the two acute angles in a right triangle, where α = 5x 3 + 20 and β = 2x 3 + 14. Find α. A) 24° B) 30° C) 60° D) 66°
Answer:
C) 60°
Step-by-step explanation:
The sum of the acute angles in a right triangle is 90°, so we have ...
α + β = 90
(5x³ +20) +(2x³ +14) = 90
7x³ = 56 . . . . . . subtract 34, collect terms
x³ = 8 . . . . . . . . divide by 7
Using this value in the expression for α, we find its measure to be ...
α = 5·8 +20 = 60
α = 60°
At a stop sign, some drivers come to a full stop, some come to a rolling stop' (not a full stop, but slow down), and some do not stop at all. We would like to test if there is an association between gender and type of stop (full, rolling, or no stop). We collect data by standing a few feet from a stop sign and taking note of type of stop and the gender of the driver. Below is a contingency table summarizing the data we collected.
Male Female
Full stop 6 6
Rolling stop 16 15
No stop 4 3
If gender is not associated with type of stop, how many males would we expect to not stop at all?
A. 6.24
B. 5.76
C. 3.64
D. 3.36
Answer:
Option C.
Step-by-step explanation:
Given information:
Male-Full stop = 6, Male-Rolling stop = 16, Male-No stop = 4
Female-Full stop = 6, Female-Rolling stop = 15, Female-No stop = 3
Using the given information we get
Total number of males = 6 + 16 + 4 = 26
Total number of males = 6 + 15 + 3 = 24
Probability of No stop is
[tex]p=\dfrac{\text{No-stop}}{Total}[/tex]
[tex]p=\dfrac{4+3}{50}[/tex]
[tex]p=\dfrac{7}{50}[/tex]
We need to find the number of males that would we expect to not stop at all.
Expected number of males = Number of males × Probability
Expected number of males = [tex]26\times \dfrac{7}{50}[/tex]
Expected number of males = [tex]3.64[/tex]
Therefore, the correct option is C.
If gender is independent of type of stop, the expected number of males who would not stop at all is calculated by the formula (Total Males * Total No Stops) / Total Observations, resulting in approximately 3.64.
Explanation:To determine the expected number of males who would not stop at all at the stop sign, we would use the formula for expected frequency in a contingency table if the two variables are independent. The formula is (row total * column total) / grand total.
From the given data, the total number of males is the sum of male full stops, rolling stops, and no stops (6 + 16 + 4 = 26). The total number of 'no stops,' regardless of gender, is 4 (male) + 3 (female) = 7. The grand total of all observations is 50 (sum of all types of stops for both genders).
So, plug these values into the formula, we get: Expected no stops (male) = (26*7)/50 = 3.64
So, if gender and stop type were not associated, we would expect approximately 3.64 males to not stop at all at the stop sign, almost matching option C from the list provided.
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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.