Suppose that Brooklyn College financial aid allots a textbook stipend by claiming that the average textbook at BC bookstore costs $63.26. You want to test this claim. The null and alternative hypothesis in symbols would be: Ο Ho: μ = 63.26 Hy:μ / 63.26 O Ho:p ≤ 63.26 H₁:p> 63.26 Ο Η: μ 63.26 H:μ < 63.26 O Ho:p> 63.26 H₁:p < 63.26 Ο Ho:μ < 63.26 H: μ > 63.26 O Ho:p 63.26 H₁:p 63.26 The null hypothesis in words would be: O The average price of textbooks in a sample is $ 63.26 The proportion of all textbooks from the store that are less than 63.26 is equal to 50% O The average price of all textbooks from the store is $63.26 O The average of price of all textbooks from the store is greater than $63.26. O The average of price of all textbooks from the store is less than $63.26. Based on a sample of 170 textbooks at the store, you find an average of 67.66 and a standard deviation of 28.2. The point estimate (statistic) for this sample is: (to 2 decimals) S The formula for Standard Error for a mean is SE = √n Percentage Confidence z*-Value 80 1.28 90 1.645 95 1.96 98 2.33 99 2.58 The 99% confidence interval (use z*) is: to (to 3 decimals) Based on this we: O Reject the null hypothesis O Fail to reject the null hypothesis Note: The answer key for the confidence interval goes out 14 decimal places. Chances are, the autograde feature will mark your answer wrong. I have to grade this part of the question by hand. If your answer is
Suppose that Brooklyn College financial aid allots a textbook stipend by claiming that the average textbook at BC bookstore costs $63.26. You want to test this claim. The null and alternative hypothesis in symbols would be: Ο Ho: μ = 63.26 Hy:μ / 63.26 O Ho:p ≤ 63.26 H₁:p> 63.26 Ο Η: μ 63.26 H:μ < 63.26 O Ho:p> 63.26 H₁:p < 63.26 Ο Ho:μ < 63.26 H: μ > 63.26 O Ho:p 63.26 H₁:p 63.26 The null hypothesis in words would be: O The average price of textbooks in a sample is $ 63.26 The proportion of all textbooks from the store that are less than 63.26 is equal to 50% O The average price of all textbooks from the store is $63.26 O The average of price of all textbooks from the store is greater than $63.26. O The average of price of all textbooks from the store is less than $63.26. Based on a sample of 170 textbooks at the store, you find an average of 67.66 and a standard deviation of 28.2. The point estimate (statistic) for this sample is: (to 2 decimals) S The formula for Standard Error for a mean is SE = √n Percentage Confidence z*-Value 80 1.28 90 1.645 95 1.96 98 2.33 99 2.58 The 99% confidence interval (use z*) is: to (to 3 decimals) Based on this we: O Reject the null hypothesis O Fail to reject the null hypothesis Note: The answer key for the confidence interval goes out 14 decimal places. Chances are, the autograde feature will mark your answer wrong. I have to grade this part of the question by hand. If your answer is
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
![Suppose that Brooklyn College financial aid allots a textbook stipend by claiming that the average textbook
at BC bookstore costs $63.26. You want to test this claim.
The null and alternative hypothesis in symbols would be:
Ο Ho: μ = 63.26
Η: μ # 63.26
O Ho:p ≤ 63.26
H₁:p> 63.26
Ho: μ ≥ 63.26
H:μ < 63.26
O Ho:p> 63.26
H₁:p < 63.26
Ho:μ < 63.26
H:μ > 63.26
O Ho:p = 63.26
H₁:p 63.26
The null hypothesis in words would be:
The average price of textbooks in a sample is $63.26
O The proportion of all textbooks from the store that are less than 63.26 is equal to 50%
The average price of all textbooks from the store is $63.26
O The average of price of all textbooks from the store is greater than $63.26.
O The average of price of all textbooks from the store is less than $63.26.
Based on a sample of 170 textbooks at the store, you find an average of 67.66 and a standard deviation of
28.2.
The point estimate (statistic)for this sample is:
(to 2 decimals)
S
The formula for Standard Error for a mean is SE
√n
Percentage Confidence
z*-Value
80
1.28
90
1.645
95
1.96
98
2.33
99
2.58
The 99 % confidence interval (use z*) is:
to
(to 3
decimals)
Based on this we:
O Reject the null hypothesis
O Fail to reject the null hypothesis
Note: The answer key for the confidence interval goes out 14 decimal places. Chances are, the autograde
feature will mark your answer wrong. I have to grade this part of the question by hand. If your answer is
correct, the points will be returned to your score.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa588cdf0-a170-4a98-943c-21d2081310a0%2Fdfb9e5fe-1903-4860-a0e0-cd7ad521c042%2Fue782d_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Suppose that Brooklyn College financial aid allots a textbook stipend by claiming that the average textbook
at BC bookstore costs $63.26. You want to test this claim.
The null and alternative hypothesis in symbols would be:
Ο Ho: μ = 63.26
Η: μ # 63.26
O Ho:p ≤ 63.26
H₁:p> 63.26
Ho: μ ≥ 63.26
H:μ < 63.26
O Ho:p> 63.26
H₁:p < 63.26
Ho:μ < 63.26
H:μ > 63.26
O Ho:p = 63.26
H₁:p 63.26
The null hypothesis in words would be:
The average price of textbooks in a sample is $63.26
O The proportion of all textbooks from the store that are less than 63.26 is equal to 50%
The average price of all textbooks from the store is $63.26
O The average of price of all textbooks from the store is greater than $63.26.
O The average of price of all textbooks from the store is less than $63.26.
Based on a sample of 170 textbooks at the store, you find an average of 67.66 and a standard deviation of
28.2.
The point estimate (statistic)for this sample is:
(to 2 decimals)
S
The formula for Standard Error for a mean is SE
√n
Percentage Confidence
z*-Value
80
1.28
90
1.645
95
1.96
98
2.33
99
2.58
The 99 % confidence interval (use z*) is:
to
(to 3
decimals)
Based on this we:
O Reject the null hypothesis
O Fail to reject the null hypothesis
Note: The answer key for the confidence interval goes out 14 decimal places. Chances are, the autograde
feature will mark your answer wrong. I have to grade this part of the question by hand. If your answer is
correct, the points will be returned to your score.
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