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Question 1 1/
points
A statistics professor recently graded final exams for students in her introductory statistics course. In a review of her grading, she found the mean score out of 100 p o i n t s was a x ¯=7 7 , with a margin of error of 10.
Construct a confidence interval for the mean score (out of 100 p o i n t s ) on the final exam.
That is correct!
open paren 67 comma 87 close paren$$ open paren 67 comma 87 close paren - correct
Answer Explanation
Correct answers:
- open paren 67 comma 87 close paren $\left(67,\ 87\right)$
A confidence interval is an interval of values, centered on a point estimate, of the form
(pointestimate−marginoferor,pointestimate+marginoferor)
Using the given point estimate for the mean, x ¯=7 7 and margin of error 10 , the confidence interval is:
( 77 − 10 , 77 +1 0 )( 67 , 87 )
By the Empirical Rule, a 95 %confidence interval corresponds to a z - score of z = 2. Substituting the given values p ′=0. 14 and σp ′=0. 02 , a confidence interval is
( p ′− z ⋅ σp ′, p ′+ z ⋅ σp ′)( 0. 14 − 2 ⋅ 0. 02 , 0. 14 +2⋅ 0. 02 )( 0. 14 − 0. 04 , 0. 14 +0. 04 ) ( 0. 10 , 0. 18 )
Question 3
1/1 points
The pages per book in a library are normally distributed with an unknown population mean. A random sample of books is taken and results in a 95 %c o n f i d e n c e interval of ( 237 , 293 )pages.
What is the correct interpretation of the 95 %c o n f i d e n c e interval?
That is correct!
We estimate with 95 %c o n f i d e n c e that the sample mean is between 237 and 293 p a g e s.
We estimate that 95 %of the time a book is selected, there will be between 237 and 293 p a g e s.
Once a confidence interval is calculated, the interpretation should clearly state the confidence level (CL), explain what population parameter is being estimated, and state the confidence interval.
We estimate with 95 %confidence that the true population mean is between 237 and 293 pages.
Question 4
1/1 points
The population standard deviation for the heights of dogs, in inches, in a city is 3. 7 i n c h e s. If we want to be 95 %confid e nt that the sample mean is within 2 inches of the true population mean, what is the minimum sample size that can be taken?
z 0. 10 1. 282 z 0. 05 1. 645 z 0. 025 1. 960 z 0. 01 2. 326 z 0. 005 2. 576
Use the table above for the z - score, and be sure to round up to the nearest integer.
That is correct!
14 dog heights$$
14 dog heights - correct
Answer Explanation
Correct answers:
- 14 dog heights $14\text{ dog heights}$
847 st u d e n t s$$ 847 st u d e n t s - incorrect
Answer Explanation
Correct answers:
- 846 st u d e n t s $846\ students$
Given the information in the question, EBP =0. 04 since 4 %=0. 04 and zα 2 = z 0. 01 = 2. 326 because the confidence level is 98 %. The values of p ′a n d q ′a r e unknown, but using a value of
- 5 for p ′w i l l result in the largest possible product of p ′ q ′, and thus the largest possible n. If p ′=0. 5 , then q ′=1− 0. 5 =0. 5. Therefore, n = z 2 p ′ q ′ EBP 2 =2. 3262 ( 0. 5 )( 0. 5 ) 0. 042 =8 45. 4 Round the answer up to the next integer to be sure the sample size is large enough. The sample should include 846 s t u d e n t s.
Question 6
1/1 points
The average score of a random sample of 87 s e n i o r business majors at a university who took a certain standardized test follows a normal distribution with a standard deviation of 28. Use Excel to determine a 90 %c o n f i d e n c e interval for the mean of the population. Round your answers to two decimal places and use ascending order.
Answer Explanation
Correct answers:
- open paren 509 point 3 0 comma 519 point 1 8 close paren $\left(509.30,\ 519.18\right)$
A 90 %confidence interval for μ is ( x ¯− zα / 2 σn ‾√, x ¯+ zα / 2 σn ‾√). Here, α =0. 1 , σ = 28 , and n = 87. Use Excel to calculate the 90 %confidence interval.
- Open Excel, enter the given data in column A, and find the sample mean, x ¯, using the AVERAGE function. Thus, the sample mean, rounded to two decimal places, is x ¯=5 14. 24.
- Click on any empty cell, enter = CONFIDENCE.NORM ( 0. 1 , 28 , 87 ), and press ENTER.
- The margin of error, rounded to two decimal places, is zα / 2 σn ‾√≈ 4. 94. The confidence interval for the population mean has a lower limit of
- 24 − 4. 94 = 509. 30 and an upper limit of 514. 24 +4. 94 =5 19. 18. Thus, the 90 %confidence interval for μ is ( 509. 30 , 519. 18 ).
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Question 20
0/1 points
From a recent company survey, it is known that the proportion of employees older than 55 a n d considering retirement is 8 %. For a random sample of size 110 , what is standard deviation for the sampling distribution of the sample proportions, rounded to three decimal places?
1/1 points
In order to estimate the average electricity usage per month, a sample of 125 r e s i d e n t i a l customers were selected, and the monthly electricity usage was determined using the customers' meter readings. Assume a population variance of 12 , 100 kWh 2. Use Excel to find the 98 % confidence interval for the mean electricity usage in kilowatt hours. Round your answers to two decimal places and use ascending order.
That is correct!
open paren 894 point 4 3 comma 940 point 2 1 close paren$$ open paren 894 point 4 3 comma 940 point 2 1 close paren - correct
Answer Explanation
Correct answers:
- open paren 894 point 4 3 comma 940 point 2 1 close paren $\left(894.43,\ 940.21\right)$
A 98 %confidence interval for μ is ( x ¯− zα / 2 σn ‾√, x ¯+ zα / 2 σn ‾√). Here, α =0. 02 , σ = 110 , and n = 125. Use Excel to calculate the 98 %confidence interval.
- Open Excel, enter the given data in column A, and find the sample mean, x ¯, using the AVERAGE function. Thus, the sample mean is x ¯=9 17. 32.
- Click on any empty cell, enter = CONFIDENCE.NORM ( 0. 02 , 110 , 125 ), and press ENTER.
- The margin of error, rounded to two decimal places, is zα / 2 σn ‾√≈ 22. 89. The confidence interval for the population mean has a lower limit of
- 32 − 22. 89 = 894. 43 and an upper limit of 917. 32 +2 2. 89 =9 40. 21. Thus, the 98 %confidence interval for μ is ( 894. 43 , 940. 21 ).
0/1 points
Hugo averages 64 w o r d s per minute on a typing test with a standard deviation of 12 w o r d s per minute. Suppose Hugo's words per minute on a typing test are normally distributed. Let X =t h e number of words per minute on a typing test. Then, X ∼ N ( 64 , 12 ).
Suppose Hugo types 69 w o r d s per minute in a typing test on Wednesday. The z - score when x =6 9 is. This z - score tells you that x = 69 is standard deviations to the (right/left) of the mean,. Correctly fill in the blanks in the statement above.
That's not right.
Suppose Hugo types 69 w o r d s per minute in a typing test on Wednesday. The z - score when x = 69 is − 0. 417. This z - score tells you that x = 69 is 0. 417 standard deviations to the left of the mean, 64.
Suppose Hugo types 69 w o r d s per minute in a typing test on Wednesday. The z - score when x = 69 is − 0. 333. This z - score tells you that x = 69 is 0. 333 standard deviations to the left of the mean, 64.
Suppose Hugo types 69 w o r d s per minute in a typing test on Wednesday. The z - score when x =6 9 is 0. 333. This z - score tells you that x =6 9 is 0. 333 standard deviations to the right of the mean, 64.
