Question

What is the probability of randomly selecting 60 females whose average height is at most 64 inches?

A.       0.28

B.      0.72

C.       0.47

D.      0.53

Suppose you want to determine the average home run percentage for the Royals baseball team. You randomly select 23 players from Royals history and determine that the mean number of home runs per season is 14.7.

Compute a 95% confidence interval for the mean home run percentage, given that the sample standard deviation was s = 3.45

A.       14.7 ± 1.10

B.      14.7 ± 1.13

C.       14.7 ± 1.16

D.      14.7 ± 1.49

Assume the population standard deviation is estimated to be σ = 3.28. What is the sample size needed to obtain a margin of error of 1.5 with 99% confidence?

A. 17

B. 32

C. 42

D. 60

Consider the situation given in Problem #26. Which of the following would produce a confidence interval with a smaller margin of error?

A. Using a confidence level of 90%				C. Using a larger sample size
B. Using a smaller estimate for σ				D. All of the above A-C

Answer 1

Suppose you want to determine the average home run percentage for the Royals baseball team. You randomly select 23 players from Royals history and determine that the mean number of home runs per season is 14.7.

n = sample size = 23

sample mean = = 14.7

sample standard deviation = s = 3.45

Margin of error ( E) is as

for c = 0.95 , and degrees of freedom = n - 1 = 23 - 1 = 22 the critical t value using excel is as

t = "=TINV(0.05,22)" = 2.074

So correct option is D.    14.7 ± 1.49

Assume the population standard deviation is estimated to be σ = 3.28. What is the sample size needed to obtain a margin of error of 1.5 with 99% confidence?

formula of sample size(n), is as follow:

= 3.28

E = 1.5

for c = 0.99 the critical z = 2.5857

So correct option is

B. 32

Consider the situation given in Problem #26. Which of the following would produce a confidence interval with a smaller margin of error?

The correct choice is D. All of the above A-C

Because as confidence level decreases the critical value also decreases

As n increases the margin of error decreases because is in denominator

As smaller estimate (s) of , E decreases because s is in numerator.