Question

1.

Denis Shapovalov is a young Canadian tennis star. We would like to estimate the true mean speed of all of Shapovalov's serves. Speeds of his serves follow a normal distribution with standard deviation 15 km/h. What sample size is required to estimate μ to within 2 km/h with 94% confidence?

Question 33 options:

	205
	199
	211
	208
	202

2.

We would like to estimate the true mean duration of all Major League Baseball games. Game durations are known to follow a normal distribution with standard deviation 22 minutes. The average duration of a random sample of 17 games was calculated to be 154 minutes. An 87% confidence interval for the true mean duration of all Major League Baseball games is:

Question 36 options:

	(145.49, 162.51)
	(146.88, 161.12)
	(147.97, 160.03)
	(146.32, 161.68)
	(145.94, 162.06)

3.

It is determined that, in order to estimate the true mean lifetime of a particular brand of battery to within 0.1 hours with 95% confidence, a sample of 500 batteries is required. What sample size would be required to estimate the true mean lifetime to within 0.25 hours with 95% confidence?

Question 37 options:

	80
	316
	1250
	200
	3125

4.

A simple random sample of size 23 is drawn from a normal population with an unknown standard deviation. The sample standard deviation is calculated to be 6.59. A confidence interval for the population mean μ is found to be 35 ± 3. What is the confidence level of the confidence interval?

Question 38 options:

	99%
	90%
	98%
	96%
	95%

Answer 1

1)

given that

SD=15

confidence level =94%

margin of error =E=2

now

so

Answer is 199

2)

we have SD=22

sample size =n=17

sample mean =m=154

now

87% confidence interval is given by

so interval is (145.92,162.08)

so in option its closed to (145.94,162.06) so this is the correct answer

3)

for

n=500 we have E=0.1

now

now for E=0.25, we have to find n

so n~80

Hence 80 is the correct answer.

4)

we have

n=23

sample SD=S=6.59

margin of error =E=3

since we dont have population standard deviation so we will use t statistics with DF=n-1=23-1=22

so

so

this gives alpha =0.04

Hence confidence level =(1-alpha)*100 =(1-0.04)*100 =96%

Hence answer is 96%