Question

Confidence Level Question 1

One hundred random samples, each of size 25, are obtained from the Normal distribution with mean 0 and standard deviation 1 using Minitab. Subsequently, the 1-Sample Z procedure in Minitab is used (with the same confidence level) to obtain a confidence interval from each sample. Out of the 100 intervals thus obtained, 89 include the number 0. Estimate the confidence level (in percentage terms) used to generate the 100 intervals using a 95% confidence interval.

	a.	(87.3, 90.7)
	b.	(86.5, 91.5)
	c.	(82.9, 95.1)
	d.	(80.1, 97.9)
	e.	(85.7, 92.3)

Confidence Level Question 2

One hundred random samples, each of size 25, are obtained from the Normal distribution with mean 0 and standard deviation 1 using Minitab. Subsequently, the 1-Sample Z procedure in Minitab is used (with the same confidence level) to obtain a confidence interval from each sample. Out of the 100 intervals thus obtained, 89 include the number 0. What is the value of the appropriate test statistic if we want to show that the confidence level used to generate the 100 intervals was less than 95%?

	a.	-0.33
	b.	-1.01
	c.	-1.11
	d.	-2.29
	e.	-2.75

Answer 1

Confidence Level Question 1

sample success x =		89
sample size          n=		100.0
sample proportion p̂ =x/n=		0.8900
std error se= √(p*(1-p)/n) =		0.0313
for 95 % CI value of z=		1.960
margin of error E=z*std error   =		0.0613
lower bound=p̂ -E                       =		0.829
Upper bound=p̂ +E                     =		0.951

option C is correct: (82.9, 95.1)

Confidence Level Question 2

std error   se =√(p*(1-p)/n) =		0.0218
sample proportion p̂ = x/n=		0.8900
test stat z =(p̂-p)/√(p(1-p)/n)=		-2.75

option E is correct