Question

1. A researcher is interested in finding a 98% confidence interval for the mean number of times per day that college students text. The study included 144 students who averaged 44.7 texts per day. The standard deviation was 16.5 texts.

a. To compute the confidence interval use a ? z t  distribution.

b. With 98% confidence the population mean number of texts per day is between  and   texts.

c. If many groups of 144 randomly selected members are studied, then a different confidence interval would be produced from each group. About  percent of these confidence intervals will contain the true population number of texts per day and about  percent will not contain the true population mean number of texts per day.

2. You want to obtain a sample to estimate how much parents spend on their kids birthday parties. Based on previous study, you believe the population standard deviation is approximately σ=40.4 dollars. You would like to be 90% confident that your estimate is within 1.5 dollar(s) of average spending on the birthday parties. How many parents do you have to sample? n =

3. You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately σ=57.5. You would like to be 95% confident that your estimate is within 0.1 of the true population mean. How large of a sample size is required?

n =

Answer 1

1)

a) z distribtuion

b)

sample mean 'x̄=	44.700
sample size    n=	144.00
std deviation σ=	16.500
std error ='σx=σ/√n=	1.3750

for 98 % CI value of z=		2.326
margin of error E=z*std error =		3.199
lower bound=sample mean-E=		41.5013
Upper bound=sample mean+E=		47.8987
from above 98% confidence interval for population mean =(41.50 ,47.90)

c) About 98 percent of these confidence intervals will contain the true population number of texts per day and about 2 percent will not contain the true population mean number of texts per day.

2)

	for90% CI crtiical Z          =	1.645
	standard deviation σ=	40.4
	margin of error E =	1.5
required sample size n=(zσ/E)2                  =		1963

3)

	for95% CI crtiical Z          =	1.960
	standard deviation σ=	57.5
	margin of error E =	0.1
required sample size n=(zσ/E)2                  =		12701129