Question

usign R solve

In a class survey, students are asked how many hours they sleep per night. In the sample of 22 students, the (sample) mean is 6.77 hours with a (sample) standard deviation of 1.472 hours. The parameter 1 of interest is the mean number of hours slept per night in the population from which this sample was drawn, and the distribution of sleep for that population follows a normal distribution.

a. Which kind of confidence interval is appropriate to use here, z-interval or t-interval?

b. What are the criteria to check in order to use the distribution you chose?

c. Please use R to find the critical value they need when constructing a 90% CI.

d. Please use R to find the 90% CI for the mean number of hours slept per night.

Answer 1

a. Which kind of confidence interval is appropriate to use here, z-interval or t-interval?

since sample std deviation is given.use t interval.also sample size is less than 30

here n=22

. What are the criteria to check in order to use the distribution you chose?

disribution to be normal

sample is random sample

to check whether distribution is normal .

if n<30 t distribution

if n>30 large sample follows normal distribution according to central limit theorem.

c. Please use R to find the critical value they need when constructing a 90% CI.

n <- 22
qt(0.95,df=n-1)

1.720743

Solutiond:

Rcode to get confidence interval for mean is

xbar <- 6.77
s <- 1.472
n <- 22
qt(0.95,df=n-1)
margn_of_error <-qt(0.95,df=n-1)*s/sqrt(n)
lowerlimit <- xbar-margn_of_error
lowerlimit
upperlimit <- xbar+margn_of_error
upperlimit

LOWER LIMIT=6.229977

UPPER LIMIT=7.310023