Question

An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 38 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level?

A. What is the p-value? (Round your answer to four decimal places.)

B. Construct a 95% confidence interval for the true mean. Label the point estimate and the lower and upper bounds of the confidence interval.

Answer 1

The question is about the population mean. We assume that the sample is random and independent. We can either use t-dist or normall distribution for the answeres. Since n > 30 and the sample SD is given we use the t-dist.

To test the claim if the mean time is longer than 4.5. So this is a one right tailed test.

= 4.5

Но : р

> 4.5

11:

n = 38   = 5.1 S_x = 1.2

T

Test Stat =

HO Sx/Vn

Where the null mean = 4.5

HO

test Stat = 3.0822

A. What is the p-value? (Round your answer to four decimal places.)

p - value = P( > |Test Stat |)   .............using t-dist tables

tn-1

=P(t₃₇ > 3.08)

p-value = 0.0019

Since p-value < 0.01

We reject the null hpothesis at 1%. There is sufficient evidence to support the claim that the mean time is longer.

B. Construct a 95% confidence interval for the true mean. Label the point estimate and the lower and upper bounds of the confidence interval.

(1- $1595828011381_blob.png$ )% is the confidence interval for population mean

SC (attn–1,0/27 n

Where

point estimate =
T
= 5.1

$1595828011351_blob.png$ =1 - 0.95 = 0.05

Therefore the C.V. =

ta/2.n-1

=t_0.025,37

= 2.0262   .............found using t-dist tables

Where Margin of error = C.v. * SE

=

SC tn-1,0/2 * n

Substituting the value

(4.7056, 5.4944)