Question

Given the following hypothesis:

H0: μ = 100 H1: μ ≠ 100

A random sample of six resulted in the following values:

118 ,120 ,107 ,115 ,115 ,107

Using the 0.02 significance level, can we conclude that the mean is different from 100?

a. What is the decision rule? (Negative answer should be indicated by a minus sign. Round the final answers to 3 decimal places.)

Reject H0: μ = 100 and accept H1: μ ≠ 100 when the test statistic is ( , ).

b. Compute the value of the test statistic. (Round the final answer to 2 decimal places.)

Value of the test statistic c. What is your decision regarding H0? H0 .

d. Estimate the p-value.

e-1. Construct a 98% confidence interval. Confidence interval is from to .

e-2. Use the results of the confidence interval to support your decision in part (c). H0 .

Answer 1

a. What is the decision rule?

Solution:

We are given

n = 6

df = n – 1 = 5

α = 0.02

Test is two tailed.

So, the critical values are given as below:

Critical value: -3.3649, 3.3649

Reject H0: μ = 100 and accept H1: μ ≠ 100 when the test statistic is outside ( -3.365, 3.365).

b. Compute the value of the test statistic.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 100

Xbar = 113.6666667

S = 5.501514943

n = 6

df = n – 1 = 5

α = 0.02

t = (113.6666667 – 100)/[ 5.501514943/sqrt(6)]

t = 6.0849

Test statistic = 6.08

Part c

Test statistic is not lies between critical values, so we reject the null hypothesis H0.

Part d

P-value = 0.0017

(by using t-table)

P-value < α = 0.02

So, we reject the null hypothesis

There is not sufficient evidence to conclude that the population mean is equal to 100.

Part e-1

Confidence interval for Population mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

From given data, we have

Xbar = 113.6666667

S = 5.501514943

n = 6

df = n – 1 = 5

Confidence level = 98%

Critical t value = 3.3649

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 113.6666667 ± 3.3649*5.501514943/sqrt(6)

Confidence interval = 113.6666667 ± 7.5576

Lower limit = 113.6666667 - 7.5576 = 106.11

Upper limit = 113.6666667 + 7.5576 = 121.22

Confidence interval = (106.11, 121.22)

Confidence interval is from 106.11 to 121.22.

Part e-2

The value of population mean 100 is not lies between the above interval, so we reject the null hypothesis. So, this interval supports our decision in part c.