Question

# Given the following hypothesis:     H0 : μ = 125     H1 : μ ≠ 125...

 Given the following hypothesis:

 H0 : μ = 125

 H1 : μ ≠ 125

 A random sample of six resulted in the following values 132, 132, 130, 143, 140, and 130.

 Using the 0.05 significance level, can we conclude the mean is different from 125?

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

 Reject H0 : μ = 125 and fail to reject H1 : μ ≠ 125 when the test statistic is (Click to select)inside the intervaloutside the interval ( ,   ).

 (b) The value of the test statistic is (Round your answer to 3 decimal places.)

 (c) What is your decision regarding H0 ? (Click to select)Fail to rejectReject

 (d) Estimate the p-value. (Click to select)Between 0.01 and 0.05Less than 0.01Greater than 0.1Between 0.05 and 0.2

Part a

Here, we have to use one sample t test for the population mean.

The null and alternative hypotheses are given as below:

H0: µ = 125 versus Ha: µ ≠ 125

This is a two tailed test.

The test statistic formula is given as below:

t = (Xbar - µ)/[S/sqrt(n)]

From given data, we have

µ = 125

Xbar = 134.5

S = 5.576737397

n = 6

df = n – 1 = 5

α = 0.05

Critical value = - 2.5706 and 2.5706

(by using t-table or excel)

 Reject H0 : μ = 125 and fail to reject H1 : μ ≠ 125 when the test statistic is outside the interval (-2.571, 2.571).

Part b

t = (Xbar - µ)/[S/sqrt(n)]

t = (134.5 – 125)/[ 5.576737397/sqrt(6)]

t = 4.1727

Value of the test statistic = 4.173

Part c

Value of the test statistic is outside the above critical values, so we reject H0.

Part d

P-value = 0.0087

(by using t-table)

(Less than 0.01)

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