Question

A pipes manufacturer makes pipes with a length that is supposed to be 19 inches. A quality control technician sampled 24 pipes and found that the sample mean length was 19.08 inches and the sample standard deviation was 0.24 inches. The technician claims that the mean pipe length is not 19 inches. What type of hypothesis test should be performed? What is the test statistic? What is the number of degrees of freedom? Does sufficient evidence exist at the α=0.01 significance level to support the technician's claim?

Answer 1

Solution :

Since, sample size is small population standard deviation is unknown and we assume that population is normally distributed, therefore one sample t-test would be most appropriate test.

And the test would be two-tailed.

Null and alternative hypotheses :

The null and alternative hypotheses would be as follows:

Test statistic :

The test statistic is given as follows :

Where, x̅ is sample mean, μ is hypothesized value of population mean under H_{​​​​​​0}, s is sample standard deviation and n is sample size.

We have,  x̅ = 19.08 , μ = 19 , s = 0.24 and n = 24

The value of the test statistic is 1.6330.

Degrees of freedom = (n - 1) = (24 - 1) = 23

P-value :

Since, our test is two-tailed test, therefore we shall obtain two-tailed p-value for the test statistic. The two-tailed p-value is given as follows :

P-value = 2.P(T > |t|)

We have, |t| = 1.6330

P-value = 2.P(T > 1.6330)

P-value = 0.1161

The p-value is 0.1161.

Decision :

Significance level = 0.01

P-value = 0.1161

(0.1161 > 0.01)

Since, p-value is greater than the significance level of 0.01, therefore we shall be fail to reject the null hypothesis (H_{​​​​​​0}).

Conclusion :

At 0.01 significance level, there is not sufficient evidence to support the technician's claim that the mean pipe length is not 19 inches.

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