91Ó°ÊÓ

Refer to Example 5 in sections 8 to 3 for the test of mean human body temperatures of\(98.6{\rm{^\circ F}}\). Assume that the standard deviation measure for a sample of 106 temperatures changes to\(2.08{\rm{^\circ F}}\).

The level of significance is 0.01 to test the claim that the standard deviation of the population is less than \(2.08{\rm{^\circ F}}\).

To test that the sample is taken from the population of body temperatures with a standard deviation less than\({2.08^ \circ }\)F, the null and alternative hypotheses are formulated as follows.

\(\begin{array}{l}{H_0}:\sigma = 2.08\\{H_1}:\sigma < 2.08\end{array}\)

Here, \(\sigma \) is the actual standard deviation for the population.

The test statistic\({\chi ^2}\)with\(\left( {n - 1} \right)\)degrees of freedom is given as follows.

\(\begin{array}{c}{\chi ^2} = \frac{{\left( {n - 1} \right){s^2}}}{{{\sigma ^2}}}\\ = \frac{{\left( {106 - 1} \right){{\left( {0.62} \right)}^2}}}{{{{\left( {2.08} \right)}^2}}}\\ = 9.329\end{array}\).

The degree of freedom for the test is computed as follows.

\(\begin{array}{c}df = n - 1\\ = 106 - 1\\ = 105\end{array}\)

The test is left-tailed.

The critical value of\(P\left( {{\chi ^2} < {\chi ^2}_{0.01}} \right) = 0.01\).

From the chi-square table, the critical value is obtained corresponding to the degree of freedom 105 as 70.065.

The decision rule states that if the critical value is greater than the test statistic, reject the null hypothesis; otherwise, fail to reject the null hypothesis.

As the test statistic is less than the critical value, the null hypothesis is rejected.

There is sufficient evidence to support the claim that the body temperatures have a standard deviation of less than\(2.08{\rm{^\circ F}}\).

In Example 5, from sections 8 to 3, the change in the sample standard deviation is unlikely to affect the results in the hypothesis test as the sample changes. This is because the test statistic observed in the example is extreme and will not change significantly to alter the test results.