91影视

From Exercise 7. 120,

The number of hospital employees who were diagnosed with a latex allergy, n=46.

The sample mean is,\(\bar x = 19.3.\)

The sample standard deviation is,\(s = 11.9.\)

The test hypothesis is \({H_0}:{\sigma ^2} = 100\) against \({H_0}:{\sigma ^2} \ne 100\).

The condition required for a valid hypothesis test for \({\sigma ^2}\) are:

• A random sample is selected from the target population.
• The population from which the sample is selected has a distribution that is approximately normal.

The test statistic is used to test the hypothesis about \({\sigma ^2}\)is, \(\chi _c^2 = \frac{{\left( {n - 1} \right){s^2}}}{{\sigma _0^2}}\).

a.

Let \({\sigma ^2}\) represent the variance in the number of latex gloves used per week by all hospital employees.

The null and alternative hypothesis are:

\(\begin{aligned}{H_0}:{\sigma ^2} = 100\\{H_a}:{\sigma ^2} \ne 100\end{aligned}\)

The significance level, \(\alpha = 0.01.\)

This is a two-tailed test. So, the rejection region for the two-tailed test is\(\chi _c^2 < \chi _{\left( {1 - \alpha /2} \right)}^2\,\,or\,\,\chi _c^2 > \chi _{\alpha /2}^2\).

The degrees of freedom is computed as:

\(\begin{aligned}\left( {n - 1} \right) &= \left( {46 - 1} \right)\\ &= 45\end{aligned}\)

The \(\chi _{\alpha /2}^2\) value obtained from the chi-square table with \(\alpha = 0.01\) and degrees of freedom 45 is given as follows:

\(\begin{aligned}{c}\chi _{\alpha /2}^2 &= \chi _{0.01/2}^2\\ &= \chi _{0.005}^2\\ &= 73.17\end{aligned}\)

The \(\chi _{1 - \alpha /2}^2\) value obtained from the chi-square table with \(\alpha = 0.01\) and degrees of freedom 45 is given as follows:

\(\begin{aligned}\chi _{1 - \alpha /2}^2 &= \chi _{1 - 0.01/2}^2\\ &= \chi _{0.995}^2\\ &= 24.31\end{aligned}\)

Hence, the rejection region is \(\chi _c^2 < 24.31\,\,or\,\,\chi _c^2 > 73.17\).