Chapter 11.6, Problem 16PPS

To determine

Identify the hypothesis and claim.

Answer to Problem 16PPS

The null hypothesis $H_0:\mu =24$ does not get rejected because $Z-statistic$ for the sample does not fall within the critical region. The company’s claim thet each package of cheese contains exactly $24$ slices is correct.

Explanation of Solution

Given information:

A company claims that each package of cheese contains exactly $24$ slices. Using a sample of $93$ packages, Mr. Matthews calculated a mean of $24.1$ slices with a standard deviation of $0.5$ . Test the hypothesis at $5\%$ significance.

Calculation:

Consider null hypothsis $H_0$ and alternative hypothesis is $H_a$ . It is given as,

  $\begin{array}{l}{H}_0:\mu =24\\ H_a:\mu \ne 24\end{array}$

The claim is $\mu =24$ it is null hypothesis and it is equality. The alternative hypothesis is $\mu \ne 24$ is competent of $\mu =24$ .

The hypothesis test for alternative $\mu \ne 24$ is a left trailed test. Now test at $5\%$ significance.

Use $TI83$ calculator for value calculation,

Press $key{2}^{nd}$ and then DISTR. Choose $3^{rd}$ option.

  

Now calculate $Z-statistic$ for sample data,

  $z=\frac{\overline{x}-\mu }{\left(\frac{s}{\sqrt{n}}\right)}\mathrm{.....}\left(1\right)$

Now put the values, mean is $24.1$ slices, standard deviation of $0.5$ , sample of $93$ packages in equation $\left(1\right)$ .

  $\begin{array}{l}z=\frac{24.1-24}{\left(\frac{0.5}{\sqrt{93}}\right)}\\ =1.9228\end{array}$

Hence, the null hypothesis $H_0:\mu =24$ does not get rejected because $Z-statistic$ for the sample does not fall within the critical region. The company’s claim thet each package of cheese contains exactly $24$ slices is correct.