Chapter 11.6, Problem 17PPS

To determine

Identify, the manufacturer claim to place a label on the can promising exactly $12$ peaches in every can.

Answer to Problem 17PPS

The null hypothesis is rejected and the company should not make the claim on the label.

Explanation of Solution

Given information:

The number of peaches in $40$ random cans is shown below. Should the manufacturer place a label on the can promising exactly $12$ peaches in every can? Explain your reasoning.

  $\begin{array}{l}13,14,13,14,12,12,12,11,15,12,13,13,14,13,14,12,15,11,11,14,\\ 13,14,14,13,12,12,12,12,13,13,11,14,14,13,14,13,13,14,12,12\end{array}$

Calculation:

Calculate the mean of given data,

  $\begin{array}{l}13,14,13,14,12,12,12,11,15,12,13,13,14,13,14,12,15,11,11,14,\\ 13,14,14,13,12,12,12,12,13,13,11,14,14,13,14,13,13,14,12,12\end{array}$

  $\begin{array}{l}Mean=\frac{\text{Sum of given numbers}}{n}\\ n=\text{Total number of observations}\\ Mean=\frac{13+14+13+14+12+12+12+11+15+12+\mathrm{....}+12}{40}\\ =\frac{516}{40}\\ =12.9\\ now,find,\text{standard deviation }\sigma \\ {\sigma }^2=\frac{{\displaystyle \sum f{\left(x-\overline{x}\right)}^2}}{n}\left[\begin{array}{l}\therefore x,is,number\\ f=frequency\end{array}\right]\end{array}$

  $\begin{array}{l}{\sigma }^2=\frac{1}{40}\left[\begin{array}{l}4\times {\left(11-12.9\right)}^2+11\times {\left(12-12.9\right)}^2+12\times {\left(13-12.9\right)}^2\\ +11\times {\left(14-12.9\right)}^2+2\times {\left(15-12.9\right)}^2\end{array}\right]\\ =\frac{1}{40}\left[14.44+8.91+0.12+13.31+8.82\right]\\ =1.14\\ \sigma =\sqrt{1.14}\\ \sigma \approx 1.07\end{array}$

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

  $\begin{array}{l}{H}_0:\mu =12\\ H_a:\mu \ne 12\end{array}$

The claim is $\mu =12$ it is null hypothesis and it is equality. The alternative hypothesis is $\mu \ne 12$ is competent of $\mu =12$ .

The hypothesis test for alternative $\mu \ne 12$ is a left trailed test. Now test at $1\%$ 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 $12.9$ , standard deviation of $1.07$ , sample of $40$ peaches in equation $\left(1\right)$ .

  $\begin{array}{l}z=\frac{12.9-12}{\left(\frac{1.07}{\sqrt{40}}\right)}\\ =5.34\end{array}$

  $H_0:\mu =12$ , $H_a:\mu \ne 12$ , the mean of the sample data is $12.9$ and standard deviation is $1.07$ . The $Z-statistic$ is about $5.34$ which is fall in the critical region at $1\%$ significance.

Hence, the null hypothesis is rejected and the company should not make the claim on the label.