Chapter 11.6, Problem 18PPS

To determine

Identify, the manufacturer place a label on the package promising exactly $20$ chips on every cookie.

Answer to Problem 18PPS

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

Explanation of Solution

Given information:

The number of chocolate chips in $40$ random cookies is given below. Should the manufacturer place a label on the package promising exactly $20$ chips on every cookie? Explain your reasoning.

  $\begin{array}{l}21,19,20,20,,19,19,18,21,19,17,19,18,18,20,20,19,18,20,19,20,\\ 21,21,19,17,17,18,19,19,20,17,22,21,21,20,19,18,19,17,17,21\end{array}$

Calculation:

Calculate the mean of given data,

  $\begin{array}{l}21,19,20,20,,19,19,18,21,19,17,19,18,18,20,20,19,18,20,19,20,\\ 21,21,19,17,17,18,19,19,20,17,22,21,21,20,19,18,19,17,17,21\end{array}$

  $\begin{array}{l}Mean=\frac{\text{Sum of given numbers}}{n}\\ n=\text{Total number of observations}\\ Mean=\frac{21+19+20+\mathrm{...}+21}{40}\\ =\frac{767}{40}\\ \approx 19.2\\ now,find,\text{standard deviation }\sigma \\ {\sigma }^2=\frac{{\displaystyle \sum f{\left(x-\overline{x}\right)}^2}}{n}\left[\begin{array}{l}\therefore x,is,number\\ f=frequency\end{array}\right]\end{array}$

  $\begin{array}{l}{\sigma }^2=\frac{1}{40}\left[\begin{array}{l}6\times {\left(17-19.2\right)}^2+6\times {\left(18-19.2\right)}^2+12\times {\left(19-19.2\right)}^2\\ +8\times {\left(20-19.2\right)}^2+7\times {\left(21-19.2\right)}^2+1\times {\left(22-19.2\right)}^2\end{array}\right]\\ =\frac{1}{40}\left[29.04+8.64+0.48+5.12+22.68+7.84\right]\\ =1.845\\ \sigma =\sqrt{1.845}\\ \sigma \approx 1.36\end{array}$

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

  $\begin{array}{l}{H}_0:\mu =20\\ H_a:\mu \ne 20\end{array}$

The claim is $\mu =20$ it is null hypothesis and it is equality. The alternative hypothesis is $\mu \ne 20$ is competent of $\mu =20$ .

The hypothesis test for alternative $\mu \ne 20$ is a two 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 $19.2$ , standard deviation of $1.36$ , sample of $40$ cookies in equation $\left(1\right)$ .

  $\begin{array}{l}z=\frac{19.2-20}{\left(\frac{1.36}{\sqrt{40}}\right)}\\ =-3.7203\end{array}$

The null hypothesis is rejected because the $Z-statistic$ for sample falls within the critical region.

Hence, the company should not make the claim on the label.