Chapter 11.6, Problem 7CYU

To determine

Identify the hypothesis and claim.

Answer to Problem 7CYU

  $H_0:\mu \ge 84\left(claim\right)$ , $H_a:\mu <84$ , do not reject $H_0$ , the manufacturer claims that discs can hold at least $84$ minutes cannot be rejected.

Explanation of Solution

Given information:

A manufacturer of blank compact discs claims that each disc can hold at least $84$ minutes of music. Using a sample of $219$ compact discs, Cayla calculated a mean time of $84.1$ minutes per disc with a standard deviation of $1.9$ minutes. Test the hypothesis at $5\%$ significance.

Calculation:

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

  $\begin{array}{l}{H}_0:\mu \ge 84\\ H_a:\mu <84\end{array}$

The claim is $H_0:\mu \ge 84$ it is null hypothesis and it is equality. $H_a:\mu <84$ alternative hypothesis is competent of $H_0:\mu \ge 84$ .

The hypothesis test for alternative $\mu <84$ is a left trailed test. Now test at $5\%$ of the distribution.

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 time of $84.1$ minutes per disc, standard deviation of $1.9$ minutes, sample of $219$ compact discs in equation $\left(1\right)$ .

  $\begin{array}{l}z=\frac{84.1-84}{\left(\frac{1.9}{\sqrt{219}}\right)}\\ =0.7788\end{array}$

Null hypothesis could not be rejected because the $Z-statistic$ is not within the critical region.

Hence, $H_0:\mu \ge 84\left(claim\right)$ , $H_a:\mu <84$ , do not reject $H_0$ , the manufacturer claims that discs can hold at least $84$ minutes cannot be rejected.