Chapter 9.3, Problem 72E

(a)

To determine

To Calculate: the standardized test statistic.

Answer to Problem 72E

t=0.855

Explanation of Solution

Given:

  $\begin{array}{l}{H}_0:\mu =19.2\\ H_1:\mu \ne 19.2\\ \alpha =0.10\\ n=75\\ \overline{x}=19.28\\ s=0.81\end{array}$

Formula used:

  $t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}$

Calculation:

The test statistic is

  $\begin{array}{c}t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}\\ =\frac{19.28-19.2}{0.81/\sqrt{75}}\\ =0.855\end{array}$

(b)

To determine

To find: and interpret the P-value.

Answer to Problem 72E

0.30 < P < 0.40 or

P = 0.39532

There is a 39.532% possibility of getting a sample mean amount of candy of 19.28 ounces in 75 bags of candies, when the population mean amount of candy is 19.2 ounces.

Explanation of Solution

Given:

  $\begin{array}{l}{H}_0:\mu =19.2\\ H_1:\mu \ne 19.2\\ \alpha =0.10\\ n=75\\ \overline{x}=19.28\\ s=0.81\end{array}$

Formula used:

  $t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}$

Calculation:

The test statistic is

  $\begin{array}{c}t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}\\ =\frac{19.28-19.2}{0.81/\sqrt{75}}\\ =0.855\end{array}$

The P-value is the probability of getting the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. $df=n-1=75-1=74$ . It required to double the boundaries of the value of the test statistic, because the test is two-tailed since $df=74$ is not available in the table, so use the nearest smaller degrees of freedom $df=60$ instead.

  $0.30=2\left(0.15\right)<P<2\left(0.20\right)=0.40$

Command Ti83/84-calculator: 2*tcdf (0.855, 1E99,74) which will return a P-value of 0.39532. Note: it could replace 1E99 by any other very large positive number. There is a 39.532% possibility of getting a sample mean amount of candy of 19.28 ounces in 75 bags of candies, when the population mean amount of candy is 19.2 ounces.

(c)

To determine

To find: the conclusion would make.

Answer to Problem 72E

There is no enough convincing proof that the mean amount of candy that the machine put in all bags filled that day differs from 19.2 ounces.

Explanation of Solution

Given:

  $\begin{array}{l}{H}_0:\mu =19.2\\ H_1:\mu \ne 19.2\\ \alpha =0.10\\ n=75\\ \overline{x}=19.28\\ s=0.81\end{array}$

Formula used:

  $t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}$

Calculation:

The test statistic is

  $\begin{array}{c}t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}}\\ =\frac{19.28-19.2}{0.81/\sqrt{75}}\\ =0.855\end{array}$

The P-value is the probability of getting the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. $df=n-1=75-1=74$ . Note: it is required to double the boundaries of the value of the test statistic, the reason is the test is two-tailed (due to the $\ne$ in the alternative hypothesis $H_1$ ).

Although $df=74$ is not available in the table, there is need to use the nearest smaller degrees of freedom $df=60$ instead.

  $0.30=2\left(0.15\right)<P<2\left(0.20\right)=0.40$

Command Ti83/84-calculator: 2*tcdf (0.855, 1E99,74) which will return a P-value of 0.39532. Note: it could replace 1E99 by any other very large positive number. If the P-value is lesser than the significance level $\alpha$ , then the null hypothesis is rejected.

  $P>0.05\Rightarrow \text{Fail to reject }{H}_0$

There is no enough convincing proof that the mean amount of candy that the machine put in all bags filled that day differs from 19.2 ounces.