Candy! In the study of the candy machine from Exercise 66, the sample mean weight for the bags of candy was 19.28 ounces and the sample standard deviation was 0.81 ounce.

a. Calculate the standardized test statistic.

b. Find and interpret the P-value.

c. What conclusion would you make?

$$\begin{gathered} H_0:\mu =19.2 \\ {H}_0:\mu notequalto19.2 \\ \alpha =0.10 \\ n=75 \\ \overline{x}=19.28 \\ s=0.81 \end{gathered}$$

We know,

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

The test statistic is

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

From part (a)

we have,

$t=0.855$

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 ≠ in the alternative hypothesis $H_1$).

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

$0.30=2(0.15)<P<2(0.20)=0.40$

Command Ti83/84-calculator: $2^{*}$tcdf $(0.855,1\mathrm{E}99,74)$which will return a P-value of $0.39532$Note: it could replace 1 E99 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.

From part (b)

We have,

$0.30=2(0.15)<P<2(0.20)=0.40$

Command Ti83/84-calculator: $2*tcdf(0.855,1\mathrm{E}99,74)$) 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 α, 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.