91Ó°ÊÓ

$$\begin{gathered} H_0:\mathrm{μ}=19.2 \\ {H}_0:\mathrm{μ}notequalto19.2 \\ \mathrm{Î\pm }=0.10 \\ n=75 \\ \stackrel{¯}{x}=19.28 \\ s=0.81 \end{gathered}$$

We know,

$t=\frac{\stackrel{¯}{x}-{\mathrm{μ}}_0}{s/\sqrt{n}}$

The test statistic is

$$\begin{gathered} t=\frac{\stackrel{¯}{x}-{\mathrm{μ}}_0}{s/\sqrt{n}} \\ =\frac{19.28-19.2}{0.81/\sqrt{75}} \\ =0.855 \end{gathered}$$

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.

$P>0.05⇒\text{Fail to reject}{H}_0$

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.