91Ó°ÊÓ

$$\begin{gathered} n=45 \\ \stackrel{¯}{x}=125.7 \\ s=29.8 \end{gathered}$$

We know,

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

To find the hypotheses, do the following:

$$\begin{gathered} H_0:\mathrm{μ}=115 \\ {H}_1:\mathrm{μ}>115 \end{gathered}$$

The test statistic is:

$$\begin{gathered} t=\frac{\stackrel{¯}{x}-{\mathrm{μ}}_0}{s/\sqrt{n}} \\ =\frac{125.7-115}{29.8/\sqrt{45}} \\ =2.409 \end{gathered}$$

$$\begin{gathered} H_0:\mathrm{μ}=115 \\ {H}_1:\mathrm{μ}>115 \\ \mathrm{Î\pm }=0.05 \\ n=45 \\ \stackrel{¯}{x}=125.7 \\ s=29.8 \end{gathered}$$

We know,

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

The test statistic is:

$$\begin{gathered} t=\frac{\stackrel{¯}{x}-{\mathrm{μ}}_0}{s/\sqrt{n}} \\ =\frac{125.7-115}{29.8\sqrt{45}} \\ =2.409 \end{gathered}$$

If the null hypothesis is true, the P-value is the probability of getting the test statistic's value or a value that is more extreme.

$df=n-1=45-1=44$Since the table is not having a row with $df=44$so will use $df=40$it instead

$0.01<P<0.02$

$$\begin{gathered} H_0:\mathrm{μ}=115 \\ {H}_1:\mathrm{μ}>115 \\ \mathrm{Î\pm }=0.05 \\ n=45 \\ \stackrel{¯}{x}=125.7 \\ s=29.8 \end{gathered}$$

We know,

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

Calculate the test statistic's value:

$$\begin{gathered} t=\frac{\stackrel{¯}{x}-{\mathrm{μ}}_0}{s/\sqrt{n}} \\ =\frac{125.7-115}{29.8\sqrt{45}} \\ =2.409 \end{gathered}$$

If the null hypothesis is true, the P-value is the probability of getting the test statistic's value or a value that is more extreme.

$df=n-1=45-1=44$Since the table is not having a row with $df=44$so will use $df=40$it instead.

$0.01<P<0.02$

Command Ti83/84-calculator: tcdf( $(2.409,1\mathrm{E}99,44)$which will return a P-value of$0.01012$ Note: it could replace 1 E99 by any other very large positive number.
If the P-value is lesser than the significance level $\mathrm{Î\pm },$then the null hypothesis is rejected.

$P<0.05⇒\text{Reject}{H}_0$

There is sufficient evidence that the average SSHA score of students at her college who are at least 30 years old exceeds $115.$