91影视

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

We know,

$t=\frac{\stackrel{炉}{x}-{渭}_0}{s/\sqrt{n}}$

To find the hypotheses, do the following:

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

The test statistic is:

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

$$\begin{gathered} H_0:渭=115 \\ {H}_1:渭>115 \\ 伪=0.05 \\ n=45 \\ \stackrel{炉}{x}=125.7 \\ s=29.8 \end{gathered}$$

We know,

$t=\frac{\stackrel{炉}{x}-{渭}_0}{s/\sqrt{n}}$

The test statistic is:

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

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

$0.01<P<0.02$

$$\begin{gathered} H_0:渭=115 \\ {H}_1:渭>115 \\ 伪=0.05 \\ n=45 \\ \stackrel{炉}{x}=125.7 \\ s=29.8 \end{gathered}$$

We know,

$t=\frac{\stackrel{炉}{x}-{渭}_0}{sl\sqrt{n}}$

Calculate the test statistic's value:

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

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 $伪,$then the null hypothesis is rejected.

$P<0.05鈬�\text{Reject}{H}_0$

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