Question

Attitudes In the study of older students’ attitudes from Exercise 65, the sample mean SSHA score was 125.7 and the sample standard deviation was 29.8.

a. Calculate the standardized test statistic.

b. Find and interpret the P-value.

c. What conclusion would you make?

Short answer

Part a) The required answer is$t=2.409$

Part b) $0.01<P<0.02$

There is a $1.012\%$possibility of getting a sample mean SSHA score of $125.7$or higher in a sample of 45 students when the population mean SSHA score is$115.$

Part c) There is enough convincing proof that the mean SSHA score in the population of students at her college who are at least 30 years old exceeds $115$

Step-by-step solution

Part a) Step 1: Given information

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

Part a) Step 2: Calculation

We know,

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

To find the hypotheses, do the following:

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

The test statistic is:

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

Part b) Step 1: Given information

$$\begin{gathered} H_0:\mu =115 \\ {H}_1:\mu >115 \\ \alpha =0.05 \\ n=45 \\ \overline{x}=125.7 \\ s=29.8 \end{gathered}$$

Part b) Step 2: Calculation

We know,

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

The test statistic is:

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

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$0.01012.$it could replace 1 E99 by any other very large positive number.
There is a $1.012\%$possibility of getting a sample mean SSHA score of $125.7$or higher in a sample of 45 students when the population means SSHA score is $115.$

Part c) Step 1: Given information

$$\begin{gathered} H_0:\mu =115 \\ {H}_1:\mu >115 \\ \alpha =0.05 \\ n=45 \\ \overline{x}=125.7 \\ s=29.8 \end{gathered}$$

Part b) Step 2: Calculation

We know,

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

Calculate the test statistic's value:

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

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 $\alpha ,$then the null hypothesis is rejected.

$P<0.05\Rightarrow \text{Reject}{H}_0$

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