Question

In the study of older students’ attitudes from Exercise 63, the sample mean SSHA score was 125.7 and the sample standard deviation was 29.8.

(a) Calculate the test statistic.

(b) Find the P-value using Table B. Then obtain a more precise P-value from your calculator.

Short answer

a. the test statistic is$2.409$

b. The P-value using Table B is$0.0101$

Step-by-step solution

Introduction

A test statistic is a statistic utilized in statistical hypothesis testing. A hypothesis test is typically specified as far as a test statistic, considered as a numerical synopsis of an informational collection that decreases the information to one worth that can be utilized to play out the hypothesis test.

Explanation Part (a)

The sample size is n = $45$

standard deviation s = $29.8$

sample mean $\stackrel{-}{x}=125.7$

Using,

role="math" localid="1652954730678" $t=\frac{\stackrel{-}{x}-\mu }{\frac{s}{\sqrt{n}}}$

$=\frac{125.7-115}{\frac{29.8}{\sqrt{45}}}=2.409$

Explanation Part (b)

calculating the degree of freedom,

$n-1=45-1=44$

The p-value is found using,

$$\begin{gathered} =P\left(t_c>\left|0.77\right|\right) \\ =0.0101 \end{gathered}$$