Question

The one-sample t statistic from a sample of n =25 observations for the two-sided test of

$\begin{array}{r}{H}_0:\mu =64\\ H_d:\mu \ne 64\end{array}$

has the value t =- 1.12.

(a) Find the P-value for this test using (i) Table B and (ii) your calculator. What conclusion would you draw at the 5% significance level? At the 1% significance level?

(b) Redo part (a) using an alternative hypothesis of $H_a:\mu <64$

Short answer

a. p-value is$0.2738$

b. p-value is$0.1369$

Step-by-step solution

Introduction

The scientific strategy requires that one can test it. Scientists for the most part base scientific speculations on previous observations that can't satisfactorily be explained with the available scientific theories

Explanation Part (a)

The hypotheses are,

$\begin{array}{r}{H}_0:\mu =64\\ H_a:\mu \ne 64\end{array}$

calculating the degree of freedom,

$$\begin{gathered} Df=n-1 \\ =25-1=24 \end{gathered}$$

Using a calculator the p-value is found to be$0.136$

For the two-sided test p-value is=$2\times 0.1369=0.2738$

localid="1654666340202" $\Rightarrow 0.2738>(\alpha =0.01\text{and}\alpha =0.05)$

Hence the p-value is $0.2738$and as it is greater than the significance levels the null hypothesis is not rejected.

Explanation Part (b)

The hypotheses,

$\begin{array}{r}{H}_0:\mu =64\\ H_a:\mu <64\end{array}$

Calculating the p-value for one-tailed test,

localid="1654666370875" $$\begin{gathered} =P(t>\mid t|) \\ =P(t>|-1.12|)=0.1369 \end{gathered}$$

Hence the p-value is $0.1369$and as it is greater than the significance levels the null hypothesis is not rejected.