Question

Two-sided test Suppose you want to perform a test of $H_0:\mu =64$

versus $H_a:\mu notequalto64$at the $\alpha =0.05$significance level. A random sample

of size $n=25$ from the population of interest yields $\overline{x}=62.8$ and $sx=5.36$

. Assume that the conditions for carrying out the test are met.

a. Explain why the sample result gives some evidence for the alternative hypothesis.

b. Calculate the standardized test statistic and P-value.

Short answer

Part a) Sample mean$62.8$differs from$64$

Part b) The required answer:

$$\begin{gathered} t=-1.119 \\ 0.20<P<0.30or \\ P=0.2742 \end{gathered}$$

Step-by-step solution

Part a) Step 1: Given information

$$\begin{gathered} H_0:\mu =64 \\ {H}_a:\mu isnotequalto64 \\ \alpha =0.05 \\ n=25 \\ \overline{x}=62.8 \\ s=5.36 \end{gathered}$$

Part a) Step 2: The objective is to explain the sample result gives some evidence for carrying out the test are met 

The sample mean $62.8$differs from $64,$which agrees with the alternative hypothesis that the mean is not $64,$and thus the sample result provides some evidence for the alternative hypothesis.

Part b) Step 1: Given information

$$\begin{gathered} H_0:\mu =64 \\ {H}_a:\mu isnotequalto64 \\ \alpha =0.05 \\ n=25 \\ \overline{x}=62.8 \\ s=5.36 \end{gathered}$$

Part b) Step 2: The objective is to calculate the standardized test static and P value. 

We know,

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

The t-test statistic is:

$$\begin{gathered} t=\frac{\overline{x}-{\mu }_0}{s/\sqrt{n}} \\ =\frac{62.8-64}{5.36/\sqrt{25}} \\ =-1.119 \end{gathered}$$

The P-value is the probability of receiving the test statistic's value, or a value more extreme, assuming the null hypothesis is true.

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

$0.20=2(0.10)<P<2(0.15)=0.30$

The command for the Ti83/84-calculator:

$2^{*}tcdf(-1\mathrm{E}99,-1.119,24)$which will return a P-value of $0.2742$