Question

One-sided test Suppose you want to perform a test of H0: μ=5 versus Ha: μ<5

at the α=0.05 significance level. A random sample of size n=20 from the population of interest yields x¯=4.7 and sx=0.74 . 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) The sample mean $4.7$is less than $5$

Part b)

$$\begin{gathered} t=-1.813 \\ 0.25<P<0.05 \\ OrP=0.04283 \end{gathered}$$

Step-by-step solution

Part a) Step 1: Given information

$$\begin{gathered} H_0:\mu =5 \\ {H}_1:\mu <5 \\ \alpha =0.05 \\ n=20 \\ \overline{x}=4.7 \\ s=0.74 \end{gathered}$$

Part a) Step 2: Calculation

The sample mean of $4.7$is less than $5,$which agrees with the alternative hypothesis that the mean is less than $5,$and thus the sample result provides some evidence for the alternative hypothesis.

Part b) Step 1: Given information

$$\begin{gathered} H_0:\mu =5 \\ {H}_1:\mu <5 \\ \alpha =0.05 \\ n=20 \\ \overline{x}=4.7 \\ s=0.74 \end{gathered}$$

Part b) Step 2: Calculation

We know,

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

The test statistic is

$$\begin{gathered} t=\frac{\overline{x}-{\mu }_0}{sl\sqrt{n}} \\ =\frac{4.7-5}{0.74/\sqrt{20}} \\ =-1.813 \end{gathered}$$

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.

$$\begin{gathered} df=n-1=20-1=19. \\ 0.25<P<0.05 \end{gathered}$$

Command for Ti83/84-calculator: $(-1\mathrm{E}99,-1.813,19)$which will return a P-value of $0.04283.$ It could replace -1E99 by any other very small negative number.