Question

P-Values. In Exercises 17–20, do the following:

a. Identify the hypothesis test as being two-tailed, left-tailed, or right-tailed.

b. Find the P-value. (See Figure 8-3 on page 364.)

c. Using a significance level of \(\alpha \) = 0.05, should we reject \({H_0}\)or should we fail to reject \({H_0}\)?

The test statistic of z = -1.94 is obtained when testing the claim that \(p = \frac{3}{8}\).

Short answer

a. The test is two-tailed.

b. The p-value is equal to 0.0524.

c. The decision of the statistical test is to fail to reject \({H_0}\).

Step-by-step solution

Given information

A test statistic value of \(z = - 1.94\) is obtained, and the claim to be tested is \(p = \frac{3}{8}\).

Identify the hypotheses and tail of the test

a.

In correspondence with the given claim, the claim cannot be considered the alternative hypothesis as it always suggests a difference in the hypothesized value and the population parameter.

The given claim is assumed to be the null hypothesis.

Thus,

Null Hypothesis: \(p = \frac{3}{8}\)

Alternative Hypothesis: \(p \ne \frac{3}{8}\)

Since there is a not-equal-to sign in the alternative hypothesis, the test is two-tailed.

P-value

b.

The test statistic to test the given claim is the z-value.

The z-value is equal to -1.94.

Using the standard normal table, the corresponding two-tailed p-value for z-score equal to -1.94 is equal to:

\(\begin{aligned}{c}2P\left( {z < - 1.94} \right) = 2\left( {0.0262} \right)\\ = 0.0524\end{aligned}\)

Thus, the p-value is equal to 0.0524.

To depict the p-value on the normal probability graph, follow the given steps:

• Plot a horizontal axis representing the z-score. Also, label it as “z-score”.
• Sketch a bell-shaped curve and draw a vertical dotted line corresponding to the value “0” on the horizontal axis
• Mark the points “-1.94” and “1.94” on the horizontal axis and then shade the area to the left of the value “-1.94” and to the right of the value “1.94” with blue as shown in the figure.
• Label the shaded regions as “p-value = 0.0262”.

The following plot shows the probability value (p-value) as the shaded area under the normal probability graph. Here, the sum of the two p-values correspond to the required two-tailed p-value (0.0524).

Decision about the test

c.

If the p-value is less than the level of significance, the null hypothesis is rejected; otherwise, not.

Here, the level of significance is equal to 0.05, and the p-value is equal to 0.0524.

Since the p-value is greater than 0.05, so the decision is to fail to reject the null hypothesis.