91影视

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

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: $p鈮�\frac{3}{8}$

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

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:

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

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 鈥渮-score鈥�.
• Sketch a bell-shaped curve and draw a vertical dotted line corresponding to the value 鈥�0鈥� on the horizontal axis
• Mark the points 鈥�-1.94鈥� and 鈥�1.94鈥� on the horizontal axis and then shade the area to the left of the value 鈥�-1.94鈥� and to the right of the value 鈥�1.94鈥� with blue as shown in the figure.
• Label the shaded regions as 鈥減-value = 0.0262鈥�.

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).

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.