Question

For a test of population proportion H0: p = 0.50, the z test statistic equals 1.05. Use 3 decimal places. (a) What is the p-value for Ha: p > 0.50? .147 Correct: Your answer is correct. (b) What is the p-value for Ha: p ≠ 0.50? .294 Correct: Your answer is correct. (c) What is the p-value for Ha: p < 0.50? (Hint: The p-values for the two possible one-sided tests must sum to 1.) .8531 Correct: Your answer is correct. (d) Which of the p-values give strong evidence against H0? Select all that apply. The p-value in (a). The p-value in (b). The p-value in (c). None of the p-values give strong evidence against H0. Incorrect: Your answer is incorrect.

Answer 1

Answer:

(a) The p-value of the test statistic is 0.147.

(b) The p-value of the test statistic is 0.294.

(c) The p-value of the test statistic is 0.8531.

(d) None of the p-values give strong evidence against the null hypothesis.

Step-by-step explanation:

The p-value is well defined as the probability,[under the null hypothesis (H₀)], of attaining a result equivalent to or greater than what was the truly observed value of the test statistic.

We reject a hypothesis if the p-value of a statistic is lower than the level of significance α.

The null hypothesis for the test of population proportion is defined as:

H₀: p = 0.50

The value of z-test statistic is,

z = 1.05

(a)

The alternate hypothesis is defined as:

Hₐ: p > 0.50

Compute the p-value of the test statistic as follows:

$p-value=P(Z>1.05)\\=1-P(Z<1.05)\\=1-0.8531\\=0.1469\\\approx 0.147$

*Use a z-table for the probability value.

Thus, the p-value of the test statistic is 0.147.

(b)

The alternate hypothesis is defined as:

Hₐ: p ≠ 0.50

Compute the p-value of the test statistic as follows:

$p-value=2\times P(Z>1.05)\\=2\times 0.1469\\=0.2938\\\approx 0.294$

*Use a z-table for the probability value.

Thus, the p-value of the test statistic is 0.294.

(c)

The alternate hypothesis is defined as:

Hₐ: p < 0.50

Compute the p-value of the test statistic as follows:

$p-value= P(Z<1.05)\\=1-P(Z>1.05)=1- 0.1469\\=0.8531$

*Use a z-table for the probability value.

Thus, the p-value of the test statistic is 0.8531.

(d)

The decision rule of the test is:

If the p-value of the test is less than the significance level α, then the null hypothesis is rejected at α% level of significance.

And if the p-value of the test is more than the significance level α, then the null hypothesis is failed to be rejected.

The most commonly used level of significance are:

α = 0.01, 0.05 and 0.10

The p-value for all the three alternate hypothesis are:

p-values = 0.147, 0.294 and 0.8531.

All the p-values are quite large compared to the α values.

Thus, none of the p-values give strong evidence against the null hypothesis.

The null hypothesis was failed to be rejected.

Answer 2

In this exercise, we have to use statistical knowledge to find that:

(a) The p-value of the test statistic is 0.147.

(b) The p-value of the test statistic is 0.294.

(c) The p-value of the test statistic is 0.8531.

(d) None of the p-values give strong evidence against the null hypothesis.

The p-value is well defined as the probability,[under the null hypothesis (H₀)], of attaining a result equivalent to or greater than what was the truly observed value of the test statistic. We reject a hypothesis if the p-value of a statistic is lower than the level of significance α. The null hypothesis for the test of population proportion is defined as:

$H_0: p = 0.50$

The value of z-test statistic is, $z = 1.05$

(a) The alternate hypothesis is defined as:

$H_a: p > 0.50$

Compute the p-value of the test statistic as follows:

$p-value= P(Z>1.05)\\=1-P(Z<1.05)\\=1-0.8531\\=0.147$

Thus, the p-value of the test statistic is 0.147.

(b) The alternate hypothesis is defined as:

$H_a: p \neq 0.50$

Compute the p-value of the test statistic as follows:

$p-value=2*P(Z>1.05)\\=2*0.1469\\=0.294$

Thus, the p-value of the test statistic is 0.294.

(c) The alternate hypothesis is defined as:

$H_a: p < 0.50$

Compute the p-value of the test statistic as follows:

$p-value= P(Z<1.05)\\=1-P(Z>1.05)= 1-0.1469\\=0.8531$

Thus, the p-value of the test statistic is 0.8531.

(d) The decision rule of the test is:

If the p-value of the test is less than the significance level α, then the null hypothesis is rejected at α% level of significance. And if the p-value of the test is more than the significance level α, then the null hypothesis is failed to be rejected. The most commonly used level of significance are:

$\alpha = 0.01, 0.05, 0.10$

The p-value for all the three alternate hypothesis are:

$p-values = 0.147, 0.294, 0.8531.$

All the p-values are quite large compared to the α values. Thus, none of the p-values give strong evidence against the null hypothesis. The null hypothesis was failed to be rejected.

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