Question

For a test of H0: p = 0.50, the z test statistic equals 1.04. Use 3 decimal places.

(a) What is the p-value for H: p > 0.50?

(b) What is the p-value for H: p 0.50?

(c) What is the p-value for H: p < 0.50? (Hint: The p-values for the two possible one-sided tests must sum to 1.) (d)Which of the p-values give strong evidence against H0?Select all that apply.

1.The p-value in (a).

2.The p-value in (b).

3.The p-value in (c).

4.None of the p-values give strong evidence against H0.

Answer 1

Concepts and reason

The parametric tests used to test the population mean of one variable are z test and t test.

Conditions:

• One sample z test is used when the population standard deviation $\left( \sigma \right)$ is known and the sample size large.

• One sample t test is used when the population standard deviation $\left( \sigma \right)$ is unknown and the sample size small.

The null hypothesis states that there is no difference in the test, which is denoted by . Moreover, the sign of null hypothesis is equal, greater than or equal and less than or equal.

The hypothesis that differs from the is called alternative hypothesis. This signifies that there is a significant difference in the test. The sign of alternative hypothesis is less than, greater than, or not equal.

The probability value related with a statistical test is named as p-value. If the p-value is less than the significance level, then the null hypothesis is rejected. The smaller p-value gives the stronger evidence to reject the null hypothesis

Fundamentals

Rejection rule using p-value:

If $p{\rm{ - value}} \le \alpha \left( { = 0.05} \right)$ , then reject the null hypothesis.

If $p{\rm{ - value}} \ge \alpha \left( { = 0.05} \right)$ , then do not reject the null hypothesis.

(a)

From the given information, the test statistic is 1.04. Since, the alternative hypothesis has greater than symbol; the right tailed test is applied.

The p-value is obtained below:

From the “Standard normal distribution table”, the area to the left of $z=1.04 $ is 0.8508.

(b)

From the given information, the test statistic is 1.04. Since, the alternative hypothesis has not equal symbol; the two tailed test is applied.

The p-value is obtained below:

From the “Standard normal distribution table”, the area to the left of $z = 1.04$ is 0.8508.

$\begin{array}{c}\\p{\rm{ - value}} = 2\left( {1 - 0.8508} \right)\\\\ = 2\left( {0.1492} \right)\\\\ = 0.2984\\\end{array}$

(c)

From the given information, the test statistic is 1.04. Since, the alternative hypothesis has less than symbol; the left tailed test is applied.

The p-value is obtained below:

$\begin{array}{c}\\p{\rm{ - value}} = P\left( {z < {z_{{\rm{test}}}}} \right)\\\\ = P\left( {z < 1.04} \right)\\\end{array}$

From the “Standard normal distribution table”, the area to the left of $z = 1.04$ is 0.8508

That is,

(d)

Consider, the level of significance is $\alpha = 0.05$ and test statistic is 1.04.

• The p-value for one tailed test (greater than case) is 0.149.

• The p-value for two tailed test (greater than case) is 0.298.

• The p-value for one tailed test (less than case) is 0.851.

All the p-values is greater than the level of significance $\left( {\alpha = 0.05} \right)$ . That is, the null hypothesis is not rejected at 5% level.

Ans: Part a

The p-value is 0.149.

Part b

The p-value is 0.298.

Part c

The p-value is 0.851.

Part d

None of the p-values give strong evidence against ${H_0}$ .