Question

Use the given information to find the P-value. Also, use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis). With H1 : p ≠ 3/5, the test statistic is z = 0.78.
1) With H1: p ≠ 3/5, the test statistic is z = 0.78.
A) 0.4354; fail to reject the null hypothesis
B) 0.4354; reject the null hypothesis
C) 0.2177 fail to reject the null hypothesis
D) 0.2177; reject the null hypothesis
2) The test statistic in a left-tailed test is z = -1.83.
A) 0.0336; reject the null hypothesis
B) 0.0672; fail to reject the null hypothesis
C) 0.9664; fail to reject the null hypothesis
D) 0.0672; reject the null hypothesis
3) The test statistic in a right-tailed test is z = 0.52.
A) 0.0195; reject the null hypothesis
B) 0.3015; reject the null hypothesis
C) 0.3015; fail to reject the null hypothesis
D) 0.6030; fail to reject the null hypothesis

Answer 1

Answer:

1) With H1: p ≠ 3/5, the test statistic is z = 0.78

The p value for this case would be given by:

$p_v = 2*P(z>0.78)=0.4354$

Best option:

A) 0.4354; fail to reject the null hypothesis

2) The test statistic in a left-tailed test is z = -1.83

The p value for this case would be given by:

$p_v = P(z<-1.83)=0.0336$

Best option:

A) 0.0336; reject the null hypothesis

3) The test statistic in a right-tailed test is z = 0.52.

The p value for this case would be given by:

$p_v = P(z>0.52)=0.3015$

Best option:

C) 0.3015; fail to reject the null hypothesis

Step-by-step explanation:

The significance level for all the cases is the same $\alpha=0.05$

Part 1

With H1: p ≠ 3/5, the test statistic is z = 0.78

The p value for this case would be given by:

$p_v = 2*P(z>0.78)=0.4354$

Best option:

A) 0.4354; fail to reject the null hypothesis

Part 2

The test statistic in a left-tailed test is z = -1.83

The p value for this case would be given by:

$p_v = P(z<-1.83)=0.0336$

Best option:

A) 0.0336; reject the null hypothesis

Part 3

The test statistic in a right-tailed test is z = 0.52.

The p value for this case would be given by:

$p_v = P(z>0.52)=0.3015$

Best option:

C) 0.3015; fail to reject the null hypothesis