Karri’s manager suspects that the batteries being produced lately don’t have enough voltage. For a left-tailed test, the critical value of z so that a hypothesis test would the null hypothesis at 5% significance level would be .

fail to reject; -3.32

reject; -2.57

fail to reject; -3.27

reject; -1.645

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)=0.4354pv=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)=0.0336pv=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)=0.3015pv=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=0.05α=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)=0.4354pv=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)=0.0336pv=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)=0.3015pv=P(z>0.52)=0.3015

Best option:

C) 0.3015; fail to reject the null hypothesis

answer:

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