osula9a

2021-12-17

For a test of Ho: p = 0.5, the z test statistic equals -1.74. Find the p-value for Ha: p < 0.5.

a. 0.0892

b. 0.0446

c. 0.0409

d. 0.9554

e. 0.0818

a. 0.0892

b. 0.0446

c. 0.0409

d. 0.9554

e. 0.0818

xandir307dc

Beginner2021-12-18Added 35 answers

Solution:- p value = 0.1286

Since we have a two-tailed test, the P-value is the probability that the z statistic is less than - 1.52 or greater than 1.52.

We use the z Distribution Calculator to find P(t < - 1.52) = 0.0643, and P(t > 1.52) = 0.0643

Thus, the P-value = 0.0643 + 0.0643 = 0.1286

Interpret results. Since the P-value (0.1286) is greater than the significance level (0.05), we cannot reject the null hypothesis.

From the test we conclude that we have sufficient evidence to support the claim that p = 0.5.

Since we have a two-tailed test, the P-value is the probability that the z statistic is less than - 1.52 or greater than 1.52.

We use the z Distribution Calculator to find P(t < - 1.52) = 0.0643, and P(t > 1.52) = 0.0643

Thus, the P-value = 0.0643 + 0.0643 = 0.1286

Interpret results. Since the P-value (0.1286) is greater than the significance level (0.05), we cannot reject the null hypothesis.

From the test we conclude that we have sufficient evidence to support the claim that p = 0.5.

Jim Hunt

Beginner2021-12-19Added 45 answers

That is incorrect answer!

nick1337

Expert2021-12-28Added 777 answers

From the given information,
Null hypothesis:
H_0:p=0.5
Alternative hypothesis:
H_0:p<0.5
That is, the alternative hypothesis is one-sided hypothesis. Hence, the p-value is obtained for left tailed test.
From the given information, the value of test statistic is –1.74.
p=P(z

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