Question

Decide whether the normal sampling distribution can be used. If it can be​ used, test the claim about the population proportion p at the given level of significance α using the given sample statistics.

​Claim: p≠ 0.27​; α=0.05​; Sample​ statistics: p =0.23​,n=100

a. Can the normal sampling distribution be​ used?

b.State the null and alternative hypotheses.

c.The critical​ value(s) is/are ( ___ , ____ )

​(Round to two decimal places as needed. Use a comma to separate answers as​ needed.)

d.Find the​ z-test statistic. Select the correct choice below​ and, if​ necessary, fill in the answer box to complete your choice.

z= ______

e. What is the result of the​ test? Fail to reject or Reject?

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P = 0.27
Alternative hypothesis: P 0.27

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.

Analyze sample data. Using sample data, we calculate the standard deviation (S.D) and compute the z-score test statistic (z).

S.D = sqrt[ P * ( 1 - P ) / n ]

S.D = 0.0444
z = (p - P) /S.D

z = - 0.901

z_alpha/2 = + 1.96

where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than -0.901 or greater than 0.901.

Thus, the P-value = 0.368

Interpret results. Since the P-value (0.368) is greater than the significance level (0.05), we cannot reject the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that there is significance difference in sample proportions.

e) Fail to reject H0.