Question

In: Statistics and Probability

A 0.1 significance level is used for a hypothesis test of the claim that when parents...

A 0.1 significance level is used for a hypothesis test of the claim that when parents use a particular method of gender​ selection, the proportion of baby girls is greater than 0.5. Assume that sample data consists of 45 girls in 81 ​births, so the sample statistic of five ninths results in a z score that is 1 standard deviation above 0. Complete parts​ (a) through​ (h) below.

A. Identify the null hypothesis and the alternate hypothesis

B. What is the value of a?

C. What is the sampling distribution of the sample statistic?

D. Is the test two tailed, left tailed, or right tailed?

E. What is the value of the test statistic?

F. What is the P-Value?

G. What are the critical values?

H. What is the area of the critical region?

Solutions

Expert Solution

Solution:-

a)

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

Null hypothesis: P < 0.50
Alternative hypothesis: P > 0.50

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected only if the sample proportion is too small.

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.

c) The sampling distribution of the sample statistic have mean 0.05 and standard deviation 0.0556.

d) This is right tailed 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 = 0.05556
E)

z = (p - P) / S.D

z = 1.0

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 one-tailed test, the P-value is the probability that the z-score is greater than 1.0

F)

Thus, the P-value = 0.159.

Interpret results. Since the P-value (0.159) is greater than the significance level (0.05), we have to accept the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that when parents use a particular method of gender​ selection, the proportion of baby girls is greater than 0.5.

G) The critical values are z = 1.645.

H) The area of the critical region z > 1.645.


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