Question

determine whether the result appear to have statistical significance and also determine whether the results would appear to have practical significance in a study of a gender selection method used to increase the likelihood of a baby being born a girl 2077 users of the method gave birth to 1021 boys and 1056 girls there's about a 23% chance of getting that many girls if the method had no effect. because there is a 23% chance of getting that many girls by Chance the method blank blank couples would likely use a procedure that rises the likelihood of a girl from approximately 50% rate expected by the chance to the blank percent produced by the method so this method blank

Answer 1

Solution:-

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.

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

σ = sqrt[ P * ( 1 - P ) / n ]

σ = 0.010971
z = (p - P) / σ

z = 0.77

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 more than 0.77. Thus, the P-value = 0.222

Interpret results. Since the P-value (0.222) 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 method has increased the likelihood of a baby being born a girl.