A 0.1 significance level is used for a hypothesis test of the claim that parents
ID: 3202593 • Letter: A
Question
A 0.1 significance level is used for a hypothesis test of the claim that parents use a particular method of gender selection, the proportion of baby girls is greater than 0.5. Assume that sample data consists of 55 girls in 100 births, so the sample statics of 11/20 results in a z score that 1 standard deviation above 0. A 0.1 significance level is used for a hypothesis test of the claim that parents use a particular method of gender selection, the proportion of baby girls is greater than 0.5. Assume that sample data consists of 55 girls in 100 births, so the sample statics of 11/20 results in a z score that 1 standard deviation above 0.Explanation / Answer
Solution:-
P = 0.5
p = 55/100 = 0.55
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P < 0.5
Alternative hypothesis: P > 0.5
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 0.1. 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 () and compute the z-score test statistic (z).
= sqrt[ P * ( 1 - P ) / n ]
= 0.05
z = (p - P) /
z = 1.00
where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.
The P-value = 0.1587
Interpret results. Since the P-value (0.1587) is more than the significance level (0.05), we have to accept the null hypothesis.
From this we do not have sufficient evidence in the favor of the claim that the proportion of baby girls is greater than 0.5.
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