A sample of middle-school students were asked whether they would prefer to wear

Question

A sample of middle-school students were asked whether they would prefer to wear uniforms. Of the 150 boys who responded, 37 said “yes”, and of the 120 girls who responded, 31 said “yes”. School officials are interested in the difference in the two groups’ opinions. (a) What statistic would be used in this case to address the officials’ question? What is the standard error of this statistic? (b) With 90% confidence, make a statement about the difference in opinions between all the boys and girls in the population of all middle-school students from which the students here were sampled. (c) Can the officials claim a significant difference? Why or why not?

Explanation / Answer

Solution:-

x1 = 37, n1 = 150

x2 = 31, n2 = 120

p1 = 37/150 = 0.2466

p2 = 31/120 = 0.2583

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

Null hypothesis: P1 = P2

Alternative hypothesis: P1 P2

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the proportion from population 1 is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.10(90% confidence interval). The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p1 * n1 + p2 * n2) / (n1 + n2)

p = 0.25185

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

SE = 0.05316

z = (p1 - p2) / SE

z = - 0.220

where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.

Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 0.22 or greater than 0.22.

We use the Normal Distribution Calculator to find P(z < - 0.22) = 0.4129, and P(z > 0.22) = 0.4129

Thus, the P-value = 0.4129 + 0.4129 = 0.8258

Interpret results. Since the P-value (0.8258) is more than the significance level (0.10), we have to accept the null hypothesis.

From the test we do not have sufficient evidence in the favor of the claim that there is a significant difference.

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