Question

136 people ages 16 to 19 were taken from of high school dropouts on Oahu is different (either way) from that or Sweetwater County? Use a 1% level of significance. County, Wyorming, and 11 were found to be high school dropouts. Do these data indicate that the population propertion 0.01 The Student's t. We assume the population distributions are The Student's t. The number of trials is sufficiently large 0.54

Answer 1

B. The standard normal. The number of trials is sufficiently large.

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 0.01. 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 = 0.0722
SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.03041
z = (p₁ - p₂) / SE
z = - 0.53812
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 -0.54 or greater than 0.54
Thus, the P-value = 0.5892
Interpret results. Since the P-value (0.5892) is greater than the significance level (0.01), we can accept the null hypothesis.