For problems 1 through 3, state the claim and its opposite and the null and alte

Question

For problems 1 through 3, state the claim and its opposite and the null and alternate hypotheses, the type of analysis that should be performed (either testing inferences about two proportions, means from independent samples, or means from matched pairs), and the formula used to calculate the test statistic. Show the values substituted into the formula, and calculate the value of the test statistic, draw and label a graph, determine the P-value, state what to do with the null hypothesis, and use the flow chart to state your conclusion. Assume that the samples are simple random samples and all requirements are satisfied.

1. In 2005 Gallup surveyed 1028 adults and found that 226 of them had smoked at least one cigarette in the past week. In 1990, Gallup asked the same question of 1028 adults and found that 278 adults had smoked at least one in the past week. Test the claim that the 2005 proportion of American adults who smoked at least one cigarette in the past week has decreased since 1990.

Explanation / Answer

Solution:-

1)

n1 = 1028, x1 = 226

p1 = 0.2195

x2 = 278, n2 = 1028

p2 = 0.2704

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

Null hypothesis: P1990< P2005

Alternative hypothesis: P1990 > P2005

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. 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 = 0.4903

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

SE = 0.02185

z = (p1 - p2) / SE

z = 2.315

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 one-tailed test, the P-value is the probability that the z-score is more than 2.315. We use the Normal Distribution Calculator to find P(z > 2.315) = 0.017. Thus, the P-value = 0.0102

Interpret results. Since the P-value (0.0102) is less than the significance level (0.05), we have reject the null hypothesis.

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