Question

A report summarizes a survey of people in two independent random samples. One sample consisted of 700 young adults (age 19 to 35) and the other sample consisted of 400 parents of children age 19 to 35. The young adults were presented with a variety of situations (such as getting married or buying a house) and were asked if they thought that their parents were likely to provide financial support in that situation. The parents of young adults were presented with the same situations and asked if they would be likely to provide financial support to their child in that situation. (a) When asked about getting married, 41% of the young adults said they thought parents would provide financial support and 43% of the parents said they would provide support. Carry out a hypothesis test to determine if there is convincing evidence that the proportion of young adults who think parents would provide financial support and the proportion of parents who say they would provide support are different. (Use α = 0.05. Use a statistical computer package to calculate the P-value. Use μyoung adults − μparents. Round your test statistic to two decimal places and your P-value to three decimal places.) z = P-value = (b) The report stated that the proportion of young adults who thought parents would help with buying a house or apartment was 0.37. For the sample of parents, the proportion who said they would help with buying a house or an apartment was 0.27. Based on these data, can you conclude that the proportion of parents who say they would help with buying a house or an apartment is significantly less than the proportion of young adults who think that their parents would help? (Use α = 0.05. Use a statistical computer package to calculate the P-value. Use μyoung adults − μparents. Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value =

Answer 1

Solution:-

a)

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 0.05. 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 = 0.4173
SE = sqrt{ p * ( 1 - p ) * [ (1/n₁) + (1/n₂) ] }
SE = 0.0288
z = (p₁ - p₂) / SE

z = - 0.694

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 -0.694 or greater than 0.694.

Thus, the P-value = 0.488

Interpret results. Since the P-value (0.488) 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 there is convincing evidence that the proportion of young adults who think parents would provide financial support and the proportion of parents who say they would provide support are different.

b)

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

Null hypothesis: P > 0.37
Alternative hypothesis: P < 0.37

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, shown in the next section, is a one-sample z-test.

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

S.D = sqrt[ P * ( 1 - P ) / n ]

S.D = 0.02414
z = (p - P) / S.D

z = - 4.14

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 less than -4.14.

Thus, the P-value = less than 0.0001

Interpret results. Since the P-value (almost 0) is less than the significance level (0.05), we have to reject the null hypothesis.

From the above test we have sufficient evidence in the favor of the claim that the proportion of parents who say they would help with buying a house or an apartment is significantly less than the proportion of young adults who think that their parents would help.