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 200 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.)

Answer 1

`z=\frac{0.41-0.43}{\sqrt{(0.41(1-0.41))/(700)+(0.43(1-0.43))/(200)}}=-0.51`

`p_v =2*P(Z<-0.51)=0.614`

And we can use the following R code to find it: "2*pnorm(-0.505)"

The p value is a very high value and using any significance given
`\alpha=0.05` always
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the two proportions are not significantly different.

Explanation:

1) Data given and notation

`n_(1)=700` sample of young adults (age 19 to 35)

`n_(2)=200` sample of children age 19 to 35

`p_(1)=0.41` represent the proportion of young adults said they thought parents would provide financial support

`p_(2)=0.43` represent the proportion of parents said they would provide support

`\alpha=0.05` represent the significance level

z would represent the statistic (variable of interest)

`p_v` represent the value for the test (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to check if the proportion for the two samples are different , the system of hypothesis would be:

Null hypothesis:
`p_(1) = p_(2)`

Alternative hypothesis:
`p_(1) \\eq p_(2)`

We need to apply a z test to compare proportions, and the statistic is given by:

`z=\frac{p_(1)-p_(2)}{\sqrt{(p_1 (1-p_1))/(n_(1))+(p_2 (1-p_2))/(n_(2))}}` (1)

3) Calculate the statistic

Replacing in formula (1) the values obtained we got this:

`z=\frac{0.41-0.43}{\sqrt{(0.41(1-0.41))/(700)+(0.43(1-0.43))/(200)}}=-0.51`

4) Statistical decision

We can calculate the p value for this test.

Since is a two tailed test the p value would be:

`p_v =2*P(Z<-0.51)=0.614`

And we can use the following R code to find it: "2*pnorm(-0.505)"

The p value is a very high value and using any significance given
`\alpha=0.05` always
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the two proportions are not significantly different.