Question

Not Answered Country Financial, a financial services company, uses surveys of adults age 18 and older to determine if personal financial fitness is changing over time (USA Today, April 4, 2012). In February of 2012, a sample of 1000 adults showed 410 indicating that their financial security was more than fair. In February of 2010, a sample of 900 adults showed 315 indicating that their financial security was more than fair. a. State the hypotheses that can be used to test for a significant difference between the population proportions for the two years.

Answer 1

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

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

`z=\frac{0.410-0.350}{\sqrt{0.382(1-0.382)((1)/(1000)+(1)/(900))}}=2.688`

`p_v =2*P(Z>2.688)= 0.0072`

Comparing the p value with the significance level assumed
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to to reject the null hypothesis, and we can say that the proportions analyzed are significantly different at 5% of significance.

Explanation:

Data given and notation

`X_(1)=410` represent the number of people indicating that their financial security was more than fair in 2012

`X_(2)=315` represent the number of people indicating that their financial security was more than fair in 2010

`n_(1)=1000` sample 1 selected

`n_(2)=900` sample 2 selected

`p_(1)=(410)/(1000)=0.410` represent the proportion estimated of people indicating that their financial security was more than fair in 2012

`p_(2)=(315)/(900)=0.350` represent the proportion estimated of people indicating that their financial security was more than fair in 2010

`\hat p` represent the pooled estimate of p

z would represent the statistic (variable of interest)

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

`\alpha=0.05` significance level given

Part a: Concepts and formulas to use

We need to conduct a hypothesis in order to check if is there is a difference between the two proportions, the system of hypothesis would be:

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

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

Hypothesis testing

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

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

Where
`\hat p=(X_(1)+X_(2))/(n_(1)+n_(2))=(410+315)/(1000+900)=0.382`

z-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the statistic

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

`z=\frac{0.410-0.350}{\sqrt{0.382(1-0.382)((1)/(1000)+(1)/(900))}}=2.688`

Statistical decision

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

`p_v =2*P(Z>2.688)= 0.0072`

Comparing the p value with the significance level assumed
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to to reject the null hypothesis, and we can say that the proportions analyzed are significantly different at 5% of significance.