Question

Country financials, a financial services company, uses surveys of adults age 18 and older to determine if personal financial fitness is changing over time. In February 2012, a sample of 1000 adults showed 410 indicating that their financial security was more that fair. In Feb 2010, a sample of 900 adults showed 315 indicating that their financial security wwas more than fair.


1.State the hypothesis that can be used to test for a significant difference between the population proportions for the two years?


2.What is the sample proportion indicating that theirfinancial security was more that fair in 2012?In 2010?


3.Conduct the hypothesis test and compute the p-value.At a .05 level of significance what is your conclusion?


4.What is the 95% confidence interval estimate of the difference between the two population proportion?

Answer 1

Answer:

1) Null hypothesis:$p_{1} = p_{2}$  

Alternative hypothesis:$p_{1} \neq p_{2}$  

2) The proportion in 2012 seems to be higher than the proportion in 2010

3) $z=2.78$    

$p_v =2*P(Z>2.78)=0.00544$  

The p value is a very low value and using the significance level given $\alpha=0.05$ always $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of people sampled in 2012 is significantly different than the porportion samples in 2010.  

4) And the 95% interval would be given by: (0.0307;0.0933)

Step-by-step explanation:

Part 1

Data given and notation  

$X_{1}=412$ represent the number of people in 2012 indicating that their financial security was more that fair

$X_{2}=315$ represent the number of people in 2010 indicating that their financial security was more that fair

$n_{1}=1000$ sample in 2012 selected

$n_{2}=900$ sample in 2010 selected  

$p_{1}=\frac{412}{1000}=0.412$ represent the proportion of people in 2012 indicating that their financial security was more that fair

$p_{2}=\frac{315}{900}=0.350$ represent the proportion of people in 2010 indicating that their financial security was more that fair

z would represent the statistic (variable of interest)  

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

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if the proportion for men with red/green color blindness is a higher than the rate for women  , the system of hypothesis would be:  

Null hypothesis:$p_{1} = p_{2}$  

Alternative hypothesis:$p_{1} \neq p_{2}$  

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

$z=\frac{p_{1}-p_{2}}{\sqrt{\frac{p_1 (1-p_1)}{n_1}}+\frac{p_2 (1-p_2)}{n_2}}}$   (1)  

Part 2

What is the sample proportion indicating that theirfinancial security was more that fair in 2012?In 2010?

$p_{1}=\frac{412}{1000}=0.412$ represent the proportion of people in 2012 indicating that their financial security was more that fair

$p_{2}=\frac{315}{900}=0.350$ represent the proportion of people in 2010 indicating that their financial security was more that fair

The proportion in 2012 seems to be higher than the proportion in 2010

Part 3: Conduct the hypothesis test and compute the p-value.At a .05 level of significance what is your conclusion?

Calculate the statistic  

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

$z=\frac{0.412-0.350}{\sqrt{\frac{0.412(1-0.412)}{1000}}}+\frac{0.35 (1-0.35)}{900}}}=2.78$    

Statistical decision

For this case we don't have a significance level provided $\alpha$, but we can calculate the p value for this test.    

Since is a one side test the p value would be:  

$p_v =2*P(Z>2.78)=0.00544$  

So the p value is a very low value and using the significance level given $\alpha=0.05$ always $p_v<\alpha$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of people sampled in 2012 is significantly different than the porportion samples in 2010.  

Part 4

What is the 95% confidence interval estimate of the difference between the two population proportion?

The confidence interval is given by this formula:

$(p_1 -p_2) \pm z_{\alpha/2} \sqrt{\frac{p_1 (1-p_1)}{n_1}+\frac{p_2 (1-p_2)}{n_2}}$

And if we replace the values given we have:

$(0.412 -0.350) \pm 1.96\sqrt{\frac{0.412 (1-0.412)}{1000}+\frac{0.35 (1-0.35)}{900}}$

$0.062\pm 0.0313$

And the 95% interval would be given by: (0.0307;0.0933)

And since the interval not contains the 0 that agrees with the result obtained on the hypothesis test.

Answer 2

Answer:

a

Step-by-step explanation: