Question

g A certain financial services company uses surveys of adults age 18 and older to determine if personal financial fitness is changing over time. A recent sample of 1,000 adults showed 410 indicating that their financial security was more than fair. Suppose that just a year before, a sample of 700 adults showed 245 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. (Let p1 = population proportion most recently saying financial security more than fair and p2 = population proportion from the year before saying financial security more than fair. Enter != for ≠ as needed.)

Answer 1

Answer:

Null hypothesis:$p_{1} = p_{2}$    

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

$z=\frac{0.410-0.35}{\sqrt{0.385(1-0.385)(\frac{1}{1000}+\frac{1}{700})}}=2.502$    

$p_v =2*P(Z>2.502)= 0.0123$    

Comparing the p value with the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to to reject the null hypothesis, and we can say that the proportion analyzed is significantly different between the two groups at 5% of significance.    

Step-by-step explanation:

Data given and notation    

$X_{1}=410$ represent the number of people indicating that their financial security was more than fair for the recent year

$X_{2}=245$ represent the number of people indicating that their financial security was more than fair for the year before

$n_{1}=1000$ sample 1 selected  

$n_{2}=700$ sample 2 selected  

$p_{1}=\frac{410}{1000}=0.410$ represent the proportion estimated of indicating that their financial security was more than fair this year

$p_{2}=\frac{245}{700}=0.35$ represent the proportion estimated of indicating that their financial security was more than fair the year before

$\hat p$ 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)  

$\alpha$ significance level given  

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:$p_{1} \neq p_{2}$    

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

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

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{410+245}{1000+700}=0.385$  

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:    

$z=\frac{0.410-0.35}{\sqrt{0.385(1-0.385)(\frac{1}{1000}+\frac{1}{700})}}=2.502$    

Statistical decision  

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

$p_v =2*P(Z>2.502)= 0.0123$    

Comparing the p value with the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to to reject the null hypothesis, and we can say that the proportion analyzed is significantly different between the two groups at 5% of significance.