Mathematics, 27.03.2020 01:24 hcameron65

The U. S. Census Bureau conducts annual surveys to obtain information on the percentage of the voting-age population that is registered to vote. Suppose that 433433 employed persons and 555555 unemployed persons are independently and randomly selected, and that 257257 of the employed persons and 227227 of the unemployed persons have registered to vote. Can we conclude that the percentage of employed workers ( p1p1 ), who have registered to vote, exceeds the percentage of unemployed workers ( p2p2 ), who have registered to vote? Use a significance level of α=0.01α=0.01 for the test.

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Answer from: Boris1002

$z=\frac{0.594-0.409}{\sqrt{0.490(1-0.490)(\frac{1}{433}+\frac{1}{555})}}=5.772$

$p_v =P(Z5.772)\approx 5.32x10^{-9}$

Comparing the p value with the significance level given $\alpha=0.01$ we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can conclude that the proportion who have registered to vote, exceeds the percentage of unemployed workers at 1% of significance

Step-by-step explanation:

Data given and notation

$X_{1}=257$ represent the number of employed registred to vote

$X_{2}=227$ represent the number of unemployed registred to vote

$n_{1}=433$ sample of employed registred to vote

$n_{2}=555$ sample of unemployed registred to vote

$p_{1}=\frac{257}{433}=0.594$ represent the proportion of employed registred to vote

$p_{2}=\frac{227}{555}=0.409$ represent the proportion of unemployed registred to vote

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

Concepts and formulas to use

We need to conduct a hypothesis in order to check if the proportion for employed is higher than the proportion for unemployed, the system of hypothesis would be:

Null hypothesis: $p_{1} - p_{2} \leq 0$

Alternative hypothesis: $p_{1} - \mu_{2} 0$

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 $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{257+227}{433+555}=0.490$

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:

$z=\frac{0.594-0.409}{\sqrt{0.490(1-0.490)(\frac{1}{433}+\frac{1}{555})}}=5.772$

Statistical decision

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

$p_v =P(Z5.772)\approx 5.32x10^{-9}$

Answer from: Quest

Ithink the answer is d i hope it

Answer from: Quest

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The U. S. Census Bureau conducts annual surveys to obtain information on the percentage of the votin...