Question

A particular poll tracks daily the percentage of Americans who approve or disapprove of the performance by President 1. Daily results are based on random telephone interviews with approximately 1300 national adults. The poll reports that 41​% of adults approve of President 1. The same poll reported an approval rating of 42​% for President 2. A news anchor remarks that​ "President 1​ doesn't even get as much approval as President 2​ did." Is there evidence that this difference is​ real? State and test the appropriate hypotheses.

Answer 1

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

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

`z=\frac{0.41-0.43}{\sqrt{0.42(1-0.42)((1)/(1300)+(1)/(1300))}}=-1.033`

`p_v =2*P(Z<-1.033)=0.302`

So the p value is a very high value and using any significance level for example
`\alpha=0.05, 0,1,0.15` 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 the proportion 1 is not significantly different from the proportion 2.

Explanation:

1) Data given and notation

n = 1300 sample size selected

`p_(1)=0.41` represent the proportion of adults approve of President1.

`p_(2)=0.42` represent the proportion of adults approve of President2.

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 1 is different from proportion 2 , 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{\hat p (1-\hat p)((1)/(n_(1))+(1)/(n_(2)))}}` (1)

Where
`\hat p=(X_(1)+X_(2))/(n_(1)+n_(2))=(0.41+0.43)/(2)=0.42`

3) Calculate the statistic

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

`z=\frac{0.41-0.43}{\sqrt{0.42(1-0.42)((1)/(1300)+(1)/(1300))}}=-1.033`

4) 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 two sided test the p value would be:

`p_v =2*P(Z<-1.033)=0.302`

So the p value is a very high value and using any significance level for example
`\alpha=0.05, 0,1,0.15` 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 the proportion 1 is not significantly different from the proportion 2.