Question

In a recent​ year, a poll asked 2362 random adult citizens of a large country how they rated economic conditions. In the​ poll, 22​% rated the economy as​ Excellent/Good. A recent media outlet claimed that the percentage of citizens who felt the economy was in​ Excellent/Good shape was 24​%. Does the poll support this​ claim? ​a) Test the appropriate hypothesis. Find a 99​% confidence interval for the proportion of adults who rated the economy as​ Excellent/Good. Check conditions. ​b) Does your confidence interval provide evidence to support the​ claim? ​c) What is the significance level of the test in​ b? Explain.

Answer 1

a) The 99% confidence interval is given by (0.198;0.242).

b) Based on the p value obtained and using the significance level assumed
`\alpha=0.01` we have
`p_v>\alpha` so we can conclude that we fail to reject the null hypothesis, and we can said that at 1% of significance the proportion of people who are rated with Excellent/Good economy conditions not differs from 0.24. The interval also confirms the conclusion since 0.24 it's inside of the interval calculated.

c)
`\alpha=0.01`

Explanation:

Data given and notation

n=2362 represent the random sample taken

X represent the people who says that they would watch one of the television shows.

`\hat p=(X)/(n)=0.22` estimated proportion of people rated as​ Excellent/Good economic conditions.

`p_o=0.24` is the value that we want to test

`\alpha` represent the significance level

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that 24% of people are rated with good economic conditions:

Null hypothesis:
`p=0.24`

Alternative hypothesis:
`p \\eq 0.24`

When we conduct a proportion test we need to use the z statistic, and the is given by:

`z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}` (1)

The One-Sample Proportion Test is used to assess whether a population proportion
`\hat p` is significantly different from a hypothesized value
`p_o`.

Part a: Test the hypothesis

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

np = 2362x0.22=519.64>10 and n(1-p)=2364*(1-0.22)=1843.92>10

Condition satisfied.

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

`z=\frac{0.22 -0.24}{\sqrt{(0.24(1-0.24))/(2362)}}=-2.28`

The confidence interval would be given by:

`\hat p \pm z_(\alpha/2)\sqrt{(\hat p (1-\hat p))/(n)}`

The critical value using
`\alpha=0.01` and
`\alpha/2 =0.005` would be
`z_(\alpha/2)=2.58`. Replacing the values given we have:

`0.22 - (2.58)\sqrt{(0.22(1-0.22))/(2362)}=0.198`

`0.22 + (2.58)\sqrt{(0.22(1-0.22))/(2362)}=0.242`

So the 99% confidence interval is given by (0.198;0.242).

Part b

Statistical decision

P value method or p value approach . "This method consists on determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided is
`\alpha=0.01`. The next step would be calculate the p value for this test.

Since is a bilateral test the p value would be:

`p_v =2*P(z<-2.28)=0.0226`

So based on the p value obtained and using the significance level assumed
`\alpha=0.01` we have
`p_v>\alpha` so we can conclude that we fail to reject the null hypothesis, and we can said that at 1% of significance the proportion of people who are rated with Excellent/Good economy conditions not differs from 0.24. The interval also confirms the conclusion since 0.24 it's inside of the interval calculated.

Part c

The confidence level assumed was 99%, so then the signficance is given by
`\alpha=1-confidence=1-0.99=0.01`