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

Answer:

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) $\alpha=0.01$

Step-by-step 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.

$\hat p=\frac{X}{n}=0.22$ 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:$p \neq 0.24$  

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

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (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:  

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

The confidence interval would be given by:

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

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:

$0.22 - (2.58)\sqrt{\frac{0.22(1-0.22)}{2362}}=0.198$

 $0.22 + (2.58)\sqrt{\frac{0.22(1-0.22)}{2362}}=0.242$

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:  

$p_v =2*P(z$  

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 $\alpha=1-confidence=1-0.99=0.01$