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.
Accepted Solution
A:
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 [tex]\alpha=0.01[/tex] we have [tex]p_v>\alpha[/tex] 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) [tex]\alpha=0.01[/tex]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.
[tex]\hat p=\frac{X}{n}=0.22[/tex] estimated proportion of people rated as Excellent/Good economic conditions.
[tex]p_o=0.24[/tex] is the value that we want to test
[tex]\alpha[/tex] represent the significance level
z would represent the statistic (variable of interest)
[tex]p_v[/tex] 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:[tex]p=0.24[/tex]
Alternative hypothesis:[tex]p \neq 0.24[/tex]
When we conduct a proportion test we need to use the z statistic, and the is given by:
[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)
The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].
Part a: Test the hypothesisCheck 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 enoughnp = 2362x0.22=519.64>10 and n(1-p)=2364*(1-0.22)=1843.92>10Condition satisfied.
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.22 -0.24}{\sqrt{\frac{0.24(1-0.24)}{2362}}}=-2.28[/tex] The confidence interval would be given by:[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]The critical value using [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex] would be [tex]z_{\alpha/2}=2.58[/tex]. Replacing the values given we have:[tex]0.22 - (2.58)\sqrt{\frac{0.22(1-0.22)}{2362}}=0.198[/tex] [tex]0.22 + (2.58)\sqrt{\frac{0.22(1-0.22)}{2362}}=0.242[/tex]So the 99% confidence interval is given by (0.198;0.242).Part bStatistical 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 [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.
Since is a bilateral test the p value would be:
[tex]p_v =2*P(z<-2.28)=0.0226[/tex]
So based on the p value obtained and using the significance level assumed [tex]\alpha=0.01[/tex] we have [tex]p_v>\alpha[/tex] 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 cThe confidence level assumed was 99%, so then the signficance is given by [tex]\alpha=1-confidence=1-0.99=0.01[/tex]