Question

In a representative sample of 1000 adult Americans, only 440 could name at least one justice who is currently serving on the U.S. Supreme Court. Using a significance level of 0.01, carry out a hypothesis test to determine if there is convincing evidence to support the claim that fewer than half of adult Americans can name at least one justice currently serving on the Supreme Court. (Round your test statistic to two decimal places and your P-value to four decimal places.)

Answer 1

`z=\frac{0.440 -0.5}{\sqrt{(0.5(1-0.5))/(1000)}}=-3.79`

`p_v =P(z<-3.79)=0.000075`

So the p value obtained was a very low value and using the significance level given
`\alpha=0.01` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of Americans that can name at least one justice currently serving on the Supreme Court is significantly lower than 0.5 or 50%.

Explanation:

Data given and notation

n=1000 represent the random sample taken

X=440 represent the people that could name at least one justice who is currently serving on the U.S. Supreme Court

`\hat p=(440)/(1000)=0.440` estimated proportion of people that could name at least one justice who is currently serving on the U.S. Supreme Court

`p_o=0.5` is the value that we want to test

`\alpha=0.01` represent the significance level

Confidence=99% or 0.99

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 true proportion of Americans can name at least one justice currently serving on the Supreme Court is less than 0.5 .:

Null hypothesis:
`p \geq 0.5`

Alternative hypothesis:
`p < 0.5`

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`.

Calculate the statistic

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

`z=\frac{0.440 -0.5}{\sqrt{(0.5(1-0.5))/(1000)}}=-3.79`

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about 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
`\alpha=0.01`. The next step would be calculate the p value for this test.

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

`p_v =P(z<-3.79)=0.000075`

So the p value obtained was a very low value and using the significance level given
`\alpha=0.01` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of Americans that can name at least one justice currently serving on the Supreme Court is significantly lower than 0.5 or 50%.