Question

A New York Times/CBS News Poll asked a random sample of U.S. adults the question, "Do you favor an amendment to the Constitution that would permit organized prayer in public schools?" Based on this poll, the 95% confidence interval for the population proportion who favor such an amendment is (0.63, 0.69). Based on this poll, a reporter claims that more than two-thirds of U.S. adults favor such an amendment. Use the confidence interval to evaluate this claim.

Answer 1

`p_v =P(z>-0.436)=0.669`

So the p value obtained was a very high value and using the significance level assumed
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is NOT significantly higher than 2/3.

We got the same conclusion just looking the confidence interval since our interval contains the value 2/3=0.667 we have enough evidence to fail to reject the null hypothesis on this case.

Explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

`p \sim N(p,\sqrt{(p(1-p))/(n)})`

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by
`\alpha=1-0.95=0.05` and
`\alpha/2 =0.025`. And the critical value would be given by:

`z_(\alpha/2)=-1.96, z_(1-\alpha/2)=1.96`

The confidence interval for the mean is given by the following formula:

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

For this case we can get an estimation for
`\hat p` like this:

`\hat p= (0.63+0.69)/(2)=0.66`

And the margin of error would be:

`ME= (0.69-0.63)/(2)=0.03`

And we can estimate the standard error since
`ME= z SE`

`SE = (0.03)/(1.96)=0.0153`

We need to conduct a hypothesis in order to test the claim that the proportion is higher than 2/3 or not:

Null hypothesis:
`p \leq (2)/(3)`

Alternative hypothesis:
`p > (2)/(3)`

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=(0.66 -(2)/(3))/(0.0153)=-0.436`

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 assumed is
`\alpha=0.05`. The next step would be calculate the p value for this test.

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

`p_v =P(z>-0.436)=0.669`

So the p value obtained was a very high value and using the significance level assumed
`\alpha=0.05` we have
`p_v>\alpha` so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of interest is NOT significantly higher than 2/3.

We got the same conclusion just looking the confidence interval since our interval contains the value 2/3=0.667 we have enough evidence to fail to reject the null hypothesis on this case.