Question

University personnel are concerned about the sleeping habits of students and the negative impact on academic performance. In a random sample of 377 U.S. college students, 209 students reported experiencing excessive daytime sleepiness (EDS).

A. Is there sufficient evidence to conclude that more than half of U.S. college students experience EDS? Use a 5% level of significance.
B. What is a 90% confidence interval estimate for the proportion of all of U.S. college students who experience excessive daytime sleepiness?

Answer 1

a)
`z=\frac{0.554 -0.5}{\sqrt{(0.5(1-0.5))/(377)}}=2.097`

`p_v =P(Z>2.097)=0.018`

If we compare the p value obtained and the significance level given
`\alpha=0.05` 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 5% of significance the proportion of students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

b) The 90% confidence interval would be given by (0.512;0.596)

Explanation:

Part a

Data given and notation

n=377 represent the random sample taken

X=209 represent the students reported experiencing excessive daytime sleepiness (EDS)

`\hat p=(209)/(377)=0.554` estimated proportion of students reported experiencing excessive daytime sleepiness (EDS)

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

`\alpha=0.05` represent the significance level

Confidence=95% or 0.95

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 the true proportion is higher than 0.5:

Null hypothesis:
`p\leq 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{0.554 -0.5}{\sqrt{(0.5(1-0.5))/(377)}}=2.097`

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=0.05`. 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>2.097)=0.018`

If we compare the p value obtained and the significance level given
`\alpha=0.05` 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 5% of significance the proportion of students reported experiencing excessive daytime sleepiness (EDS) is significantly higher than 0.5 or the half.

Part b

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 90% of confidence, our significance level would be given by
`\alpha=1-0.90=0.1` and
`\alpha/2 =0.05`. And the critical value would be given by:

`t_(\alpha/2)=-1.64, t_(1-\alpha/2)=1.64`

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)}`

If we replace the values obtained we got:

`0.554 - 1.64\sqrt{(0.554(1-0.554))/(377)}=0.512`

`0.554 + 1.64\sqrt{(0.554(1-0.554))/(377)}=0.596`

The 90% confidence interval would be given by (0.512;0.596)