Answer :
Answer:
a) [tex]z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097[/tex]
[tex]p_v =P(Z>2.097)=0.018[/tex]
If we compare the p value obtained and the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] 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)
Step-by-step 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)
[tex]\hat p=\frac{209}{377}=0.554[/tex] estimated proportion of students reported experiencing excessive daytime sleepiness (EDS)
[tex]p_o=0.5[/tex] is the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level
Confidence=95% or 0.95
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 the true proportion is higher than 0.5:
Null hypothesis:[tex]p\leq 0.5[/tex]
Alternative hypothesis:[tex]p > 0.5[/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].
Calculate the statistic
Since we have all the info requires we can replace in formula (1) like this:
[tex]z=\frac{0.554 -0.5}{\sqrt{\frac{0.5(1-0.5)}{377}}}=2.097[/tex]
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 [tex]\alpha=0.05[/tex]. The next step would be calculate the p value for this test.
Since is a right tailed test the p value would be:
[tex]p_v =P(Z>2.097)=0.018[/tex]
If we compare the p value obtained and the significance level given [tex]\alpha=0.05[/tex] we have [tex]p_v<\alpha[/tex] 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
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
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 [tex]\alpha=1-0.90=0.1[/tex] and [tex]\alpha/2 =0.05[/tex]. And the critical value would be given by:
[tex]t_{\alpha/2}=-1.64, t_{1-\alpha/2}=1.64[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
If we replace the values obtained we got:
[tex]0.554 - 1.64\sqrt{\frac{0.554(1-0.554)}{377}}=0.512[/tex]
[tex]0.554 + 1.64\sqrt{\frac{0.554(1-0.554)}{377}}=0.596[/tex]
The 90% confidence interval would be given by (0.512;0.596)