Answer to Question #152042 in Statistics and Probability for JG

Question

Based on a survey of 150 AUS students, we found 36 of them do not eat breakfast. Describe the sampling distribution of the sample proportion, and find the point estimate and standard error.
a) Construct an 89% confidence interval of the percentage of all AUS students who eat breakfast and state the confidence statement.
b) Do we have enough evidence to conclude that less than 25% of AUS students do not eat breakfast at 9% level of significance? (Use both critical value method and p-value method)

Answer 1

The population proportion is denoted p and the sample proportion is denoted $\hat{p}$

$\hat{p}=\frac{36}{150}=0.24$ = point estimate

SE = estimated standard error = $\sqrt{\hat{p}*(1-\hat{p})/n}=\sqrt{0.24*(1-0.24)/150}=0.0349$

a) Finding z-value:

P(-z>x>z)=0.89

z=1.598

$\hat{p}=1-0.24=0.76$

Confidence interval:

$\hat{p}±z*\sqrt{\hat{p}*(1-\hat{p})/n}$

$0.76±0.056$

Stating the confidence statement:

The population proportion is greater than 0.704 and less than 0.816 with 89% confidence.

b) Critical value method:

H₀: p=0.25

H₁: p<0.25

Finding critical value:

P(x<z) = 0.09

z=-1.341

Computing test value:

Z = $\frac{\hat{p}-p}{SE}=\frac{0.24-0.25}{0.0349}=-0.287$

We can't reject null hypothesis since test value>critical value.

We do not have enough evidence to conclude that less than 25% of AUS students do not eat breakfast at 9% level of significance.

p-value method:

p-value = P(x<-0.287) = 0.387

0.387>0.09

We can't reject null hypothesis since p-value>level of significance.

We do not have enough evidence to conclude that less than 25% of AUS students do not eat breakfast at 9% level of significance.