We have to perform an hypothesis test of a proportion.
The claim is that the sample has a different proportion than the population.
Then, the null and alternative hypothesis are:
[tex]\begin{gathered} H_0\colon\pi=0.24 \\ H_a\colon\pi\neq0.24 \end{gathered}[/tex]The significance level is 0.05.
The sample has a size n=86.
The sample proportion is p=0.349.
[tex]p=X/n=30/86=0.349[/tex]The standard error of the proportion is:
[tex]\begin{gathered} \sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt[]{\dfrac{0.24\cdot0.76}{86}} \\ \sigma_p=\sqrt{0.002121}=0.046 \end{gathered}[/tex]Then, we can calculate the z-statistic as:
[tex]z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.349-0.24-0.5/86}{0.046}=\dfrac{0.103}{0.046}=2.241[/tex]This test is a two-tailed test*, so the P-value for this test is calculated as:
[tex]\text{P-value}=2\cdot P(z>2.241)=0.025[/tex]* We use a two-tailed test because we are looking for difference above or below the population proportion.
As the P-value (0.025) is smaller than the significance level (0.05), the effect is significant.
The null hypothesis is rejected.
At a significance level of 0.05, there is enough evidence to support the claim that the sample has a different proportion than the population.
Answer:
1) The null and alternative hypothesis are:
[tex]\begin{gathered} H_0\colon\pi=0.24 \\ H_a\colon\pi\neq0.24 \end{gathered}[/tex]2) The test statistic is z=2.241.
3) The P-value is 0.025. The value in the standard normal distribution is:
4) As the effect is significant (the P-value is less than the significance level), there is evidence to reject the null hypothesis.
The conclusion is that this sample has a proportion that is significantly different from that from the population.
