Question

A survey was conducted in a large city to investigate public opinion on banning the use of trans fats in restaurant cooking. A random sample of 230 city residents with school-age children was selected, and another random sample of 341 city residents without school-age children was also selected. Of those with school-age children, 94 opposed the banning of trans fats, and of those without school-age children, 147 opposed the banning of trans fats. An appropriate hypothesis test was conducted to investigate whether there was a difference between the two groups of residents in their opposition to the banning of trans fats. Is there convincing statistical evidence of a difference between the two population proportions at the significance level of 0.05

Answer 1

Answer:

There is no enough evidence to claim that there is a difference between the two population proportions.

Step-by-step explanation:

We have to perform an hypothesis testing for a difference between two population proportions.

The null hypothesis will state that both proportions are the same, and the alternative hypothesis will state that they differ. This would be than a two-side hypothesis test.

We can write this as:

$H_0: \pi_1-\pi_2=0\\\\H_a:\pi_1-\pi_2\neq0$

The significance level for this test is 0.05.

The sample of city residents with school-age children has a sample size n1=230 and a sample proportion p1=0.41

$p_1=X_1/N_1=94/230=0.41$

The sample of city residents without school-age children has a sample size n2=341 and a sample proportion p2=0.51

$p_2=X_2/N_2=147/341=0.43$

The weighted p, needed to calculate the standard error, is the weighted average of both sample proportions:

$p=\dfrac{n_1p_1+n_2p_2}{n_1+n_2}=\dfrac{230*0.41+341*0.51}{230+341}=\dfrac{94+147}{571} =\dfrac{241}{571}=0.42$

The standard error of the difference of proportions can now be calculated as:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n_1}+\dfrac{p(1-p)}{n_2}}=\sqrt{\dfrac{0.42*0.58}{230}+\dfrac{0.42*0.58}{341}}\\\\\\\sigma_p=\sqrt{\dfrac{0.2436}{230}+\dfrac{0.2436}{341}}=\sqrt{0.00106+0.00071}=\sqrt{0.00177}\\\\\\\sigma_p=0.042$

The test statistic z is:

$z=\dfrac{p_1-p_2}{\sigma_p}=\dfrac{0.41-0.43}{0.042}= \dfrac{0.020}{0.042}= 0.476$

The P-value for this two side test and this value of the z-statistic is:

$P-value=2*P(z>0.476)=0.634$

The P-value is bigger than the significance level, so the effect is not significant. The null hypothesis failed to be rejected.

There is no enough evidence to claim that there is a difference between the two population proportions.