Answer to Question #132175 in Statistics and Probability for Keagan

Question

2.2 In a recent poll, 800 voters were randomly sampled and asked whether they would approve of
the building of an incinerator in the state. According to the poll, 450 voters approved and 350
voters disapproved. Is there enough evidence to argue that more than 50 % of the voters support
the building of the incinerator?
Use a 5 % level of significance
2.3 Suppose the Acme Drug Company develops a new drug, designed to prevent colds. The company
states that the drug is equally effective for men and women. To test this claim, they choose a
simple random sample of 100 women and 200 men from a population of 100,000 volunteers.
At the end of the study, 38% of the women caught a cold; and 51% of the men caught a cold.
Based on these findings, can we reject the company's claim that the drug is equally effective for
men and women? Use a 0.05 level of significance.

Answer 1

2.2

The following information is provided: The sample size is  $N=800,$ the number of favorable cases is $X=450,$ and the sample proportion is 

$$\hat{p}=\dfrac{X}{N}=\dfrac{450}{800}=0.5625,$$

and the significance level is $\alpha=0.05.$

The following null and alternative hypotheses need to be tested:

$H_0:p\leq0.5$

$H_1:p>0.5$

This corresponds to a right-tailed test, for which a z-test for one population proportion needs to be used.

Based on the information provided, the significance level is $\alpha=0.05,$ and the critical value for a right-tailed test is $z_c=1.645.$

The rejection level for this right-tailed test is $R=\{z:z>1.645\}$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p}-p_0}{\sqrt{p_0(1-p_0)/N}}$$

$$z=\dfrac{0.5625-0.5}{\sqrt{0.5(1-0.5)/800}}\approx3.5355$$

Since it is observed that $z=3.5355>1.645=z_c,$ then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population proportion $p$ is greater than $p_0=0.5,$ at the $\alpha=0.05$ significance level.

Using the P-value approach: The p-value is $p=0.000204,$ and since $p=0.000204<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p$ is greater than $p_0=0.5,$ at the $\alpha=0.05$ significance level.

2.3

For sample 1, we have that the sample size is $N_1=100,$ the number of favorable cases is $X_1=38,$ and the sample proportion is $\hat{p_1}=0.38.$

For sample 2, we have that the sample size is $N_2=200,$ the number of favorable cases is $X_2=102,$ and the sample proportion is $\hat{p_2}=0.51.$

The value of the pooled proportion is computed as

$$\bar{p}=\dfrac{X_1+X_2}{N_1+N_2}=\dfrac{38+102}{100+200}=\dfrac{7}{15}$$

Also, the given significance level is $\alpha=0.05.$

The following null and alternative hypotheses need to be tested:

$H_0:p_1=p_2$

$H_1:p_1\not=p_2.$

This corresponds to a two-tailed test, for which a z-test for two population proportions needs to be conducted.

Based on the information provided, the significance level is $\alpha=0.05,$ and the critical value for a two-tailed test is $z_c=1.96.$

The rejection level for this two-tailed test is $R=\{z:|z|>1.96\}$

The z-statistic is computed as follows:

$$z=\dfrac{\hat{p_1}-\hat{p_2}}{\sqrt{\bar{p}(1-\bar{p})(1/N_1+1/N_2)}}=$$

$$=\dfrac{0.38-0.51}{\sqrt{\dfrac{7}{15}(1-\dfrac{7}{15})(1/100+1/200)}}\approx-2.1276$$

Since it is observed that $|z|=2.1276>1.96=z_c,$ then concluded that the null hypothesis is rejected. Therefore, there is enough evidence to claim that the population proportion $p_1$ is different than $p_2,$ at the $\alpha=0.05$ significance level.

Using the P-value approach: The p-value is $p=0.000204,$ and since $p=0.03337<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population proportion $p_1$ is different than $p_2,$ at the $\alpha=0.05$ significance level.