Solutions

Please note that the steps show rounded numbers, but that the final answers to the problems are calculated without rounding.

Problem Part Solution
1 - Pie
2 - \(n_1 \times \hat{p}_1 \geq 10 \text{ and } n_1 \times (1 - \hat{p}_1) \geq 10\)
\(261\times(0.038) = 10 \geq 10 \text{ and } 261\times(1 - 0.038) = 251\geq 10\)

\(n_2 \times \hat{p}_2 \geq 10 \text{ and } n_2 \times (1 - \hat{p}_2) \geq 10\)
\(1614\times(0.103) = 166 \geq 10 \text{ and } 1614\times(1 - 0.103) = 1448\geq 10\)

The requirements are met.
3 - ( -0.092 , -0.037 ) We are 95 % confident that the true difference of the proportions of city children with hay fever and rural children with hay fever is between -0.092 and -0.037 .

If you swapped the definition of groups 1 and 2, then you would get the same values with opposite signs: ( 0.037 , 0.092 ). This is also correct.
4 - No. This means that it is plausible that the likelihood of a child contracting hay fever is different in the city than in rural areas.
5 - \(n_1 \times \hat{p}_1 \geq 10 \text{ and } n_1 \times (1 - \hat{p}_1) \geq 10\)
\(200,000\times(0.0002) = 33 \geq 10 \text{ and } 200,000\times(1 - 0.0002) = 199,967\geq 10\)

\(n_2 \times \hat{p}_2 \geq 10 \text{ and } n_2 \times (1 - \hat{p}_2) \geq 10\)
\(200,000\times(0.0006) = 115 \geq 10 \text{ and } 200,000\times(1 - 0.0006) = 199,885\geq 10\)

The requirements are met.
6 - ( -0.00053 , -0.00029 ) We are 95 % confident that the true difference of the proportions of vaccinated children who developed polio and non-vaccinated children who developed polio is -0.00053 and -0.00029 .

If you swapped the definition of groups 1 and 2, then you would get the same values with opposite signs: ( 0.00029 , 0.00053 ). This is also correct.
7 - Pie

Bar
8 - \(n_1 \times \hat{p}_1 \geq 10 \text{ and } n_1 \times (1 - \hat{p}_1) \geq 10\)
\(117\times(0.624) = 73 \geq 10 \text{ and } 117\times(1 - 0.624) = 44\geq 10\)

\(n_2 \times \hat{p}_2 \geq 10 \text{ and } n_2 \times (1 - \hat{p}_2) \geq 10\)
\(168\times(0.518) = 87 \geq 10 \text{ and } 168\times(1 - 0.518) = 81\geq 10\)

The requirements are met.
9 - \(H_0: p_1 = p_2\)
\(H_a: p_1 > p_2\)
10 - \(\hat{p_1} = 0.624\)
\(\hat{p_2} = 0.518\)
11 - \(z = 1.775\)
12 - \(\text{p-value} = 0.038\)
13 - applet
14 - reject the null hypothesis
15 - There is sufficient evidence to suggest that the proportion of men who cheat in college is greater than the proportion of women who cheat in college.
16 - The p-value would double and be equal to 0.076 . This p-value is not significant and we would fail to reject the null hypothesis. With a two sided test we would not have sufficient evidence to conclude that there is a difference between the proportion of women and men who cheat in college.
17 - \(n_1 \times \hat{p}_1 \geq 10 \text{ and } n_1 \times (1 - \hat{p}_1) \geq 10\)
\(1655\times(0.03) = 50 \geq 10 \text{ and } 1655\times(1 - 0.03) = 1605\geq 10\)

\(n_2 \times \hat{p}_2 \geq 10 \text{ and } n_2 \times (1 - \hat{p}_2) \geq 10\)
\(1652\times(0.019) = 31 \geq 10 \text{ and } 1652\times(1 - 0.019) = 1621\geq 10\)

The requirements are met.
18 - \(H_0: p_1 = p_2\)
\(H_a: p_1 \neq p_2\)
19 - \(\hat{p_1} = 0.03\)
\(\hat{p_2} = 0.019\)
20 - \(z = 2.129\)
21 - \(\text{p-value} = 0.033\)
22 - applet
23 - reject the null hypothesis
24 - There is sufficient evidence to suggestthat the proportion of Clarinex subjects with dry mouth is different than the proportion of placebo subjects with dry mouth.