Question

You record the number of sunny days out of a sample of 90 randomly-selected days in City A and City B. In City A’s sample, there were 54 sunny days. In City B’s sample, there were 63 sunny days.

Is City B significantly sunnier than City A? Use a significance level of 0.05.

A) z approximately equals 2.00 for a P-value of 0.02275. Since the P-value is less than 0.05, fail to reject the null hypothesis that there is no difference in the commute times.
B) z approximately equals -1.41 for a P-value of 0.078652. Since the P-value is greater than 0.05, fail to reject the null hypothesis that there is no difference in the commute times.
C) z approximately equals 1.41 for a P-value of 0.039326. Since the P-value is less than 0.05, reject the null hypothesis that there is no difference in the commute times.
D) z approximately equals -2.00 for a P-value of 0.02275. Since the P-value is less than 0.05, reject the null hypothesis that there is no difference in the commute times.

Answer 1

To determine whether City B is significantly sunnier than City A, we can perform a hypothesis test for the difference between two proportions.

**Step 1: Define the hypotheses.**
- Null hypothesis (\(H_0\)): \(p_A = p_B\) (the proportion of sunny days is the same in both cities).
- Alternative hypothesis (\(H_1\): \(p_A < p_B\) (City B has a larger proportion of sunny days).

**Step 2: Calculate the sample proportions.**
- In City A, the proportion of sunny days \( \hat{p_A} = \frac{54}{90} = 0.6\).
- In City B, the proportion of sunny days \( \hat{p_B} = \frac{63}{90} \approx 0.7\).

**Step 3: Calculate the pooled proportion.**
\[
\hat{p} = \frac{x_A + x_B}{n_A + n_B} = \frac{54 + 63}{90 + 90} = \frac{117}{180} \approx 0.65
\]

**Step 4: Calculate the standard error (SE) of the difference in proportions.**
\[
SE = \sqrt{\hat{p}(1 - \hat{p}) \left( \frac{1}{n_A} + \frac{1}{n_B} \right)} = \sqrt{0.65 \times 0.35 \left( \frac{1}{90} + \frac{1}{90} \right)} = \sqrt{0.65 \times 0.35 \times \frac{2}{90}}
\]
Calculating this gives approximately:
\[
SE \approx \sqrt{0.65 \times 0.35 \times \frac{2}{90}} \approx \sqrt{0.65 \times 0.35 \times 0.0222} \approx \sqrt{0.005049} \approx 0.0711
\]

**Step 5: Compute the z-statistic.**
\[
z = \frac{\hat{p_B} - \hat{p_A}}{SE} = \frac{0.7 - 0.6}{0.0711} \approx \frac{0.1}{0.0711} \approx 1.41
\]

**Step 6: Find the P-value.**
Since this is a one-tailed test, we will look up the z-score of 1.41,
which corresponds to a P-value of approximately 0.079 (using a z-table or calculator).

**Step 7: Make a decision.**
Here, the significance level is \( \alpha = 0.05\).
Since the P-value (0.079) is greater than 0.05, we fail to reject the null hypothesis.

Given this analysis, the correct conclusion from the options provided is:

**B) z approximately equals 1.41 for a P-value of 0.078652. Since the P-value is greater than 0.05, fail to reject the null hypothesis that there is no difference in the sunny days.**