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.