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.
o 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.
o 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.
o 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.
o 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.
Please show your work.
To determine if City B is significantly sunnier than City A, we can use a z-test for two proportions.
First, calculate the proportions of sunny days in each city:
City A: 54/90 = 0.60
City B: 63/90 = 0.70
Next, calculate the pooled proportion:
p = (54 + 63) / (90 + 90) = 0.65
Then, calculate the standard error:
SE = sqrt( p(1-p) * (1/90 + 1/90) )
SE = sqrt( 0.65*0.35 * 0.0222 )
SE = 0.0514
Now, calculate the z-score:
z = (0.70 - 0.60) / 0.0514
z = 1.94
Using a Z-table or statistical software, we find that the P-value is approximately 0.0266, which is less than the significance level of 0.05. Therefore, we reject the null hypothesis and conclude that City B is significantly sunnier than City A.