Question

Elliot read a report from a previous year saying that 6 % 6%6, percent of adults in his city biked to work. He wanted to test whether this had changed, so he took a random sample of 240 240240 adults in his city to test H 0 : p = 0.06 H 0 ​ :p=0.06H, start subscript, 0, end subscript, colon, p, equals, 0, point, 06 versus H a : p ≠ 0.06 H a ​ :p  ​ =0.06H, start subscript, start text, a, end text, end subscript, colon, p, does not equal, 0, point, 06, where p pp is the proportion of adults in Elliot's city that bike to work. The sample results showed 21 2121 adults who biked to work, and the corresponding test statistic was z ≈ 1.79 z≈1.79z, approximately equals, 1, point, 79. Assuming that the necessary conditions are met, what is the approximate P-value for Elliot's significance test?

Answer 1

Answer: To find the P-value, we first need to determine the direction of the alternative hypothesis. Since the alternative hypothesis is two-tailed, we need to divide the significance level (α) by 2 before finding the critical values and the rejection region. Therefore, we have:

H0: p = 0.06

Ha: p ≠ 0.06

α = 0.05/2 = 0.025

The test statistic is z = 1.79, which represents the number of standard errors that the sample proportion is from the hypothesized population proportion under the null hypothesis. We can find the P-value by looking up the area in the standard normal distribution table beyond the test statistic in both tails:

P-value = P(Z ≤ -1.79 or Z ≥ 1.79)

= P(Z ≤ -1.79) + P(Z ≥ 1.79)

= 0.0364 + 0.0364

= 0.0728

Therefore, the approximate P-value for Elliot's significance test is 0.0728. Since the P-value is greater than the significance level of 0.05, we fail to reject the null hypothesis and conclude that there is not enough evidence to suggest that the proportion of adults who bike to work in Elliot's city has changed significantly from the previous year.

Explanation: