Alocal eat-in pizza restaurant wants to investigate the possibility of starting to deliver pizzas. the owner of the store has determined that home delivery will be successful only if the average time spent on a delivery does not exceed 25 minutes. the owner has randomly selected 21 customers and delivered pizzas to their homes in order to test whether the mean delivery time actually exceeds 25 minutes. suppose the p-value for the test was found to be .0284. state the correct conclusion. question 22 options: at ? = .025, we fail to reject h0.at ? = .05, we fail to reject h0.at ? = .03, we fail to reject h0.at ? = .02, we reject h0.

At α = .025, we fail to reject

Step-by-step explanation:

The null and the alternative hypothesis

H0:  μ ≤ 25  

H1:  μ  > 25  

When testing this hypothesis, the decision rule is to reject the null hypothesis if the p-value is less than the given level of significance (α), otherwise do not reject.

The p-value for the test was found to be .0284. At α = .025, the p-value is greater than the level of significance, therefore, we fail to reject H0 and conclude that the mean time does not exceed 25 minutes.

answer: h

step-by-step explanation:

= 5

step-by-step explanation:

to find the 15th term substitute n = 15 into the nth term formula

= (- 3 × 15 ) + 50 = - 45 + 50 = 5