The quality-control manager at a compact fluorescent light bulb (cfl) factory needs to determine whether the mean life of a large shipment of cfls is equal to 7450 hours. the population standard deviation is 1 comma 350 hours. a random sample of 81 light bulbs indicates a sample mean life of 7 comma 240 hours. a. at the 0.05 level of significance, is there evidence that the mean life is different from 7450 hours question mark b. compute the p-value and interpret its meaning. c. construct a 95% confidence interval estimate of the population mean life of the light bulbs. d. compare the results of (a) and (c). what conclusions do you reach

Let be the population mean .

By observing the given information, we have :-

Since the alternative hypotheses is two tailed so the test is a two-tailed test.

We assume that the  life of a large shipment of CFLs is normally distributed.

(a) Given : Sample size :  n=81 , since n>30 so we use z-test.

Sample mean :

Standard deviation :

Test statistic for population mean :-

The critical value (two-tailed) corresponds to the given significance level :-

Since the observed value of z (-1.4) is less than the critical value (1.96) , so we do not reject the null hypothesis.

Hence, we conclude that we have enough evidence to accept that the mean life is different from 7450 hours .

(b) The p-value : , it means that the probability that the life of CFLs less than 7240 and greater than 7240 is 0.1615.

(c) The confidence interval for population mean is given by :-

,

81 seats

step-by-step explanation:

if you multiply 0.65 times 125 then you get 81.5 seats. well since you can't have half of a seat you have to round down and get 81 seats in the no children zones. that leave 44 seats outside of the no children zone.