D" size batteries produced by mnm corporation have had a life expectancy of 87 hours. because of an improved production process, the company believes that there has been an increase in the life expectancy of its "d" size batteries. a sample of 36 batteries showed an average life of 88.5 hours. assume from past information that it is known that the standard deviation of the population is 9 hours.

(a) use a 0.01 level of significance, and test to determine if there has been an increase in the life expectancy of the "d" size batteries.
(b) what is the p-value associated with the sample results? what is your conclusion, based on the p-value?

a) There is no enough evidence of an increase in life expectancy.

b) P-value=0.07.

The P-value indicates that this sample result is not rare to happen (is greater than the significance level) even if the mean of the population is still 87 hours.

Step-by-step explanation:

We have to perform a hypothesis test on the mean, with known standard deviation of the population.

The null hypothesis, which we will reject or not, states that the life expectancy has not increase (it stays equal or less than 87 hours).

The null and alternative hypothesis are

The significance level is α=0.01.

The z-value is

The P-value for z is

The P-value (0.07) is greater than the significance level (0.01), so the effect is not significant. We fail to reject the null hypothesis.

There is no enough evidence of an increase in life expectancy.

answer: 4 meters

step-by-step explanation: to you understand 2/3 better you can do 1 divided by 3 = . then multiply that by the 2 which is .66666666666666667 then you would do 6 x .66666666666666667 which is 4.      

hope this .

you didn't post any answer choices, so its not 100% clear what your teacher is looking for

however, log(15/7) is equivalent to log(15) - log(7)

i'm using the rule log(x/y) = log(x) - log(y)