Recently many companies have been experimenting with telecommuting, allowing employees to work at home on their computers. among other things, telecommuting is supposed to reduce the number of sick days taken. suppose that at one firm, it is known that over the past few years employees have taken a mean of 5.4 sick days. this year, the firm introduces telecommuting. management chooses a simple random sample of 80 employees to follow in detail, and, at the end of the year, these employees average 4.4 sick days with a standard deviation of 2.8 days. let μ represent the mean number of sick days for all employees of the firm.

(a). find the p-value for testing h0 : ? ? 5.4 versus h1 : ? < 5.4.
(b). do you believe it is plausible that the mean number of sick days is at least 5.4, or are you convinced that it is less than 5.4? explain your reasoning.

We conclude that the telecommuting reduced the number of sick days.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 5.4

Sample mean, = 4.4

Sample size, n = 80

Alpha, α = 0.05

Sample standard deviation, s = 2.8

First, we design the null and the alternate hypothesis

We use One-tailed z test to perform this hypothesis.

Formula:

Putting all the values, we have

Now,

We calculate the p-value from the z-table

P-value = 0.000702, which is very low.

b) Since we have a very low p-value that is it is less than the significance level, we fail to accept the null hypothesis and reject it at significance level of 5%. We would require a very small value of alpha for us to accept the null hypothesis thus we would conclude that there is statistically significant evidence to reject the null hypothesis that the mean was equal to 5.4.

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐

i believe the correct answer is b.