Sales at a fast-food restaurant average $6,000 per day. the restaurant decided to introduce an advertising campaign to increase daily sales. to determine the effectiveness of the advertising campaign, a sample of 49 days of sales were taken. they found that the average daily sales were $6,300 per day. from past history, the restaurant knew that its population standard deviation is about $1,000. if the level of significance is 0.01, have sales increased as a result of the advertising campaign? multiple choice

a. reject the null hypothesis and conclude that the mean is equal to $6,000 per day.
b. fail to reject the null hypothesis.
c. reject the null hypothesis and conclude the mean is lower than $6,000 per day.
d. reject the null hypothesis and conclude the mean is higher than $6,000 per day.

Option B) Fail to reject the null hypothesis.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = $6,000

Sample mean, = $6,300

Sample size, n = 49

Alpha, α = 0.01

Population standard deviation, σ = $1,000

First, we design the null and the alternate hypothesis

We use one-tailed(right) z test to perform this hypothesis.

Formula:

Putting all the values, we have

Now,

Since,  

We fail to reject the null hypothesis and accept the null hypothesis. Thus, we conclude that sales have not increased as a result of the advertising campaign

Option B) Fail to reject the null hypothesis.

no this is a nonlinear negative

answer 1 : option "c"

answer 2 : option "d"

answer 3 : option "b"

answer 4 : option "d"

answer 5 : option "b"

answer 6 : option "a"

answer 7 : option "c"

answer 8 : option "d"

answer 9 : option "c"

answer 10 : option "d"

answer 11 : option "c"

answer 12 option "b"

answer 13 option "c"

answer 14 option "c"

answer 15 option "d"

answer 16 option "a"

answer 17 option "b"

answer 18 option "c"

answer 19 option "a"

answer 20 option "a"

answer 21 option "c"

d