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.

x+14

step-by-step explanation:

x is greater than 5

step-by-step explanation:

2x+7 is greater than 17

    -7                           -7

2x is greater than 10

divide by 2 each side

x is greater than 5