Question

A grocery store has an average sales of $8000 per day. The store introduced several advertising campaigns in order to increase sales. To determine whether or not the advertising campaigns have been effective in increasing sales, a sample of 64 days of sales was selected. It was found that the average was $8300 per day. From past information, it is known that the standard deviation of the population is $1200. The correct null hypothesis for this problem is

Answer 1

We need to conduct a hypothesis in order to check if the true mean for sales is significantly higher than 8000, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 8000`

Alternative hypothesis:
`\mu > 8000`

`z=(8300-8000)/((1200)/(√(64)))=2`

`p_v =P(z>2)=0.0228`

Explanation:

Data given

`\bar X=8300` represent the sample mean for the sales

`\sigma=1200` represent the population standard deviation

`n=64` sample size

`\mu_o =8000` represent the value that we want to test

z would represent the statistic (variable of interest)

System of hypothesis

We need to conduct a hypothesis in order to check if the true mean for sales is significantly higher than 8000, the system of hypothesis would be:

Null hypothesis:
`\mu \leq 8000`

Alternative hypothesis:
`\mu > 8000`

The statistic to check this hypothesis is given by:

`z=(\bar X-\mu_o)/((\sigma)/(√(n)))` (1)

Calculate the statistic

`t=(8300-8000)/((1200)/(√(64)))=2`

P-value

Since is a one right tailed test the p value would be:

`p_v =P(z>2)=0.0228`