Question

The sales of a grocery store had an average of $8,000 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 $8,300 per day. From past information, it is known that the standard deviation of the population is $1,200. The value of the test statistic is:

The p-value is, (round to 4 decimal places):

Answer 1

Null hypothesis:
`\mu \leq 8000`

Alternative hypothesis:
`\mu > 8000`

`z=(8300-8000)/((1200)/(√(64)))=2`

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

Explanation:

1) Data given and notation

`\bar X=8300` represent the sample mean

`\sigma=1200` represent the population standard deviation

`n=64` sample size

`\mu_o =800` represent the value that we want to test

`\alpha` represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

`p_v` represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 8000, the system of hypothesis are :

Null hypothesis:
`\mu \leq 8000`

Alternative hypothesis:
`\mu > 8000`

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

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

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

`z=(8300-8000)/((1200)/(√(64)))=2`

4) P-value

Since is a one-side upper test the p value would given by:

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

5) Conclusion

If we compare the p value and the significance level assumed, for example
`\alpha=0.05` we see that
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is significantly higher than $8000.