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

Answer:

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:$\mu > 8000$  

$z=\frac{8300-8000}{\frac{1200}{\sqrt{64}}}=2$    

$p_v =P(z>2)=0.0228$  

Step-by-step 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:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$  (1)  

Calculate the statistic

$t=\frac{8300-8000}{\frac{1200}{\sqrt{64}}}=2$    

P-value

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

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