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

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dfbustos dfbustos · Answer 1 · 2020-05-04T14:30:25+02:00

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:[tex]\mu \leq 8000[/tex]

Alternative hypothesis:[tex]\mu > 8000[/tex]

[tex]z=\frac{8300-8000}{\frac{1200}{\sqrt{64}}}=2[/tex]

[tex]p_v =P(z>2)=0.0228[/tex]

Step-by-step explanation:

Data given

[tex]\bar X=8300[/tex] represent the sample mean for the sales

[tex]\sigma=1200[/tex] represent the population standard deviation

[tex]n=64[/tex] sample size

[tex]\mu_o =8000[/tex] 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:[tex]\mu \leq 8000[/tex]

Alternative hypothesis:[tex]\mu > 8000[/tex]

The statistic to check this hypothesis is given by:

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)

Calculate the statistic

[tex]t=\frac{8300-8000}{\frac{1200}{\sqrt{64}}}=2[/tex]

P-value

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

[tex]p_v =P(z>2)=0.0228[/tex]