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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 A. µ <= 8000. B. µ <= 8300. C. µ = 8000. D. µ > 8300.

Respuesta :

Answer:

C)  µ = 8000.

Step-by-step explanation:

Explanation:-

Given data A grocery store has an average sales of $8000 per day

mean of the Population μ = $ 8000

sample size 'n' = 64

mean of the sample x⁻ = $ 8300

Null Hypothesis : H₀ : μ = $ 8000

Alternative Hypothesis : H₁: μ > $ 8000

Test statistic

[tex]Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }[/tex]

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

Z = 2

Level of significance : ∝ = 0.05

Z₀.₀₅ = 1.96

The calculated value Z = 2 > 1.96 at 0.05 level of significance

Null hypothesis is rejected

Alternative hypothesis is accepted at 0.05 level of significance

Conclusion :-

The advertising campaigns have been effective in increasing sales

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