Agrocery 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.(provide mathematical steps and explanations in detail to receive full credit.) the researcher would like to test the following hypothesis : ? ? ? 8,000 : ? ? > 8,000 (a) calculate the value of the test statistic. let ? =0.05 (b) what is the conclusion based on the critical value approach? let ? =0.05 (c) what is the conclusion based on the p-value value approach? let ? =0.05

a) Z= 0.0228

b) based on the critical value, it is a one tailed test

c) Since the p-value is less than the level of significance (α= 0.05) so we reject the null hypothesis.

This implies that the advertising campaign has been effective in increasing sales.

Step-by-step explanation:

The null hypothesis is H₀ : µ = 8000

The alternative hypothesis is H₁ : µ ˃ 8000

Mean (µ) = 8000

Standard deviation (σ) = 1200

n = 64

We will use the Z test to test the hypothesis

Z = (X - µ)/ (σ/√n)

Z = (8300 – 8000)/ (1200/√64)

Z = 300/ (1200/8)

Z = 300/ 150

Z= 2

From the normal distribution table,

2 = 0.4772

Φ(z) = 0.4772

Since Z is positive,

P(x˃a) =0.5 - Φ(z)

= 0.5 – 0.4773

= 0.0228

The required P-value = 0.0228

The P-value of one tail Z test at α= 0.05 level of significance

P-value = p(Z˃2.5)

= 0.0228

Since the p-value is less than the level of significance (α= 0.05) so we reject the null hypothesis.

This implies that the advertising campaign has been effective in increasing sales.

the answer to this question is 3/19

step-by-step explanation: