Answer:
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.
