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.
The second one (straight line parallel to the y axis)
We are asked to find 7 is what percent to 30.
We will use proportions to solve our given problem.
Let us assume that 7 is n% of 30. So we can set a proportion as:
Let us solve for n.
Therefore, 7 is 
of 30.
Answer:
U =35.5
Step-by-step explanation:
Since this is a right angle, we can use trig functions
tan theta = opp/ adj
tan U = 5/7
Take the inverse of each side
tan ^-1 tan U = tan ^-1 (5/7)
U = 35.53767779
Round to the nearest tenth
U =35.5
