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.
Answer:
C. A graph titled Years Collecting Stamps versus Stamps in Collection has years collecting stamps on the x-axis and stamps in collection on the y-axis. Points are at (2, 100), (3, 100), (3, 125), (4, 150), (4, 175), (5, 175)
Step-by-step explanation:
In the x-axis goes the values from the column 'Number of years collecting stamps' of the table. And In the y-axis goes the values from the column 'Number of stamps in collection' of the table.
To make the graph identify each pair of values in the plane and mark it.
Answer:
Option A and B are correct
Step-by-step explanation:
We have to add here
6x+5y=1.
6x-5y=7
sow e will get 12x=8 , x=8/12=2/3
Option A is correct.
If We substract bottom from furst then
6x+5y=1
-6x+5y=-7
10y=-6
y=-6/10=-3/5
we put y=-3/5 in any equation and find x
6x+5(-3/5)=1
6x=4
x=4/6=2/3
So Option B also correct.
I guess this will help you
Step-by-step explanation:
30 , 60 , 90 , 120, 150
,LCM of 12 and 30 is 60
