Answer:
a) critical values ≈ -3.182 and 3.182 shows that there is difference in average prices
b) H0 : μd = 0
Ha : μd ≠ 0
Step-by-step explanation:
<u>a) Determine if there is any significant difference in the average prices </u>
using ∝ = 0.05
and μ( advertiser ) - μ ( competitor ) = μd
test statistic ( t ) = -3.011
df = 3
hence the critical values ( two tailed test ) = -3.182 and 3.182. this shows that there is sufficient evidence to show that there is difference in the average prices
<u>b) state the hypothesis</u>
H0 : μd = 0 ( null )
Alternate hypothesis : Ha : μd ≠ 0
It represents the thing the avadocate needs to be muiltiplied by
First, we need to work out the total number of students who were being surveyed.
We know that half of the students has two pets. The rest of the students make up the other half. So, we have 3 students + 2 students + 8 students = 13 students that make half of the sample population
That means total number of students being surveyed is 13+13=26 students
Then we work out the probability
P(One pet) = 8/26 = 4/13
P(Two pets) = 1/2
P(Three pets) = 3/26
P( Four pets) = 2/26 = 1/13
The probability distribution is shown in the table below. Let

be the number of pets and

is the probability of owning the number of pets
Answer
its the third one.
Step-by-step explanation: