Answer:
Step-by-step explanation:
a)
Test statistic:
here test statistic lie in rejection region,that why null hypothesis fails
so Yes, its significant.
b)
Test statistic:
c)
sample variability increases, therefore likelihood of rejecting the null hypothesis decreases.
Square's side range is (√0,√100) =(0,10); this more than 0 and less than 10.
Answer:
Option D.
Step-by-step explanation:
In the given graph x-axis represents the number of owners and y-axis represents yearly cost of pets, in hundreds.
It is given that the graph passes through the points (1,15), (3,7) and (10,0).
We need to check whether (4.5, 6) is a realistic solution for the function or not. (4.5,6) point represents
Number of owners = 4.5
yearly cost of pets = 6
It is realistic that owners spend $600 a year on their pet(s).
Number of owners can not be a fraction value. So, it is not realistic to have 4.5 owners.
Therefore, the correct option is D.
For the answer to the question above
Retail price of steel-belted radial tire = $89.50
Discount offered = 10% = $8.95
Value before federal tax = $89.50 - $8.95 = $80.55
Federal tax = 12%=$80.55(0.12) =$9.67
Local sales tax = 5% =$80.55(0.05)=$4.03
Selling price = $80.55 +$9.67+$4.03 =$94.25
I hope my answer helped you. Feel free to ask more questions. Have a nice day!
Answer:
There is no significant evidence which shows that there is a difference in the driving ability of students from West University and East University, <em>assuming a significance level 0.1</em>
Step-by-step explanation:
Let p1 be the proportion of West University students who involved in a car accident within the past year
Let p2 be the proportion of East University students who involved in a car accident within the past year
Then
p1=p2
p1≠p2
The formula for the test statistic is given as:
z=
where
- p1 is the <em>sample</em> proportion of West University students who involved in a car accident within the past year (0.15)
- p2 is the <em>sample</em> proportion of East University students who involved in a car accident within the past year (0.12)
- p is the pool proportion of p1 and p2 (
) - n1 is the sample size of the students from West University (100)
- n2 is the sample size ofthe students from East University (100)
Then we have z=
≈ 0.6208
Since this is a two tailed test, corresponding p-value for the test statistic is ≈ 0.5347.
<em>Assuming significance level 0.1</em>, The result is not significant since 0.5347>0.1. Therefore we fail to reject the null hypothesis at 0.1 significance
