Answer:
$6.4
Step-by-step explanation:
You just do 32 × .2 to give you the discount of 6.4
You multiple by .2 because 20% as a decimal is .2 ( you do 20% divided by 100 to give you .2)
Answer:
(E) The bias will decrease and the variance will decrease.
Step-by-step explanation:
Given that researchers working the mean weight of a random sample of 800 carry-on bags to e the airline.
We have to find out the effect of increasing the sample size on variance and bias.
We know as per central limit theorem, sample mean follows a normal distribution with mean = sample mean
and std deviation of sample mean = std error = 
Thus std error the square root of variance is inversely proportional to the square root of sample size.
Also whenever we increase sample size the chances of bias would decrease as the sample size becomes larger
So correct answer is both bias and variation will decrease.
(E) The bias will decrease and the variance will decrease.
I guess no solution
4x +3 = 4X + 18
4X - 4X = 18 - 3
0 = 15
Answer:
The difference in the sample proportions is not statistically significant at 0.05 significance level.
Step-by-step explanation:
Significance level is missing, it is α=0.05
Let p(public) be the proportion of alumni of the public university who attended at least one class reunion
p(private) be the proportion of alumni of the private university who attended at least one class reunion
Hypotheses are:
: p(public) = p(private)
: p(public) ≠ p(private)
The formula for the test statistic is given as:
z=
where
- p1 is the sample proportion of public university students who attended at least one class reunion (
)
- p2 is the sample proportion of private university students who attended at least one class reunion (
)
- p is the pool proportion of p1 and p2 (
)
- n1 is the sample size of the alumni from public university (1311)
- n2 is the sample size of the students from private university (1038)
Then z=
=-0.207
Since p-value of the test statistic is 0.836>0.05 we fail to reject the null hypothesis.