Answer:
-243
Step-by-step explanation:
Starting salaries of 90 college graduates who have taken a statistics course have a mean of $43,993 and a standard deviation of
$9,144. Using 99% confidence level, find the following:
A. The margin of error E:________
B. The confidence interval for the mean: μ__________< μ<_____________
A. The margin of error E:________
B. The confidence interval for the mean: μ__________< μ<_____________
1 answer:
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Solution :
Given :
Sample mean,
Sample size, n = 129
Sample standard deviation, s = 8.2
a. Since the population standard deviation is unknown, therefore, we use the t-distribution.
b. Now for 95% confidence level,
α = 0.05, α/2 = 0.025
From the t tables, T.INV.2T(α, degree of freedom), we find the t value as
t =T.INV.2T(0.05, 128) = 2.34
Taking the positive value of t, we get
Confidence interval is ,
(32.52, 35.8)
95% confidence interval is (32.52, 35.8)
So with confidence of the population of the mean number of the pounds per person per week is between 32.52 pounds and 35.8 pounds.
c. About of confidence intervals which contains the true population of mean number of the pounds of the trash that is generated per person per week and about
that doe not contain the true population of mean number of the pounds of trashes generated by per person per week.