Avg. Monthly Expenditure for Cellphone Service 1. Below is the data in the month
ID: 3266235 • Letter: A
Question
Avg. Monthly Expenditure for Cellphone Service
1. Below is the data in the monthly dollar amounts for the cellphone service for a random sample of 8 households in your community.
$47
$36
$41
$49
$35
$52
$42
$49
$38
$51
a. Find a 95% confidence interval for , the average monthly expenditure for cellphone service.
Hints: For 95% confidence interval, the t-statistic from the “t distribution critical values” is 2.262
b. Find the sample mean x
c. Find the standard deviation s
d. Find the standard error SE =s/n
e. Find the Margin of error m= 2.262*SE
95% confidence interval is x ±m
( Please make sure that you identify the value of t-statistic and show your calculations)
1. Why do we use t-statistic in this example and not the z-score?
2. Find the 80% confidence interval for the population mean
3. Find the 90% confidence interval for the population mean
4. Find the 95% confidence interval for the population mean
5. Find the 99% confidence interval for the population mean
6. Find the 99.9% confidence interval for the population mean
7. Describe how will you select a random sample of eight households (in the above problem)
$47
$36
$41
$49
$35
$52
$42
$49
$38
$51
Explanation / Answer
a)
Here, mean = 44 , std.deviation = 6.377 , n =10
t value at 95% CI = 2.262
CI = mean +/- t * ( s/sqrt(n))
= 44 +/- 2.262 * ( 6.377 / sqrt(10))
= (39.438 , 48.562)
b)
Sample mean = ( 47 + 36 + 41 + 49 + 35 + 52 + 42 + 49 + 38 + 51)/10 = 44
c)
standard deviation = sqrt((summation of x - mean)^2 /n -1)
= sqrt ( (47 - 44 )^2 + ( 36 - 44)^2 + ( 41 - 44)^2 + ( 49 - 44)^2 + ( 35-44)^2 + ( 52 - 44)^2 + ( 42 - 44)^2 + ( 49 - 44)^2 + (38 -44)^2 + ( 51 -44)^2/ 9 )
= 6.377
d)
Standard error= s / sqrt(n)
= 6.377 / sqrt(10)
= 2.016
e) ME = 2.262 * SE
= 2.262 * 2.016
= 4.560
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