For a sample of 60 University of Florida students, the mean number of times per
ID: 3174605 • Letter: F
Question
For a sample of 60 University of Florida students, the mean number of times per week reading a newspaper was found to be 4.1 with a standard deviation of 2.5 a. Construct and interpret a 95% confidence interval for the population mean of the number of times per week people read a newspaper. b. Would a 99% confidence interval be wider, narrower or the same? Why? c. Suppose the standard deviation had been 6.0 instead of 2.5. Find a 95% confidence interval for the population mean and compare it to the one obtained in part (a). How has it changed? Why did it change? d Suppose that the sample size had been 3600 instead of 60 (using the original standard deviation of 2.5). Find the 95% confidence interval for^population mean, and compare it to the one obtained in part (a). Why is it different?Explanation / Answer
a.
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=4.1
Standard deviation( sd )=2.5
Sample Size(n)=60
Confidence Interval = [ 4.1 ± t a/2 ( 2.5/ Sqrt ( 60) ) ]
= [ 4.1 - 2.001 * (0.323) , 4.1 + 2.001 * (0.323) ]
= [ 3.454,4.746 ]
b.
it would be wider
c.
Mean(x)=4.1
Standard deviation( sd )=6
Sample Size(n)=60
Confidence Interval = [ 4.1 ± t a/2 ( 6/ Sqrt ( 60) ) ]
= [ 4.1 - 2.001 * (0.775) , 4.1 + 2.001 * (0.775) ]
= [ 2.55,5.65 ]
d.
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=4.1
Standard deviation( sd )=2.5
Sample Size(n)=3600
Confidence Interval = [ 4.1 ± t a/2 ( 2.5/ Sqrt ( 3600) ) ]
= [ 4.1 - 1.961 * (0.042) , 4.1 + 1.961 * (0.042) ]
= [ 4.018,4.182 ]
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.