A school principal is interested in assessing the performance of her students on

Question

A school principal is interested in assessing the performance of her students on the Totally Oppressive Standardized test (TOST). She selects a simple random sample of 16 of her students and finds the following set of scores: Assume that the sample is drawn from a population with a standard deviation sigma = 7.20. What is the mean score for the sample of students? _ Find and interpret in words the 95% confidence interval for the mean. _ Find and interpret In words the 99% confidence interval for the mean. _ Why a the margin of error smaller for the 95% confidence interval in question b than for the 99% confidence interval In question c? _

Explanation / Answer

a.
Mean score of the sample of students(x)=13
b.
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=13
Standard deviation( sd )=7.2
Sample Size(n)=16
Confidence Interval = [ 13 ± Z a/2 ( 7.2/ Sqrt ( 16) ) ]
= [ 13 - 1.96 * (1.8) , 13 + 1.96 * (1.8) ]
= [ 9.47,16.53 ]
c.
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=13
Standard deviation( sd )=7.2
Sample Size(n)=16
Confidence Interval = [ 13 ± Z a/2 ( 7.2/ Sqrt ( 16) ) ]
= [ 13 - 2.58 * (1.8) , 13 + 2.58 * (1.8) ]
= [ 8.36,17.64 ]
Interpretations:
1) We are 99% sure that the interval [8.36 , 17.64 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean  

d.
When confidence interval decreases, the value of z-value be down & it pull downs the margin of error value

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