A sample of randomly selected students from a local elementary school yielded th

Question

A sample of randomly selected students from a local elementary school yielded the following scores on a standardized test. Assume that the population distribution of scores is approximately normal. 78 100 93 95 93 86 105 92 85 81 92 96 92 81 95 106 106 96 85 92 a. Calculate a 95% confidence interval for a population mean score. (Round to two decimal places.) b. Calculate a 95% prediction interval for the score of a single student randomly selected from this population. (Round to two decimal places.) c. Calculate an interval that includes at least 99% of the scores in the population distribution using a confidence level of 95%. (Round to two decimal places.)

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)=92.45
Standard deviation( sd )=8.0947
Sample Size(n)=20
Confidence Interval = [ 92.45 ± t a/2 ( 8.0947/ Sqrt ( 20) ) ]
= [ 92.45 - 2.093 * (1.81) , 92.45 + 2.093 * (1.81) ]
= [ 88.66,96.24 ]

b.

Interpretations:
1) We are 95% sure that the interval [88.66 , 96.24 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean  
  
  
  
  
  
  
  

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