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home / study / questions and answers / math / statistics and probability / in the following situation, assume the random variable ... Question in the following situation, assume the random variable is normally distributed and use a normal distribution or a t-distribution to construct a 90% confidence interval for the population mean. If convenient use a calculator. A) In a random sample of 10 adults from a nearby county, the mean waste generated per person per day was 3.71 pounds and standard deviation was 1.78 pounds B) repeat part (A), assuming the statistics came from a sample size of 500. (A) For the sample of 10 adults, the 90% confidence level is: ______,_______? (B) For the sample of 500 adults, the 90% confidence level is: _____,____?

Solution

A)
For the sample of 10 adults, 90% 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)=3.71
Standard deviation( sd )=1.78
Sample Size(n)=10
Confidence Interval = [ 3.71 ± t a/2 ( 1.78/ Sqrt ( 10) ) ]
= [ 3.71 - 1.833 * (0.563) , 3.71 + 1.833 * (0.563) ]
= [ 2.678,4.742 ]

B)
For the sample of 500 adults, 90% Confidence Interval
Confidence Interval = [ 3.71 ± t a/2 ( 1.78/ Sqrt ( 500) ) ]
= [ 3.71 - 1.648 * (0.08) , 3.71 + 1.648 * (0.08) ]
= [ 3.579,3.841 ]