Find the confidence level that correspmds to a value of or = .02 How much area i
ID: 3160357 • Letter: F
Question
Find the confidence level that correspmds to a value of or = .02
How much area is under the standard normal curve from -3 to+ 3 standard deviations?
The IQs of a random sample city have a mean of 200 highschool students living in a large city have a mean of 107 and a standard devation of 12.4. Find a 95% confidende interval for the true average I.Q. of all the highchool stduents.
A university dean wishes to estimate the average number at hours his part time instrcutors teach per week. The standard devation from a previous study, is 2.6 hours. How large a sample must be selcted if he wants 99% confident of finding weather the true mean differs from the smaple mean by 1 hour or less.
Explanation / Answer
The IQs of a random sample city have a mean of 200 highschool students living in a large city have a mean of 107 and a standard devation of 12.4. Find a 95% confidende interval for the true average I.Q. of all the highchool stduents.
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)=107
Standard deviation( sd )=12.4
Sample Size(n)=200
Confidence Interval = [ 107 ± t a/2 ( 12.4/ Sqrt ( 200) ) ]
= [ 107 - 1.972 * (0.877) , 107 + 1.972 * (0.877) ]
= [ 105.271,108.729 ]
A university dean wishes to estimate the average number at hours his part time instrcutors teach per week. The standard devation from a previous study, is 2.6 hours. How large a sample must be selcted if he wants 99% confident of finding weather the true mean differs from the smaple mean by 1 hour or less.
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.01% LOS is = 1.972 ( From Standard Normal Table )
Standard Deviation ( S.D) = 2.6
ME =1
n = ( 1.972*2.6/1) ^2
= (5.127/1 ) ^2
= 26.288 ~ 27
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