QUESTION : A sample of 10 observations is selected from a normal population for

Question

QUESTION : A sample of 10 observations is selected from a normal population for which the population standard deviation is known to be 5. The sample mean is 20.
• Determine the standard error of the mean. (5 points)

• Explain why we can use the standard formula to determine the 95 percent condence interval even though the sample is less than 30. (7 points)

•Determine the 95 percent condence interval for the population mean.(7 points)

please it is very important

• Determine the 90 percent condence interval for the population mean.(7 points)

Explanation / Answer

• Determine the standard error of the mean. (5 points)

SE = signa/sqrt(n) = 5/sqrt(10) = 1.58113883 [ANSWER]

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• Explain why we can use the standard formula to determine the 95 percent condence interval even though the sample is less than 30. (7 points)

It is becasue the population is normal, so th sampling distribution is normal even for small sample size.

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•Determine the 95 percent condence interval for the population mean.(7 points)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    20          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    5          
n = sample size =    10          
              
Thus,              
Margin of Error E =    3.098975162          
Lower bound =    16.90102484          
Upper bound =    23.09897516          
              
Thus, the confidence interval is              
              
(   16.90102484   ,   23.09897516   ) [ANSWER]

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• Determine the 90 percent condence interval for the population mean.(7 points)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.05          
X = sample mean =    20          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    5          
n = sample size =    10          
              
Thus,              
Margin of Error E =    2.600741939          
Lower bound =    17.39925806          
Upper bound =    22.60074194          
              
Thus, the confidence interval is              
              
(   17.39925806   ,   22.60074194   ) [ANSWER]

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