Question

(S 9.2) Recall that a confidence interval for the sample mean can be calculated using the interval

x¯?tn?1?sn??????x¯+tn?1?sn???

Thus, the margin of error is tn?1?sn???

We can recover the margin of error from an interval constructed on the calculator using algebra.

Suppose a random sample of size 16 was taken from a normally distributed population, and the sample standard deviation was calculated to be s = 6.3. We'll assume the sample mean is 10 for convenience.

a) Calculate the margin of error for a 90% confidence interval for the population mean.

Round your response to at least 3 decimal places.

b) Calculate the margin of error for a 95% confidence interval for the population mean.

Round your response to at least 3 decimal places.

NOTE both these values are over 2. Suppose we want a smaller margin of error.
c) Approximately how large of a sample size is needed to construct a 90% confidence interval with a margin of error less than 1.5 given an estimate for the standard deviation of 6.3?

d) Approximately How large of a sample size is needed to construct a 95% confidence interval with margine of error less than 1.5 given an estimate for the standard deviation of 6.3?

Answer 1

) std error =std deviation/sqrt(n)=6.3/sqrt(10)

a)for (n-1=9) degree of freedom and 90% CI ; critical t =1.833

hence margin of error =t*std error=3.652

b)

for (n-1=9) degree of freedom and 95% CI ; critical t =2.262

hence margin of error =t*std error=4.506 ( please try 4.507 if this comes wrong due to rounding error)

c)

for 90 % CI value of z=	1.645
standard deviation =	6.30
margin of error E =	1.5
required sample size n=(z/E)2    =	48.0

d)

for 95 % CI value of z=	1.960
standard deviation =	6.30
margin of error E =	1.5
required sample size n=(z/E)2                                         =	68.0