Homework Sourse3 We want to estimate the mean capacitance for a set of capa
3. We want to estimate the mean capacitance for a set of capacitors. The nominal rating for the units is 9 F. A sample of 10 units finds that the sample mean and variance are X¯ = 9.4 F and s 2 = 0.53 F 2 .
(a) Calculate the 95% confidence interval for the mean capacitance of these capacitors.
(b) Based on your results above would you reject the null hypothesis H0 : µ = 9?
(c) Now consider that you collected identical measurements for X¯ and s 2 but from a sample size of 20. Are your results the same as in part b. If not why?
Solution
a)
Note that
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 9.4
t(alpha/2) = critical t for the confidence interval = 2.262157163
s = sample standard deviation = 0.728010989
n = sample size = 10
df = n - 1 = 9
Thus,
Lower bound = 8.879212311
Upper bound = 9.920787689
Thus, the confidence interval is
( 8.879212311 , 9.920787689 ) [ANSWER]
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B)
No, as u = 9 is inside the confidence interval.
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c)
Note that
Lower Bound = X - t(alpha/2) * s / sqrt(n)
Upper Bound = X + t(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 9.4
t(alpha/2) = critical t for the confidence interval = 2.093024054
s = sample standard deviation = 0.728010989
n = sample size = 20
df = n - 1 = 19
Thus,
Lower bound = 9.059280369
Upper bound = 9.740719631
Thus, the confidence interval is
( 9.059280369 , 9.740719631 )
Now, u = 9 is no longer inside the confidence interval. Thus, we now reject Ho: u = 9.
It is not the same becasue the standard error of the mean is now lower, so we are \"more sure\" of our data. As we can see, the number of standard errors from the mean became larger because of that.
