Question

(1 point) a) A random sample of 13 cans of peach halves has a mean weight of 16 ounces and standard deviation of 0.4 ounces. Find a 90% confidence interval for a true standard deviation of the weights of all cans of peach halves. Confidence interval: b) What would be the confidence interval for a true standard deviation if the sample size was 45? Confidence interval:(

Answer 1

(A)

Here we have given that

n= Number of cans of peach halves = 13

= sample mean = 16 ounces

T

s= standard deviation = 0.4 ounces

Now we want to find the 90% confidence interval for population standard deviation of weights of all cans of peach halves $\sigma$

(n − 1)S2 XL <o< (n − 1) S2 X²R

1st we calculate the critical values

c= confidence level = 0.90

$\alpha$ = level of significance = 1-c = 1-0.90 = 0.10

degress of freedom = n-1 = 13-1=12

y?L= x°C, D.F)

              = $\chi ^2(\frac{0.10}{2},12)$

              = $\chi ^2(0.05,12)$

              =21.03 Using EXCEL software = CHIINV(probability =0.05, D.F=12)

$\chi ^2 R =\chi ^2 ((1-(\frac{\alpha}{2}),D.F)$

               = $\chi ^2 ((1-(\frac{0.10}{2}),12)$

               = $\chi ^2 (0.95,12)$

               = 5.23   Using EXCEL software = CHIINV(probability =0.95, D.F=12)

Now, the 90% confidence interval for population standard deviation of weights of all cans of peach halves $\sigma$

(n − 1)S2 XL <o< (n − 1) S2 X²R

$\sqrt {\frac{(13-1)0.4^2}{21.03}} < \sigma < \sqrt {\frac{(13-1)0.4^2}{5.23}}$

$0.30 < \sigma < 0.61$

Confidence interval (0.30, 0.61)

(b)

Now, we want to find the CI for n=sample size=45

1st we calculate the critical values

c= confidence level = 0.90

$\alpha$ = level of significance = 1-c = 1-0.90 = 0.10

degress of freedom = n-1 = 45-1=44

y?L= x°C, D.F)

              = $\chi ^2(\frac{0.10}{2},44)$

              = $\chi ^2(0.05,44)$

              = 60.48 Using EXCEL software = CHIINV(probability =0.05, D.F=44)

$\chi ^2 R =\chi ^2 ((1-(\frac{\alpha}{2}),D.F)$

               = $\chi ^2 ((1-(\frac{0.10}{2}),44)$

               = $\chi ^2 (0.95,44)$

               = 29.79 Using EXCEL software = CHIINV(probability =0.95, D.F=44)

Now, the 90% confidence interval for population standard deviation of weights of all cans of peach halves $\sigma$

(n − 1)S2 XL <o< (n − 1) S2 X²R

$\sqrt {\frac{(45-1)0.4^2}{60.48}} < \sigma < \sqrt {\frac{(45-1)0.4^2}{29.79}}$

$0.34 < \sigma < 0.49$

Confidence interval (0.34, 0.49)

Interpretation:

We can conclude that we are 90% confident that population standard deviation will fall within this confidence interval.