Question

1. a sample of 30 boxes of cereal has a sample standard deviation of 0.81 ounces. construct a 96% confidence interval to estimate the true standard deviation of the filling process for the boxes of cereal.

a) (0.709, 0.709)
b) (0.638, 1.105)
c) (0.520, 1.100)
d) (0.407, 1.222)
e) none of the above.

Explanation:

Step1: Find the confidence level

For a 96% confidence level, the significance level \(\alpha = 1 - 0.96=0.04\). Then \(\alpha/2 = 0.02\).

Step2: Determine the critical value \(z_{\alpha/2}\)

Using the standard normal distribution table or a calculator, \(z_{0.02}\approx 2.05\) (since \(P(Z>z_{0.02}) = 0.02\) and \(P(Z < z_{0.02})=0.98\)).

Step3: Calculate the margin of error \(E\)

The formula for the margin of error for a population standard deviation (assumed known) is \(E = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\). Given \(\sigma = 0.81\), \(n = 30\).

$$ E=2.05\times\frac{0.81}{\sqrt{30}}\approx2.05\times\frac{0.81}{5.477}\approx2.05\times0.148\approx0.303 $$

Step4: Assume the sample mean (not given, but if we assume some wrong - approach check)

If we assume a wrong - way (but since no sample mean is given, maybe there is a typo and it's a confidence interval for the standard deviation which is wrong. But if we assume it's a confidence interval for the mean with wrong values)
Let's re - check the formula for confidence interval for the mean \(\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\).
If we calculate \(z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\) for each option:

• Option A: \((0.709,0.709)\) is a point, not an interval (invalid as a confidence interval for non - degenerate case)
• Option B: Margin of error \(E_1=1.105 - 0.638=0.467\), \(z=\frac{0.467\sqrt{30}}{0.81}\approx\frac{0.467\times5.477}{0.81}\approx3.15\) (not \(z\) for 96% confidence)
• Option C: Margin of error \(E_2 = 1.100 - 0.520 = 0.58\), \(z=\frac{0.58\sqrt{30}}{0.81}\approx\frac{0.58\times5.477}{0.81}\approx3.9\) (not \(z\) for 96% confidence)
• Option D: Margin of error \(E_3=1.222 - 0.407 = 0.815\), \(z=\frac{0.815\sqrt{30}}{0.81}\approx\frac{0.815\times5.477}{0.81}\approx5.5\) (not \(z\) for 96% confidence)

Answer:

E. None of the above.