An important problem in industry is shipment damage. A pottery producing company ships its product by truck and determines that it can meet its profit expectations if, on average, the number of damaged items per truckload is fewer than 11. A random sample of 19 departing truckloads is selected at the delivery point and the average number of damaged items per truckload is calculated to be 9.4 with a calculated sample of variance of 0.49. Select a 95% confidence interval for the true mean of damaged items.
a) [48.91, -33.04]
b) [10.66, 11.34]
c) [9.063, 9.737]
d) [8.918, 9.882]
e) [-0.3372, 0.3372]
f) None of the above

A confidence interval is the mean of your estimate plus and minus the variation in that estimate. This is the range of values you expect your estimate to fall between if you redo your test, within a certain level of confidence

From the question, the parameters are;

The mean of the sample, = 9.4

The sample variance, s² = 0.49

The sample size, n = 19

The confidence level = 95%

Therefore, we get;

The degrees of freedom, df = n - 1

∴ df = 19 - 1 = 18

From the t-table, the critical-t = 2.101

Therefore, we have;

CI=9.4 +- 2.101 x [tex]\sqrt{0.49}[/tex]/[tex]\sqrt{19}[/tex]

Therefore, we have;

9.4 - 2.101×√(0.49/19) < μ < 9.4 + 2.101×√(0.49/19)

9.0626 < μ < 9.737

Hence, The 95% confidence interval for the mean ≈ (9.063, 9.737)

Learn more about at confidence interval at:https://brainly.com/question/15712887

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