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.

The 95% confidence interval for the mean is approximately;

(9.063, 9.737)

Step-by-step explanation:

The given 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;

Therefore, we have;

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

9.0626 < μ < 9.737

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

y=3n + 1

step-by-step explanation:

this is wrong because it says the second number is one less and i added one instead

0.32

step-by-step explanation:

2.24/7