Answer to Question #263384 in Statistics and Probability for samkelo

Question

Sales staff for GoodsDistributors submit weekly reports listing the customer contacts made during each week. A sample of 11 weekly reports showed a sample mean of 15.3 customer contacts per week. Assume that the population of contacts is N(μ, σ²) with a sample variance of 10.89 and determine a 90% two-sided confidence interval for the population mean. Do not round any of the values when used in the final calculation. Use a DECIMAL POINT in your answer.

• z or t score = Answer

 (rounded to three decimal places)

• standard error (rounded to three decimals) = Answer

• confidence limits (rounded to one decimal) = [Answer

, Answer

] customers

Answer 1

1. The critical value for $\alpha = 0.1$ and $df = n-1 = 10$  degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 1.812461.$

$t_c=1.812$

2.

$$SE=t_c\times\dfrac{s}{\sqrt{n}}=1.812461\times\dfrac{10.89}{\sqrt{11}}\approx5.951$$

standard error $=5.951$

3.

$$CI=(\bar{X}-SE, \bar{X}+SE)$$

$$=(15.3-5.951, 15.3+5.951)$$

$$=(9.3, 21.3)$$

confidence limits $(9.3, 21.3)$ customers.