Question

A carpet company advertises that it will deliver your carpet within 15 days of purchase. A sample of 49 past customers is taken. The average delivery time in the sample was 16.2 days. Assume the population standard deviation is known to be 5.6 days. a. Using a critical value, test to determine if their advertisement is legitimate. Let alpha = .0392. b. What s the type II error if the actual mean of carpet delivery is mean 17? show all working

Answer 1

a)

TS = (Xbar - mu)/(sd/sqrt(n))

= (16.2 - 15)/(5.6/Sqrt(49))

= 1.5

critical value at alpha = 0.0392

is 2.0621

since TS < critical value

we fail to reject the null hypothesis

Their advertisement is legitimate

b)

type ii error = fail to reject the null when null is not true

P( |xbar - 15|/((5.6/Sqrt(49)) < 2.0621 | mu = 17)

= P( 15 - 2.0621 * (5.6/Sqrt(49)) < Xbar < 15 + 2.0621 *(5.6/Sqrt(49)) |mu =17)

= P( (15 - 2.0621 * (5.6/Sqrt(49)) - 17)/ (5.6/Sqrt(49))   < Z < (15 + 2.0621 * (5.6/Sqrt(49)) -17) / (5.6/Sqrt(49)))

= P( -2.0621 - 2/( (5.6/Sqrt(49))) < Z < 2.0621 -2/( (5.6/Sqrt(49)))

= P( -4.5621 < Z< -0.4379 )

= 0.3307