Question

A large manufacturing plant has analyzed the amount of time required to produce an electrical part and determined that the times follow a normal distribution with mean time μ = 45 hours. The production manager has developed a new procedure for producing the part. He believes that the new procedure will decrease the population mean amount of time required to produce the part. After training a group of production line workers, a random sample of 25 parts will be selected and the average amount of time required to produce them will be determined. If the switch is made to the new procedure, the cost to implement the new procedure will be more than offset by the savings in manpower required to produce the parts. Use the hypotheses: H_o: μ ≥ 45 hours and H_a: μ < 45 hours. Determine the p-value of the test statistic if the sample mean amount of time is x̄ = 43.118 hours with the sample standard deviation s = 5.5 hours.

a. 0.95818

b. 0.04354

c. 0.04973

d. 0.04999

Answer 1

Given :

s = 5.5

xbar =43.118

n = 25

df = n - 1 = 25 -1 = 24

Calculate p value,

Step 1 calculate t test statistic

Formula:

t = (xbar - ) / s/ )

t = ( 43.118 - 45 ) / (5.5 /25)

t = -1.882 / 1.1

t = -1.711 ( Rounding 3 decimal)

Hence we get , t test statistic = -1.711

Step : 2 , Now, using excel command calculate p value

=Tdist( | t test statistic | , df , 1 )

=Tdist( 1.711 , 24 , 1 )

=0.04999

Hence we get p value = 0.04999

Answer : d) 0.04999