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 hypothesis: H0: μ ≥ 45 hours and Ha: μ < 45 hours. If the sample mean amount of time is = 43.118 hours with the sample standard deviation s = 5.5 hours, give the appropriate conclusion, for α = 0.025.

We are given that 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.

A random sample of 25 parts will be selected and the average amount of time required to produce them will be determined. The sample mean amount of time is = 43.118 hours with the sample standard deviation s = 5.5 hours

Let [tex]\mu[/tex] = population mean amount of time required to produce an electrical part using new procedure

SO, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \geq[/tex] 45 hours {means that the new procedure will remain same or increase the population mean amount of time required to produce the part}

Alternate Hypothesis, [tex]H_a[/tex] : [tex]\mu[/tex] < 45 hours {means that the new procedure will decrease the population mean amount of time required to produce the part}

The test statistics that will be used here is One-sample t test statistics because we don't know about the population standard deviation;

T.S. = [tex]\frac{\bar X -\mu}{{\frac{s}{\sqrt{n} } } }[/tex] ~ [tex]t_n_-_1[/tex]

where, [tex]\mu[/tex] = sample mean amount of time = 43.118 hours

s = sample standard deviation = 5.5 hours

n = sample of parts = 25

So, test statistics = [tex]\frac{43.118-45}{{\frac{5.5}{\sqrt{25} } } }[/tex] ~ [tex]t_2_4[/tex]

= -1.711

Now at 0.025 significance level, the t table gives critical value of -2.064 at 24 degree of freedom for left-tailed test. Since our test statistics is more than the critical value of t so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the new procedure will remain same or increase the population mean amount of time required to produce the part.