A large manufacturing plant has analyzed the amount of time required to produce an electrical part and determined that the times

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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.

Mathematics

1 answer:

Inessa [12.5K] · Answer 1 · 2026-04-02T20:39:09+03:00

Answer:

We determine that the new procedure will not result in a decrease in the average time required to produce the part.

Step-by-step explanation:

A large manufacturing facility has examined the time taken to create an electrical component and confirmed that the times conform to a normal distribution with a mean of μ = 45 hours.

A random sample of 25 parts will be taken, and the average production time will be calculated. The mean time calculated for the sample is = 43.118 hours, and the sample standard deviation is s = 5.5 hours

Let $\mu$ denote the average time required to produce the electrical part using the new method

Therefore, Null Hypothesis, $H_0$ : $\mu \geq$ 45 hours {indicating the new procedure will not change or will increase the average time needed for production}

Alternative Hypothesis, $H_a$ : $\mu$ < 45 hours {implying the new procedure will reduce the average time required to produce the part}

For this situation, we apply the One-sample t test statistic since the population standard deviation is unknown; T.S. = $\frac{\bar X -\mu}{{\frac{s}{\sqrt{n} } } }$ ~ $t_n_-_1$

where, $\mu$ = sample mean amount of time = 43.118 hours

s = sample standard deviation = 5.5 hours

n = sample size of parts = 25

Thus, test statistics = $\frac{43.118-45}{{\frac{5.5}{\sqrt{25} } } }$ ~ $t_2_4$

= -1.711

At the 0.025 significance level, the t-table indicates a critical value of -2.064 at 24 degrees of freedom for a left-tailed test. Since our test statistic exceeds the critical value of t, we lack sufficient evidence to reject the null hypothesis as it does not fall within the rejection area.

Consequently, we conclude that the new procedure will either maintain or extend the average time required to produce the part.