Question

When it is operating properly, a chemical plant has a daily production rate that is normally distributed with a mean of 885 tons/day and a standard deviation of 42 tons/day. During an analysis of period, the output is measured with random sampling on 60 consecutive days, and the mean output is found to be x=875 tons/day. The manager claims that at least 95 % probability that the plant is operating properly. Is he right? Justify your answer!

Answer 1

Answer:

The test statistic Z = 1.844 < 1.96 at 0.05 level of significance

Null hypothesis is accepted

Yes he is right

The manager claims that at least 95 % probability that the plant is operating properly

Step-by-step explanation:

Explanation:-

Given data Population mean

                                           μ     = 885 tons /day

Given random sample size

                                          n = 60

mean of the sample

                                        x⁻  = 875 tons/day

The standard deviation of the Population

                                      σ = 42 tons/day

Null hypothesis:- H₀: The manager claims that at least 95 % probability that the plant is operating properly

Alternative Hypothesis :H₁: The manager do not claims that at least 95 % probability that the plant is operating properly

Level of significance =  0.05

The test statistic

 $Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }$

$Z = \frac{875 -885}{\frac{42}{\sqrt{60} } }$

$Z = \frac{-10}{5.422} = -1.844$

|Z| = |-1.844| = 1.844

The tabulated value

                         $Z_{\frac{0.05}{2} } = Z_{0.025} = 1.96$

The calculated value Z = 1.844 < 1.96 at 0.05 level of significance

Null hypothesis is accepted

Conclusion:-

The manager claims that at least 95 % probability that the plant is operating properly