Question

At a certain manufacturing plant, a machine produced ball bearings that should have a diameter of 0.79mm. If the machine produces ball bearings that are either too small or too large, the ball bearings must be scrapped. Every hour, a quality control manager takes a random sample of 31 ball bearings to test to see if the process is "out of control" (i.e. to test to see if the average diameter differs from 0.79 mm). Assume the population is normally distributed. Conduct an appropriate hypothesis test when your sample yields an average of 0.83 and a standard deviation of 0.055. Find the t-statistic and the appropriate conclusion at the 0.1 level of significance.

Answer 1

Answer:

The value of t test statistics is 4.05.

We conclude that the average diameter differs from 0.79 mm at the 0.1 level of significance.

Step-by-step explanation:

We are given that at a certain manufacturing plant, a machine produced ball bearings that should have a diameter of 0.79 mm.

A random sample of 31 ball bearings yields an average of 0.83 and a standard deviation of 0.055.

Let $\mu$ = average diameter of ball bearings.

So, Null Hypothesis, $H_0$ : $\mu$ = 0.79 mm     {means that the average diameter equals to 0.79 mm}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 0.79 mm     {means that the average diameter differs from 0.79 mm}

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

                        T.S. =  $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$  ~ $t_n_-_1$

where, $\bar X$ = sample average diameter = 0.83 mm

            s = sample standard deviation = 0.055 mm

            n = sample of ball bearings = 31

So, test statistics  =  $\frac{0.83-0.79}{\frac{0.055}{\sqrt{31} } }$  ~ $t_3_0$  

                               =  4.05

The value of t test statistics is 4.05.

Now, at 0.1 significance level the t table gives critical values of -1.697 and 1.697 at 30 degree of freedom for two-tailed test.

Since our test statistics does not lie within the range of critical values of t, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the average diameter differs from 0.79 mm.