Given a sample of variance 127 from a set of nine observations, construct a 95% confidence interval for the population variance

1) Given a sample of variance 127 from a set of nine observations, construct a 95% confidence interval for the population variance.

We know that under the null hypothesis,

~

So the 95% confidence interval for variance is given by,

( , )

Here, S2 = 127, n =9, α=0.05 thus,2.18, = 17.53

So here the 95% confidence interval for variance is,

( , ) = (57.957787, 466.055046)

2) An operations manager feels that the output rate of experienced employees is greater than that of new employees but he does not expect the variability in output rates to differ for the two groups. Previous studies indicate that the average unit output per hour for new employees at a particular type of work is 20 units per hour with a variance of 56 units squared. For a group of 20 employees with 5 years’ experience, the average output for this same type of work is 30 units per hour with a sample variance of 28 units squared. Does the variability in output appear to differ at the two experience levels? Test the hypotheses at the .05 significance level.

Here the hypothesis of interest is,

H0: against Ha:

Where population variance for old employees.

To test this hypothesis we need to use the test statistic. Using usual notation the test statistic to test the above hypothesis is given by,

~under the null hypothesis.

From the given data we have,

S2 =28, n=20.

Thus,

Critical values =

We reject the null hypothesis at 100α% significance level if,

0r

So we can see that the test statistic does not lie in the critical region thus we are failing to reject the null hypothesis at 5% significance level and concluding that at 5% significance level the variability in output does not appear to differ at the two experience levels.

3) A quality control manager for an automotive firm is concerned with uniformity in the number of defects in cars coming off the assembly line. If one assembly line has significantly more variability in the number of defects, then changes have to be made. The quality control manager has collected the following data:

	Number of Defects
	Assembly Line 1	Assembly Line 2
Mean	10	11
Variance	9	25
Sample Size	20	16

Does assembly line 2 have significantly more variability in the number of defects? Test at the .05 significance level.

Here the hypothesis of interest is,

H0: against Ha:

Where population variance for Assembly line 1 and population variance for Assembly line 2.

To test this hypothesis we need to use the F test statistic. Using usual notation the test statistic to test the above hypothesis is given by,

F = ~F(n1-1,n2-1) under the null hypothesis.

Since it’s a left tail test thus we reject the null hypothesis at 100α% significance level if,

Observed F <F(n1-1,n2-1,1-α).

Here from the given data we can see,

S12 = 9, S22 = 25,n1 = 20, n2 = 16.

Thus the value of the test statistic is,

F = 9/25 = 0.36.

Critical value = 0.4476

As, observed F = 0.36 < Critical value = 0.4476 so we are rejecting the null hypothesis at 5% significance level and concluding that the data gives enough evidence to conclude assembly line 2 have significantly more variability in the number of defects.

4) A recruitment firm subjects candidates to interviews by three different executives and then seeks to obtain a consensus evaluation for each candidate. Each executive gives the candidate a positive or a negative rating. The interview results of 100 candidates are as given below:

Possible Positive Ratings from Three Interviews	Number of Candidates Receiving Each of These Ratings
0	18
1	47
2	24
3	11

The recruitment manager feels that the interview process can be approximated by a binomial distribution with p = 0.40 (i.e. a 40% chance of any candidate receiving a positive rating on any one interview). Test this hypothesis at the .20 level of significance.

Here the hypothesis of interest is,

H0: The interview process can be approximated by a binomial distribution with p = 0.40,

Against,

Ha: H0 is not true.

Here the Vi-sq goodness of fit test is the appropriate test statistic to use.

Thus the test statistic is given by,

Chi-sq = ~Chi-sq(df) under the null hypothesis.

So we reject the null hypothesis at 100α% significance level if,

Observed test statistic > Critical value.

Here,

Df= number of groups-1 = 4-1 =3,

Significance level = 0.20,

Thus critical value = Chi-sq(3,0.2) = 4.642.

Using the given information and binomial pmf we have the following table,

Possible Positive Ratings from Three Interviews	Number of Candidates Receiving Each of These Ratings (Observed frequency)	Expected Frequency	(O-E)^2/E
0	18	21.6	0.600
1	47	43.2	0.334
2	24	28.8	0.800
3	11	6.4	3.306
Total	100	100	5.041

So from the table we can see that,

Test statistic = 5.041.

As test statistic > Critical value thus we are rejecting the null hypothesis at 20% significance level and concluding that the interview process cannot be approximated by a binomial distribution with p = 0.40.

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