## Question 648:

1## Answer:

No answer provided yet.Q1:For this test we'll be conducting a 2-sample t-test.

- The null hypothesis is there is no difference in the mean increase between growth and income mutual funds. The alternative hypothesis is there IS a difference in the mean increase between growth and income mutual funds (one is greater than the other).
- The decision rule is to reject the Null hypothesis if the test statistic exceeds the critical value. The critical value for a 2-tailed t-test for a significance level of .01 can be found using a t-table of critical value with 73 degrees of freedom (n1 + n2-2). We obtain the value 2.65. You can also lookup this value using the Inverse t distribution calculator http://www.usablestats.com/calcs/tinv.
- The test statistic is calculated as the difference between the two means (1100-10900 = 10, divided by the standard deviation of the difference (SE). The SE is computed as the square root of var1/n1 + var2/n2. Recall that the variance is just the square of the standard deviation, which we were given. This gets us the equation SQRT(2025/35 +3025/40) = 11.553. This gets us a test statistic of 10/11.553 = .86554
- The p-value can be found using the excel function =TDIST(.86554,72,2), where the parameters are the test statistic, the degrees of freedom and 1 or 2 sided test. You can also lookup these values using the Percentiles from the t-distribution calculator. http://www.usablestats.com/calcs/tdist We get the p-value of .3896
- Since the test statistic is below the critical value we would NOT reject the null hypothesis. We cannot conclude given this data that there is a difference between the average increase between funds.

2. Is the mean salary of accountants who have reached partnership status higher than that for accountants who are not partners? A sample of 15 accountants who have the partnership status showed a mean salary of $82,000 with a standard deviation of $5,500. A sample of 12 accountants who were not partners showed a mean of $78,000 with a standard deviation of $6,500. At the 0.05 significance level can we conclude that accountants at the partnership level earn larger salaries?

Q2:

We would conduct a 1-directional 2-sample t-test.

- The null hypothesis is that the mean partner salary is equal to or less than the non-partner accountants salary. The alternative hypothesis is the mean salary for partner IS greater than non-partner.
- The decision rule is to reject the Null hypothesis if the test statistic exceeds the critical value. The critical value for a 1-tailed t-test for a significance level of .05 can be found using a t-table of critical values with 25 degrees of freedom (n1 + n2-2). We obtain the value 1.7081. You can also lookup this value using the Inverse t distribution calculator http://www.usablestats.com/calcs/tinv (select 1-sided area).
- The test statistic is calculated as the difference between the two means (82000-78000 = 4000, divided by the standard deviation of the difference (SE). The SE is computed as the square root of (var1/n1 + var2/n2). Recall that the variance is just the square of the standard deviation, which we were given. This gets us the equation SQRT(30250000/15 +42250000/40) = 2353.189. This gets us a test statistic of 4000/2353.189= 1.699.
- The p-value can be found using the excel function =TDIST(1.699,25,1), where the parameters are the test statistic, the degrees of freedom and 1 or 2 sided test. You can also lookup these values using the Percentiles from the t-distribution calculator. http://www.usablestats.com/calcs/tdist We get the p-value of .5197.
- Since the test statistic is just slightly below the critical value we would NOT reject the null hypothesis. We cannot conclude given this data that there is a difference between the average salaries of partner or non-partner accountants.

An important note on these computations. We assumed that the population variances are not equal. An alternative approach is to assume the variances ARE equal. The only difference to the computations above is to use a pooled estimate of the standard deviation in computing the Standard Error of the Mean. Usually there isn't a difference when we assume or don't assume equal variances. However, in this case, when we assume equal variances we get the test statistic of 1.73, which exceeds the critical value. The p-value associated with this is .04774 which is less than .05. That would led to rejecting the null hypothesis and concluding that partner accountants have a higher mean salary than non-partner accounts. Which is correct: reject or not reject? It depends on the assumptions, and we weren't told to assume equal or unequal variances. The more conservative choice is to choose unequal variances (which is what I did). This also illustrates why using hard rules of reject and not-reject, especially when the p-value hovers between .04 and .06 is a bit arbitrary. But that is a discussion for elsewhere. Usually these questions are written for you to think through things, so if you were exposed to equal vs. unequal variances in your course, I suggest you change the test statistic to 1.73, the p-value to .04774 and the decision to reject the Null. In the end, you can make a case for either approach.

Q3:

We would conduct a 2-proportion test here.

- The Null hypothesis is that the absenteeism rate for afternoon employees is equal to or less than the absenteeism rate of day shift employees. The alternative hypothesis is the absenteeism rate for afternoon employees is higher than day shift employees.
- We will reject the Null hypothesis is the test statistic exceeds the critical value (which is also reflected in a p-value being less than .01).

The samples are large enough that we can use the Normal Approximation to the Binomial to answer the difference between two proportions (the check is if np & nq > 5 which are 15 and 62 for the lowest values so we're fine). We want to know if the difference in proportions (15/120 – 18/80) = .10 is greater than what we'd expect to see from chance alone.

We divide this difference by the square-root of a denominator which accounts for the chance which will provide you with a z-score and becomes the test statistic. The denominator is the square root of:

1 1

-- + -- * PQ

n_{1} n_{2}

Where P = (x1 + x2)/(n1+n2) Q is 1-P.

P = (15+18)/(120+80) = .165

Q = 1-.165 = .835

PQ = .165*.835 = .1377

1/n1 + 1/n2 = 1/120 + 1/80 = .020833

So multiply .1377 * .020833= 0.00287

Now the square root of this is SQRT(0.00287) = .053575

So the equation is the observed difference .10/.053575= 1.866 which is the test-statistic.

d. You now look this value up using a table of normal values or the z-score to percentile calculator using the 1-sided area. You should get 3.098 % or a p-value of .0309.

e. Since this is above .01 and the test statistic is above the critical value of 2.32 we do not reject the Null hypothesis and cannot conclude the absenteeism rate is higher for afternoon workers.

Q4:

For this test we will conduct a 1-Way ANOVA since we have more than 2 means.

- The null hypothesis is that there is no difference between any of the mean company response times. The alternative hypothesis is that at least one mean time is different.
- The decision rule is to reject the NULL hypothesis if the test statistic exceeds the critical value (and the p-value is less than .05).
- The test statistic F is calculated as the MS Treatment (Mean Squares) divided by the MS Error. This ratio computes the variability between groups with the variability within each group. Ideally we want most variability to occur between the groups. While the computations are best left for software, the F-statistic is 1.48125/.26 = 5.77. The p-value associated with an F of 5.77 on 2 and 13 degrees of freedom is .016.
- Since the p-value is less than .05 we would reject the null hypothesis. We would conclude that there is at least one difference in the mean response times between fire companies.