Answers to End-Of-Chapter Exercises

Chapter 2

ANSWERS TO END-OF-CHAPTER EXERCISES

1. The null hypothesis states that mean sales, per representative are $24,000. The alternative hypothesis is that mean sales are not $24,000.

Since the sample standard deviation is reported, the appropriate test statistic follows a

t distribution:

However, because the sample size of 100 is so large, we can apply the standard normal table. Since the alternative is one of inequality, with level of significance being .05, we need the area making up two and one-half percent of the distribution in each tail. From the Standard Normal Table we see that the probability that a Z-variable exceeds 1.96 is .025. Accordingly, our rejection region is for values of the test statistic that lie either above 1.96 or below -1.96.

Using the sample mean, sample standard deviation, and sample size, the calculated value of the test statistic is:

Since this lies outside the rejection region, we fail to reject the null at the 95% level of confidence and conclude that the reported mean level of sales is statistically indistinguishable from $24,000.

2. State the null hypothesis as the inventory level is the same as the industry average, and the alternative as one of inequality:

Since both the population mean and variance are unknown, the appropriate test statistic is based on the t-distribution:

The critical region for a 5% level of significance with sample size 120 for a two-tailed alternative is given by:

Given that n = 120, = 310, and s = 72; the absolute value of the test statistic under the null is:

Since 2.28 > 1.96, we can reject the null and conclude the manufacturers mean tire inventory is significantly different from the industry norm.

3.a) The mean is the arithmetic average of all the numbers in the data set. The median is the value that splits the respondents into two equal parts when they are arrayed from the lowest to the highest value. The mode is the value that occurs most frequently.

Using FORECASTXTM, the following descriptive statistics were calculated:

Statistics for Solution
Credit Hours
Mean / 8.3
Median / 9
Mode / 9
Standard Deviation / 2.6378
Sample Variance / 6.9579
Range / 11
Minimum / 2
Maximum / 13
Standard Error / 0.5898

b) Since the claim is that business graduate students take fewer credit hours than the average graduate student, we state the null such that its rejection merits the claim. Specifically, the null states that the mean number of credit hours taken by the business graduate student is equal to or larger than the mean number of credit hours taken by typical graduate students. We state the alternative as mean credit hours taken is less than the mean credit hours taken by a typical graduate student:

Since both the population mean and variance are unknown, the appropriate test statistic is based on the t-distribution.

The critical region for a one-tailed alternative at the 5% level of significance with sample size 20 is given by:

Given that n = 20, = 8.3, and s = 2.638, we can calculate the sample value of the test statistic under the null hypothesis:

Since 1.356 < 1.725, we fail to reject the null hypothesis and conclude that business graduate students do not typically take less credits per quarter than non-business graduate students.

4.The issue is whether or not ACC’s sales staff is comparable to those of other producers in the same industry. Accordingly, state the null as ACC’s mean sales per salesperson equals that of other producers, with the alternative of inequality:

The following statistics are used to test this assertion.

Statistics for the Solution
Sales
Mean / 219,017.5625
Median / 180,142.0000
Mode / None
Standard Deviation / 76,621.2783
Sample Variance / 5,870,820,285.1958
Standard Error / 19,155.3196
Range / 229,726.0000
Minimum / 110,027.0000
Maximum / 339,753.0000

Since both the population mean and variance are unknown, the appropriate test statistic is based on the t-distribution:

The critical region for a 5% level of significance, when the sample size is 16 under a two-tailed alternative, is given by:

Given that n = 16, = 219,018, and s = 76621.3, we can calculate the sample value of the test statistic under the null hypothesis:

Since 1.878 < 2.131, we cannot reject the null and conclude that ACC’s sales staff are comparable to that of other producers.

5.a) Since the mean is 205 pounds, and the normal distribution is symmetric about the mean, half of the population should lie above the mean. Therefore 50 percent of the players would weigh more than 205 pounds.

b) To make statistical inference from the probability distribution of football player weights, we use the standard normal distribution with the appropriate transformation. The appropriate Z value for 250 pounds is:

Z = (250 - 205)/30 = 1.5.

Accordingly, what percent of players weigh less than 250 pounds is the same as asking:

P(Z < 1.5) = P[Z ≤ 0] + P[0 ≤ Z < 1.5] = .5000 + .4332 = .9332.

Using the relative frequency interpretation of probabilities this would imply that 93.3% of the players would weigh less than 250 pounds.

c) From Table 2-4: Z.10 = -1.285. Accordingly, 90% of the area under the standard normal density lies above the Z-value of –1.285:

P[Z > -1.285] = [(X - 205)/30 > -1.285] = .90

P[X - 205 > -38.55] = P[X > 166.45] = .90.

Therefore 90% of the players would weigh more than 166.45 pounds.

d) P[180 ≤ X ≤ 230] = P[180 - 205 ≤ X - 205 ≤ 230 - 205]

= P[-25 ≤ X - 205 ≤ 25] = P[-25/30 ≤ Z ≤ 25/30] = P[-.8333 ≤ Z ≤ .8333]

= P[-.8333 ≤ Z ≤ 0] + P[0 ≤ Z ≤ .8333] = .2967 + .2967 = .5934

Therefore, 59.34% of the players would weigh between 80 and 230 pounds.

6.a) Ms. Wharton’s hope is that her bank is viewed more favorably than the average bank, which has an approval rating of 7.01. We formulate the null such that its rejection merits Ms. Wharton’s hopes:

Since data were derived from a market research survey, we view both the population mean and variance as having been estimated. Accordingly, the appropriate test statistic is based upon the t-distribution:

To find the calculated value of the test statistic based upon this sample, we note the following information: Sample Mean = 7.25, 0 = 7.01, s = 2.51, and n = 400.

The critical region for significance level .05 and sample size 400, under a one-tailed alternative, is given by:

P[t399 > 1.645] = .05.

Since our calculated t-value falls into the critical region, we can reject the null hypothesis at the 5% level of significance.

c) The calculated value of our t-statistic would now be:

Hence, we cannot reject the null with this smaller sample, other things held constant.

Why? The key is to examine the variance of the sampling distribution of the sample mean, which depends on the sample size. As sample size increases, we can be more confident in our estimate of the population mean.

7.a) Using FORECASTXTM, the following descriptive statistics about class size were obtained.

Statistics for Solution
Number of students
Mean / 40.1600
Median / 42.0000
Mode / 20.0000
Standard Deviation / 13.1424
Sample Variance / 172.7233
Standard Error / 2.6285
Range / 54.0000
Minimum / 10.0000
Maximum / 64.0000

b) The standard error of the sample mean is 2/n. Estimating the population variance by the sample variance and noting the sample size, the standard error of the estimated sample mean is (172.72/25)1/2 = 2.628.

c) The sampling distribution of the sample mean shows that the sample mean is an unbiased estimator of the population mean. Accordingly, 40.16 is our point estimate for the population class size.

d) Since the population mean and variance are unknown, we use the following confidence interval involving the t-distribution to make probability statements about the unknown population mean:

For a 95 percent confidence interval, using t24,.025 = 2.064, we get:

P[34.74 <  < 45.58] = .95.

For 90% confidence interval, using t24,.05 = 1.711, we get:

P[35.66 <  < 44.66] = .90.

The 95% confidence interval is a wider since we are statistically more confident.

8.a) To examine whether there has been an upward trend in annual larceny thefts in the United States, a time-series plot of annual data from 1972 through 1994 was prepared.

As shown in the time-series plot, there is a positive trend over the sample period.

b) The ACF and PACF estimates and correlograms for THEFTS are reported below.

rk = 2/(sqrt n) so in this case rk = 2/(sqrt 23) = .417.
ACF Values For Larceny Thefts
Obs / ACF /
/ Upper Limit / Lower Limit
1 / .7986 / .4087 / -.4087
2 / .5594 / .4087 / -.4087
3 / .4045 / .4087 / -.4087
4 / .3298 / .4087 / -.4087
5 / .2844 / .4087 / -.4087

To test the significance of individual autocorrelation parameters, we use the following 95% confidence approximation, which states that the critical value for rk is: where n is the sample size. For estimated autocorrelation coefficients in excess of rk in absolute value, we reject the null hypothesis of zero autocorrelation assuming a two-tailed alternative at the .05 level of significance.

Since sample size is 23, the appropriate critical value for testing the null of zero autocorrelation is .417. Examining the autocorrelation function, we see that we can reject the null of zero autocorrelation at lags 1, and 2, since their coefficient estimates exceed .417. Autocorrelations at lags 3 throgh 5 are not statistically different from zero. Accordingly, because of trend the data are nonstationary.

c) To account for trend, Holt’s exponential smoothing or a regression trend model should be considered.

9.A time-series plot of mobile home shipments (MHS) is shown below.

As indicated by the time-series plot, there is significant seasonality to mobile home sales as shown by the regular periodic variation in the data arising at the same time each year. Specifically, the plot shows seasonal downturns in quarters four and one compared to quarters two and three. The plot also reveals periods of upward trends followed by downward trends, presumably related to business cycle factors such as interest rates and unemployment.

The autocorrelation structure of MHS and correlogram is reported below.

Obs / ACF / PACF
1 / .7779 / .7779
2 / .5593 / -.1161
3 / .6165 / .5678
4 / .6911 / .0537
5 / .4623 / -.5250
6 / .2359 / -.0206
7 / .2494 / -.0345
8 / .3154 / .1172
9 / .1103 / -.2919
10 / -.1118 / -.0385
11 / -.0954 / .0099
12 / -.0136 / -.0064

Using the 95% approximation for testing the null of zero autocorrelation at lag k, we can reject the null if the estimated autocorrelation coefficient exceeds

With our sample size of 60, the critical value for rk is .258199. Examining the estimated autocorrelations in the table above show that we can reject the null of zero autocorrelation at lags 1, 2, 3, 4, 5, and 8; it is not until lag six that the autocorrelation function falls below .258199 (r6 = .235934). Accordingly, the autocorrelation results indicate a significant trend in the data.

In addition, the relatively large autocorrelation coefficients for lags of 4 and 8 quarters (r4 = .691087 and r8 = .315445) indicate significant seasonality in the MHS series.

Forecasting methods that might be suggested as good candidates for MHS, based on Table 2-1 in the text would include Winters’ exponential smoothing, time series decomposition, and a causal regression model.

10.a) Private housing starts data (PHS) are plotted below.

The time-series plot of private housing starts (PHS) shows no significant trend and significant seasonality.

b)Using FORECASTXTM, the estimated autocorrelation coefficients and correlogram for PHS are reported below.

ACF Values For Total Houses Sold (000) Per Quarter
Obs / ACF /
/ Upper Limit / Lower Limit
1 / .8258 / .2450 / -.2450
2 / .6887 / .2450 / -.2450
3 / .7155 / .2450 / -.2450
4 / .7613 / .2450 / -.2450
5 / .6044 / .2450 / -.2450
6 / .4660 / .2450 / -.2450
7 / .4898 / .2450 / -.2450
8 / .5411 / .2450 / -.2450

With our sample size of 64, the critical value for rk is .25. Examining the estimated autocorrelations in the table above show that we can reject the null of zero autocorrelation at lags 1throgh 8 using the approximate 95% confidence rule.

The autocorrelation results indicate trend in the data. On the other hand, the relatively large autocorrelation coefficients for lags of 4 and 8 quarters (r4 = . 76 and r8 = . 54) indicate significant seasonality in the PHS series.

c) Using FORECASTXTM, the estimated autocorrelation coefficients and correlograms of the first-differenced series (DPHS) are reported below (see next page).

Since we lose a data point in the first-differenced series relative to the original, our critical value of rk is now .252. Accordingly, we can reject the null for lags 2, 4, 6, 8, 10, and 12. These large autocorrelations suggests the PHS data is highly seasonal which continues into the first-differenced series.

ACF @ PACF - 12282005 Values For DTHS
Obs / ACF / Upper Limit
/ Lower Limit
1 / -.1475 / .2469 / -.2469
2 / -.5529 / .2469 / -.2469
3 / -.0726 / .2469 / -.2469
4 / .7088 / .2469 / -.2469
5 / -.1158 / .2469 / -.2469
6 / -.5272 / .2469 / -.2469
7 / -.0685 / .2469 / -.2469
8 / .6052 / .2469 / -.2469
9 / -.0591 / .2469 / -.2469
10 / -.4676 / .2469 / -.2469
11 / -.1032 / .2469 / -.2469
12 / .5803 / .2469 / -.2469

11.A time-series plot of the Japanese-yen U.S.-dollar exchange rate (EXRJ) is shown below.

As shown by the time-series plot, there appears to be no significant trend or seasonality in exchange-rate data.

The autocorrelation structure and correlograms for EXRJ are reported below:

Obs / ACF / PACF
1 / .8157 / .8157
2 / .5383 / -.3797
3 / .2733 / -.0798
4 / .0340 / -.1550
5 / -.1214 / .0408
6 / -.1924 / -.0112
7 / -.2157 / -.0537
8 / -.1978 / -.0036
9 / -.1215 / .1120
10 / -.1217 / -.3281
11 / -.1823 / -.1047
12 / -.2593 / -.1248

The approximate 95% critical value for rejecting the null of zero autocorrelation at lag k with a sample size of 24 is .408. Since the autocorrelation coefficients fall to below the critical value after just two periods, we can conclude that there is no trend in the data.

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