Solutions to Spring 2000 exam

Question 1

a.

A stem and leaf display can be used to sort the data and to observe the general shape of the data. In order to be able to use this display as a graphical display, it is imperative that the leaves appear directly underneath each other, and the stems have no gaps. A scale is also particularly important to provide information about the data in the display.

Scale: 5|6 = 56 cases

5 / 6
6 / 1 / 6 / 9
7 / 1 / 1 / 6 / 6 / 7
8 / 1 / 1 / 1 / 3 / 4 / 4 / 6 / 6 / 6 / 8 / 9
9 / 1 / 1 / 6 / 9
10 / 8
11 / 1 / 1 / 8
12
13 / 1 / 3

b. After examining the stem and leaf plot it appears that the number of cases purchased appears to be slightly positively skewed, as the display appears to have a slightly longer tail in the positive direction. This is due to the two weeks in which a larger than normal number of cases were purchased.

c.  (from calculator)

mode = 81 and 86 (ie bimodal)

d.  The mean (88.03) > the median (85). This is consistent with positively skewed data. Therefore it is likely that the distribution of the number of cases of salad dressing will be slightly positively skewed.

e.  Since both the stem and leaf display and the our answer to part d. are consistent with the data being positively skewed, the median would provide the best prediction for the number of cases of salad dressing to order for next week. The mean would not provide the best estimate since when the data are positively skewed, the mean would overestimate the the required number.

## Question 2

a. Let C = event a customer will buy a computer

Let S = event a customer will buy software

i.

ii.

b. The distribution of the number of doctors that recommend a certain product follows a binomial distribution where:

n = 20

## X = no. doctors that recommend the product = 0, 1, 2, 3, . . . , 20

p = probability a doctor recommends the product = 0.4

i.

Therefore the probability only two doctors recommend the product is 0.003, ie very small.

ii. ‘two or fewer’ means 0, 1, 2 ie

Therefore the probability two or fewer doctors recommend the product is 0.004. Also very small.

c. Let X = final grades

i.

We cannot attempt to find this probability without first drawing a diagram to represent the problem.

-1.5 0 z

Therefore 6.68% of students failed.

ii.

0 0.58 z

Therefore the probability a randomly selected student scores a credit or better is 0.2810.

iii.  The best place to start when solving problems of this type, is to first draw a diagram.

0.4 0.1

Therefore, the cut-off for an HD is 1.28 standard deviations above the mean.

Therefore the cut-off grade for an HD is 83.36% or 84%

Question 3.

a.

Therefore the department should sample 139 workers.

Note: We always round up to ensure the required precision and level of confidence are achieved.

b.

i.

Since we do not know s, if we assume that heights are normally distributed, we can use

Therefore the population average height of male QBM117 students is between 177.9 cm and 182.1 cm.

ii.

Step 1

Step 2

Step 3

0.05

Step 4

Reject if t > 1.711

Step 5

Step 6

Since 5 < 1.711 we reject .

There is sufficient evidence, at a = 0.05, to conclude that the average height of males in QBM117 has increased.

iii.

Question 4.

a. Step 1

Step 2

Therefore will be approximately normally distributed and hence

test statistic

Step 3

0.05

Step 4

Reject if z > 1.645

Step 5

Step 6

Since 1.33 < 1.645 we do not reject .

There is insufficient evidence, at a = 0.05, to conclude that more than 10% of TV sets need repair ie we cannot dispute the claim.

b. i.

ii.

For each extra year of experience an employee has, the salary will increase by an average of \$283.26.

iii.

Reject if p-value < 0.05

Since we reject .

Therefore there is sufficient evidence at a = 0.05 to conclude that salary is linearly related to years of experience.

We can perform the same hypothesis test using the t statistics instead of the p-value. The solutions using this method follows. Only one of the other is necessary when testing these hypotheses.

Step 1

Step 2

test statistic

Step 3

Step 4

Reject if

Step 5

Step 6

Since 9.342 > 2.009 we reject .

Therefore there is sufficient evidence at a = 0.05 to conclude that salary is linearly related to years of experience

iv.  The histogram of residuals resembles an approximate bell shape. Therefore the error variable appears to be normally distributed.

In the residual plot there appears to be a random scatter in the residuals. There is no apparent relationship or pattern between residuals and years of experience (x).

The residual plot shows no evidence of non-constant variance. The residuals do not appear to be increasing or decreasing for increasing values of x.

Therefore, we can conclude that the linear model is appropriate.

#### Part B

1. E

Sales figures would be generally be monetary values in thousands or millions, depending on the size of the company. Therefore these values and quantitative (numerical) and have an absolute zero ie \$0 means the company sold nothing.

2. D

A.  would not be appropriate as there are too many classes and much of the information about the distribution of the data may be lost;

B.  would not be appropriate since the classes are overlapping;

C.  would not be appropriate since the minimum value would not be included in any of the classes chosen;

E.  would not be appropriate as there are too few classes and much of the information about the distribution may be lost;

3. B

It is clear that if the salaries increase by 4%, the variability of the salaries will increase. To verify that the standard deviation increases by 4%, choose a small data set. Increase each value by 4% and then compare the standard deviations.

4. A

The vertical scale in an ogive has a maximum value of 1 (or 100%). Consequently, any percentile could be very easily determined from this graphical display (refer Australian Business Statistics fig2.8 p28).

5. E

This histogram has two obvious peaks, therefore the distribution shape is bimodal (refer Australian Business Statistics fig2.6 p27).

6. C

Let U = event the film has been used

= event the film has not been used

7. A

z 0 2.3

8. E

Ask yourself the following questions when trying to determine what type of probability distribution we have here.

Does the question say that the number of cars arriving at the service station follows a normal distribution? No

Does the question include information on the mean and standard deviation number of cars arriving at the service station? No

Therefore the number of cars arriving at the service station is not normally distributed.

Does the question provide an average rate ie the average number of cars arriving within a given time? Yes

On average 10 cars arrive every ten minutes……

Therefore the number of cars arriving at the service station is Poisson distributed where:

X = no. cars arriving = 0, 1, 2, 3, . . .

m = 10 cars per 10 minutes

Wherever possible, Poisson probabilites should be determined from the Poisson tables in the Appendix at the rear of the text. We use Excel for determining Poisson probabilities not found in these tables.

### .

9. E

Before finding the required probability we must first recalculate the average number of cars arriving. Since the interval of time has changed from per 10 minutes to per 5 minutes, we must adjust the average number of cars accordingly.

ie

### 'fewer than 5' means '4 or less' ie

10. D

A.  is not a random sample and may lead to biased results.

B.  would definitely not give a reliable estimate since all the items selected are defective.

C.  would only provide a sample size of 2.

E.  would overestimate the proportion of defectives since all defective transmission have been included in the sample.

11. B

Only the standard normal distribution has a mean of zero and a standard deviation of 1.

12. B

This is because the size of the rejection region increases from 0.05 to 0.1 as the level of significance increases from 5% to 10%. Therefore there is an increased probability that the test statistic falls in the rejection region.

13. E

Since , then will not be normally distributed.

14. C

Since we will reject the null hypothesis if our sample mean is significantly bigger than 100.

15.  B

## Since s is known and n > 30, the appropriate form of the test statistic is

16.  B

An alternative form of the decision rule for a hypothesis test is

Reject if p - value < a ie if p - value < 0.05

Since 0.15 > 0.05 we do reject .

17.  E

18.  C

19.  A

As read directly from the Excel output.

20.  B

Critical t would be

7