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BUSI1074 Quantitative Methods 1b – A.A. 2021/22

Worksheet 4 Solutions

Question 1. The bell curve (Smith, 2022)

Suppose that a maths class is made up of an equal number of Blue and Green Students. Within each group, marks are distributed normally, but blues are better at maths, with a mean of 60 compared to 55 for green students. Blue students are also less erratic, with a standard deviation of 5 compared to a standard deviation of 10 for green students.

(a) What proportion of blue students get more than 70?

(b) What proportion of green students get more than 70?

Answer:

(a)

We want to find the probability that the mark is over 70. For Blue students

Z = = 2

so the probability of a mark over 70 is P(Z > 2) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228 or 2.28%. The 0.9772 came from the table of areas under a normal distribution.

(b)

For Green Students

Z = = 1.5

so the probability of a mark over 70 is P(Z > 1:5) = 1 - P(Z < 1:5) = 1 - 0.9332 = 0.0668 or 6.68%. The 0.9332 came from the table of areas under a normal distribution.

(c)

In a large class with equal number of blue and green students, 4.48% of all students, (2.28+6.68)/2, would get over 70. The proportion of those that are blue is 25% (=2.28/(2.28+6.68)), the proportion that are green is 75% (=6.68/(2.28+6.68)).

Even though the question says blues are better at maths and it is true that their average is higher, three-quarters of the top group in maths are green (as are three-quarters of the bottom group). The lesson is to think about the whole distribution, not just the averages or parts of the distribution (e.g. the top of the class), and try not to be influenced by value-laden descriptions: ‘better’ or ‘less erratic’ .

Question 2. The square root rules (Page, 2018)

Howard Wainer refers to the formula for the standard error (of the average) the “most dangerous equation in the world.” For example, in the 1990s the Gates Foundation and other nonprofits advocated breaking up schools into smaller schools based on evidence that the best schools were small (Wainer, 2009). To see the flawed reasoning, imagine that schools come in two sizes—small schools with 100 students and large schools with 1,600 students—and that student scores at both types of schools are drawn from the same distribution with a mean score of 100 and a standard deviation of 80.

(a) What is the standard error for small schools?

(b) What is the standard error for large schools?

(c) If we assign the label “high-performing” to schools with means above 110 and the label “exceptional” to schools with means above 120, work out the probabilities that the small and large schools will meet either threshold. What do you learn?

Answer:

(a)

At small schools, the standard error equals 8 (the standard deviation of the student scores, 80, divided by 10, the square root of the number of students).

(b)

At large schools, the standard deviation of the mean equals 2.

(c)

For the small schools, an average score of 110 is 1.25 standard deviations above the mean; such events occur about 10% of the time. A mean score of 120 is 2.5 standard deviations above the mean; an event of that size should occur about once in 150 schools.

When we do these same calculations for large schools, we find that the “high-performing” threshold lies five standard deviations above the mean, and the “exceptional” threshold lies ten standard deviations above the mean. Such events would, in practice, never occur.

Only small schools will meet either threshold.

Thus, the fact that the very best schools are small is not evidence that smaller schools perform better. The very best schools will be small even if size has no effect solely because of the square root rules.

Question 3. Hypothesis test (interpret the p-value and calculate the power of the test) Suppose a lightbulb manufacturing plant produces bulbs with a mean life of 1000 hours and a standard deviation of 100 hours. An inventor claims to have developed an improved process that produces bulbs with a longer mean life and the same standard deviation.

(a) The plant manager randomly selects 50 bulbs produced by the improved process; the average sample life of bulbs is 1033 hours. Use the p-value approach to test the inventor’s claim at a 1% significance level.

(b) Suppose the new bulb is, in fact, better and has a mean bulb life of 1042 hours. Find the Type II error. Interpret this value.

(d) What level of sample mean life of the bulbs should the plant manager use to reject the null hypothesis if she wants the significance of her test to be 5%? What happens to Type II error and the power of the test? What could be a good way to increase the power of the test?

Answer:

(a)

H0 : u = 1000

H1 : u > 1000 (2 marks)

The p-value of the test is

Pr( > 1033|u = 1000)

= Pr ( > )

= Pr(z > 2.33)

= 1 − Pr(z < 2.33)

= 1 − 0.9901

= 0.0099 (6 marks)

The p-value of the test is 0.99%. Therefore, we reject the null hypothesis at a 1% level of significance.

If the null hypothesis is true, the likelihood of obtaining a test statistic more extreme than the observed sample value is 0.99% which is very small. If we tolerate 1% of Type I error, we would reject the null hypothesis.

The measure of the fit of the hypothesis being tested to the data is 0.99%; so we can reject the hypothesis at 1% significance level. (2 marks)

(b) Based on the definition of F, one needs to find out the probability of the non-rejection region conditional to the true value of the population mean. We know that C = µ +

za =0.01 = 1000 + 2.3263 ∙ 14. 1421 = 1032.8989, where µ = 1000 to indicate that this

computation is under the null hypothesis µ = 1000.

So:

F = P( ≤ 1032.8989|u∗ = 1042) = P (Z ≤ )= P(Z ≤ −0.64)

= P(Z > 0.64) = 1 − P(Z ≤ 0.64) = 1 − 0.7389 = 0.2611 (6 marks)

If the new bulb has a mean bulb life of 1042 hours, the probability of accepting the null hypothesis when it is false is 26. 11%. That is, there is a 26. 11% probability of failing to reject a false null hypothesis. (2 marks)

Thus the probability of rejecting the null hypothesis when indeed it is false is 73.89%.(2 marks)

(d) For a test of 5%, the rejection region for the null hypothesis contains those values of the

test statistic exceeding 1.6449:

> 1000 + 1.6449 × 14. 14 = 1023.26(3 marks)

The manager should believe the inventor’s claim if the sample mean life of the new product is greater than 1023.26 hours if she is willing to tolerate 5% chance of Type I error. (1 mark)

A higher Type I error (5% vs. 1%) implies a lower Type II error, thus a higher power of the test. (1 mark)

One way to raise the power of the test would be to increase the sample size. This would reduce the margin of error and C , where we do not reject the null hypothesis. Therefore, the area of Type II error would fall, raising the power of the test. Using a large enough sample to have sufficient power to detect a plausible alternative hypothesis is common practice in planning medical studies (Spiegelhalter, 2019). (2 marks)