10.30 In Dallas, some fire trucks were painted yellow (instead of red) to heighten their visibility. During a test period, the fleet of red fire trucks made 153,348 runs and had 20 accidents, while the fleet of yellow fire trucks made 135,035 runs and had 4 accidents. At α = .01, did the yellow fire trucks have a significantly lower accident rate? (a) State the hypotheses. (b) State the decision rule and sketch it. c) Find the sample proportions and z test statistic. (d) Make a decision. (e) Find the p-value and interpret it.

(f) If statistically significant, do you think the difference is large enough to be important? If so, to whom, and why? (g) Is the normality assumption fulfilled? Explain.

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Statistic

Red Fire Trucks

Yellow Fire Trucks

Number of Accidents

= 20

= 4

Number of Fire Runs

= 153348

= 135035

Accident Rate

= 0.000130422

= 0.000029622

= 0.000083223

Step 1: State the Hypotheses

Because it is suspected that red fire trucks have a significantly higher accident rate, we will do a right-tailed test for equality of proportions. We test only at the equality because rejecting would imply rejecting the entire class of hypotheses .

and

Step 2: State the Decision Rule

At α = .01 the right-tail critical value is , which yields the decision rule

Reject if z > 2.326.

Otherwise, do not reject

Step 3: Calculate the Test Statistic

Pooled Sample Proportion:

Assuming normality (i.e., large samples), the test statistic is:

= 2.961

Step 4: Make the Decision

Since the test statistic (z = 2.961) exceeds the critical value ( = 2.326) we reject the null hypothesis and conclude that . Since we are able to reject the hypothesis , we can also reject the entire class of hypotheses at α = .01, that is, red fire trucks have a significantly higher accident rate or yellow fire trucks have a significantly lower accident rate .

p-Value

Using excel function (1-NORMSDIST) p-value = 0.001533266

A smaller p-value indicates a more significant difference.

If statistically significant, is the difference is large enough to be important?

A smaller p-value indicates a more significant difference. Here in this case, the difference for accidents rate for red fire trucks and yellow fire trucks is large enough that makes use of yellow fire trucks for reducing accident cases..

Checking Normality

We have assumed a normal distribution for the statistic . This assumption can be checked. For a test of two proportions, the criterion for normality is nπ ≥ 10 and n(1 − π) ≥ 10 for each sample, using each sample proportion in place of π:

The normality requirement is comfortably fulfilled in this case. Ideally, these numbers should exceed 10 by a comfortable margin, as they do in this example. Since the samples are pooled, this guarantees that the pooled proportion .

10.44 Does lovastatin (a cholesterol-lowering drug) reduce the risk of heart attack? In a Texas study, researchers gave lovastatin to 2,325 people and an inactive substitute to 2,081 people (average age 58). After 5 years, 57 of the lovastatin group had suffered a heart attack, compared with 97 for the inactive pill (a) State the appropriate hypotheses. (b) Obtain a test statistic and p-value. Interpret the results at α = .01. (c) Is normality assured?(d) Is the difference large enough to be important?(e) What else would medical researchers need to know before prescribing this drug widely?(Data are from Science News 153 [May 30, 1998], p. 343.)

Statistic

Lovastatin Takers

Inactive Substitute takers

Number of Heart Attack

= 57

= 97

Number of People

= 2325

= 2081

Heart Attack Rate

= 0.024516129

= 0.046612206

= 0.034952338

Step 1: State the Hypotheses

Because it is suspected that lovastatin reduces the risk of heart attack, we will do a left-tailed test for equality. We test only at the equality because rejecting would imply rejecting the entire class of hypotheses .

and

Step 2: State the Decision Rule

At α = .01 the left-tail critical value is , which yields the decision rule

Reject if z <- 2.326

Otherwise, do not reject

Step 3: Calculate the Test Statistic

Pooled Sample Proportion:

Assuming normality (i.e., large samples), the test statistic is:

= -3.985

Using excel function NORMSDIST p-value = 0.00003374

Step 4: Make the Decision

Since the test statistic (z = -3.985) is less than the critical value ( = -2.326) we reject the null hypothesis and conclude that . Since we are able to reject the hypothesis , we can also reject the entire class of hypotheses at α = .01, that is, lovastatin reduces the risk of heart attack.

Checking Normality

We have assumed a normal distribution for the statistic . This assumption can be checked. For a test of two proportions, the criterion for normality is nπ ≥ 10 and n(1 − π) ≥ 10 for each sample, using each sample proportion in place of π:

The normality requirement is comfortably fulfilled in this case. Ideally, these numbers should exceed 10 by a comfortable margin, as they do in this example. Since the samples are pooled, this guarantees that the pooled proportion .

Is the difference large enough to be important?

A smaller p-value indicates a more significant difference. Here in this case, the difference for heart attack cases using Lovastatin and using Inactive Substitute is large enough that makes use of Lovastatin important for reducing heart attack.

What else would medical researchers need to know before prescribing this drug widely?

From hypothesis test, it is found that as compared to inactive substitute, use of lovastatin reduces the risk of heart attack. Yet, researchers need to know other side affects of this drug in long term, before prescribing it widely.

10.46 To test the hypothesis that students who finish an exam first get better grades, Professor Hardtack kept track of the order in which papers were handed in. The first 25 papers showed a mean score of 77.1 with a standard deviation of 19.6, while the last 24 papers handed in showed a mean score of 69.3 with a standard deviation of 24.9. Is this a significant difference at α = .05? (a) State the hypotheses for a right-tailed test. (b) Obtain a test statistic and p-value assuming equal variances. Interpret these results. (c) Is the difference in mean scores large enough to be important? d) Is it reasonable to assume equal variances? (e) Carry out a formal test for equal variances at α = .05, showing all steps clearly.

Step 1: Formulate the Hypotheses

Both samples are smaller than 30, therefore, the t -statistic will be used. The hypothesis for a right tailed test is:

Ho: = and Ha: >

Step 2: State the Decision Rule

Assuming equal variances, for the pooled-variance t- test, degrees of freedom are = 25 + 24 − 2 = 47.

From the t table we obtain with α= 0.05.

The decision rule is illustrated in below figure.

Step 3: Calculate the Test Statistic

The sample statistics are

and

The value of Pooled sample variance ( ) is

Using the test statistic is:

The p-value can be calculated using Excel’s one-tail function =TDIST(1.221,47,1) which gives p = 0.114088.

Step 4: Make the Decision

The test statistic value 1.221 is significantly less than the and is to the left of it. I, therefore, cannot reject the Ho.

There is no significant difference at a =.05 i.e. it cannot be said that there is differences in grades for finishing exam earlier or later.

This p-value (p = 0.114088) says that a result this extreme would happen by chance about 11.4 percent of the time if = .

The difference in sample means seems to be well within the realm of chance and is not large enough to be important.

Assuming Unequal Variances

The sample variances in this example are similar, so the assumption of equal variances is reasonable. However, if we assume unequal variances the test statistic is

The formula for adjusted degrees of freedom for the Welch-Satterthwaite test is

The adjusted degrees of freedom are rounded to the next lower integer, to be conservative.

For the unequal-variance t- test with degrees of freedom ν’= 43, from the t table we obtain with α= 0.05.

The test statistic value 1.215 is significantly less than the and is to the left of it. I, therefore, cannot reject the Ho.

There is no significant difference at a=.05 i.e. it cannot be said that there is differences in grades for finishing exam earlier or later.

Therefore, from above calculations we can say that, it is reasonable to assume equal variances.

10.56 A sample of 25 concession stand purchases at the October 22 matinee of Bride of Chucky showed a mean purchase of $5.29 with a standard deviation of $3.02.For the October 26 evening showing of the same movie, for a sample of 25 purchases the mean was $5.12 with a standard deviation of $2.14.The means appear to be very close, but not the variances. At α = .05, is there a difference in variances? Show all steps clearly, including an illustration of the decision rule. (Data are from a project by statistics students Kim Dyer, Amy Pease, and Lyndsey Smith.)

The sample statistics are

and

Step 1: Formulate the Hypotheses

For a two-tailed test for equality of variances, the hypotheses are

and

Step 2: State the Decision Rule

Degrees of freedom for the F test are

Numerator: = − 1 = 25 − 1 = 24

Denominator: = − 1 = 25− 1 = 24

For a two-tailed test, we split the α risk and put α /2 in each tail.

F (Right) = 2.2693

F (Left) = 1/2.2693 = 0.4407

Reject if F < 0.4407 or if F > 2.2693

Otherwise, do not reject

Step 3: Calculate the Test Statistic

The test statistic is

Step 4: Make the Decision

Since F = 1.992, we cannot reject the hypothesis of equal variances in a two-tailed test at α = .05. In other words, the ratio of the sample variances does not differ significantly from 1.

11.24 In a bumper test, three types of autos were deliberately crashed into a barrier at 5 mph, and the resulting damage (in dollars) was estimated. Five test vehicles of each crash as described below. Research question: Are the mean crash damages the same for these three vehicles.

Goliath

Varmint

Weasel

1600

1290

1090

760

1400

2100

880

1390

1830

1950

1850

1250

1220

950

1920

Step 1: State the Hypotheses

The hypotheses to be tested are

(The means are the same)

: Not all the means are equal (at least one mean is different)

Step 2: State the Decision Rule

There are c = 3 groups and n = 15 observations, so degrees of freedom for the F test are

Numerator: d.f.1 = c − 1 = 3 − 1 = 2 (between treatments, factor)

Denominator: d.f.2 = n − c = 15 − 3 = 12 (within treatments, error)

We will use α = .05 for the test. The 5 percent right-tail critical value from Appendix F is . Using Excel’s function = FINV(0.05, 2, 12) which yields . This decision rule is illustrated in below Figure.

Step 3: Perform the Calculations

Using Excel for the calculations, we obtain the results as shown in below table.

Anova: Single Factor

SUMMARY

Groups

Count

Sum

Average

Variance

Goliath

5

6410

1282

246320

Varmint

5

6880

1376

103580

Weasel

5

8190

1638

195170

ANOVA

Source of Variation

SS

df

MS

F

P-value

F crit

Between Groups

340360

2

170180

0.93665

0.418802

3.885294

Within Groups

2180280

12

181690

Total

2520640

14

Step 4: Make the Decision

Since the test statistic F = 0.94 is less than the critical value , we cannot reject the hypothesis of equal means. In addition, the p-value (p = 0.42) is greater than the level of significance (α = .05) which confirms that we should not reject the hypothesis of equal crash damage means.

There is not enough evidence to reject that, there is no significant difference in the mean crash damages for Goliath, Varmint and Weasel.