STAT 203 SAMPLE QUESTIONS

1.   The National Survey of Salaries and Wages in Public Schools shows that the mean salary of teachers is $40,133 and the standard deviation is about $8,000.

 If you took a random sample of 64 teachers you expect their total salary to be mu_{SUM} = n(mu) = 64(40133) = 2,568,512 ,give or take sigma_{SUM} = sqrt[64](8000) = 64,000, or so. 

 

Their mean salary is expected to be mu_{X-bar} = mu = 40133, give or take sigma_{X-bar} = sigma/sqrt[n] = 8000/sqrt[64] = 1000, or so. 

 

Find the probability that the mean salary of these 64 teachers is over $42,000  Pr(X-bar > 42000) = Pr(Z > (42000 – 40133)/1000) = Pr(z > 1.867) = .0310.

 

Do you need to assume that public school teacher salaries follow the normal curve to answer this last question?  NO (the sample is large)

 

2.  A brand of water-softener salt comes in packages marked “net weight 40 lb.”  The company that makes the salt claims that the bags contain an average of 40 lb of salt and that the standard deviation is 1.5 lb.  Assume the weights follow the normal curve. 

Obtain the probability that the weight of one bag of water-softener salt will be 39 lb or less, if the company’s claim is true  Pr(X <= 39) = Pr(Z <= (39 – 40)/1.5) = Pr(Z <= -.6667) = .2525

 

 Determine the probability that the mean weight of 10 randomly selected bags of water-softener salt will be 39 lb or less, if the company’s claim is true.  

Pr(X-bar <= 39) = Pr(Z <= (39 – 40)/(1.5/sqrt[10])) = Pr(Z <= -2.108) = .0175

 

If you bought one bag of water-softener salt and it weighed 39 lb, would you consider this sufficient evidence that the company’s claim is incorrect?   NO

Explain your answer.  H0: mu = 40, HA: mu < 40, P-value = .2525.  There is insufficient evidence (P-value = .2525) to conclude that mean fill is less than 40 lbs.

 

If you bought 10 bags of water-softener salt and their mean weight was 39 lb, would you consider this sufficient evidence that the company’s claim is incorrect? YES

Explain your answer.  H0: mu = 40, HA: mu < 40, P-value = .0175.  There is sufficient evidence (P-value = .0175) to conclude that mean fill is less than 40 lbs.

 

3.  Playing hooky.  A poll was taken of 1010 U.S. employees to find out what percentage call in sick when in fact they are not.  They were asked whether in the last year they call in sick at least once a year when they simply need time to relax.   Of the 1010 U.S. employees surveyed, 202 responded “yes.”   Here p = 202/1010 = .20, s = sqrt[p(1-p)] = sqrt[(.20(.80)] = .40

 

Find a 95% confidence interval pi = p +/- (1.96)(s/sqrt[n]) = .20 +/- (1.96)(.40/sqrt[1010]) = .20 +/- .02467 

Or  95% C.I.:  17.53% < pi < 22.47%

 

Give a symbol for the parameter being estimated  pi

 

4.    A Louis Harris Poll was taken of 1250 U.S. adults about their views of banning handgun sales.  Of those sampled, 650 favored a ban.  At the 5% significance level, does the poll provide sufficient evidence to conclude that a majority of the U.S. adults (i.e., more than 50%) favor banning handgun sales?  Let pi be the population percentage of U.S. adults who favor banning handgun sales.

 

State the null hypothesis H0:  pi = .50  State the alternative HA:  pi > .50

 

Do we need to assume normality of the population to use this test? NO (Variable is dichotomous)  Test statistic zs = (650/1250 - .50)/sqrt[(.5)(1-.50)/1250] = 1.414

 

P-value (be as specific as possible)  P-value = Pr(z > zs) = Pr(z > 1.414) = .078 Conclusion of test   Retain the Null  

State your conclusion in context of the problem  There is insufficient evidence (P-value = .078) to conclude that the majority of the U.S. adults favor banning handgun sales.

 

5. Car sales.  The World Almanac reports that in 1990 the type of cars bought in the U.S. had the following frequencies.

 

Type of car

Small

Midsize

Large

Luxury

Percentage

32.8%

44.8%

9.4%

13.0%

 

A sample of 500 car sales last year showed the following data   Compute the Expected Number using the percentages above:  32.8% of 500 is 164, etc. 

 

Type of car

Small

Midsize

Large

Luxury

TOTAL

Obs Number

133

249

47

71

500

Exp Number

164

224

47

65

500

O – E

-31

+25

0

+6

    0

 

At the 5% level of significance, perform a goodness-of-fit test to decide whether type-of-car sales have changed since 1990.

State the null hypothesis  H0:  Type-of-car sales has not changed  State the alternative HA:  Type-of-car sales has changed

 

Do we need to assume normality of the population to use this test? NO (variable is categorical) Degrees of freedom df = c – 1 = 3

Test statistic Chi-Sq = (-31)^2/164 + (25)^2/224 + 0^2/47 + (6)^2/65 = 961/164 + 625/224 + 0 + 36/65 = 9.204

 

P-value (be as specific as possible)   0.025 < P-value < .05 ( = alpha)   Conclusion of test   Reject the Null    

Detailed conclusion (be specific in context of the problem):   There is sufficient evidence (.025 < P-value < .05) to conclude that Type-of-car sales has changed. 

 

6.  Smoking and low birthweight babies.  In a study done in 1971 on smoking  among pregnant women and low birthweight, the following data was collected. 

 

Low birthweight

Normal birthweight

Total

Smokers

185 (143.81)

1891 (1932.19)

2076

Non-smokers

134 (175.19)

2395 (2353.81)

2529

TOTAL

319

4286

4605

 Expected counts are shown above in parentheses.

Perform a one-sided chi-square test at the 1% level of significance to determine whether low birthweights are associated with smoking. 

 

State the null hypothesis H0:  Smoking and BirthWeight are independent  State the alternative HA:  Smoking and BirthWeight are related

 

Do we need to assume normality of the population to use this test?  NO (variable is categorical) Degrees of freedom df = (r-1)(c-1) = 1

Test statistic Chi-Sq = (185 – 143.81)^2/143.81 + …… + (2395 – 2353.81)^2/2353.81 = 11.798 + 0.8781 + 9.685 + 0.7208 = 23.081

 

P-value (be as specific as possible)   Off Chart:  P-vale < .001/2 = .0005 Conclusion of test Reject the Null     

Detailed conclusion (be specific in context of the problem):  There is strong evidence (P-value < .0005) to conclude that smoking and birthweight are related. 

 

7.  TV viewing.   Ten married couples are randomly selected.  Their weekly viewing times, in hours, are as follows

 

   Couple

1

2

3

4

5

6

7

8

9

10

Mean

SD

Husband

23

56

34

30

41

35

26

38

27

30

34.0

9.52

Wife

26

55

52

34

32

38

38

45

29

41

39.0

9.49

   Difference

+3

-1

+18

+4

-9

+3

+12

+7

+2

+11

50.0

7.51

 

+

–

+

+

–

+

+

+

+

+

 

 

At the 5% level of significance, use the sign test to decide whether married men watch less TV, on average, than their wives.  

Let pi = population percentage of husbands who watch more than wives.  Here N+ = 7, N- = 2, N0 = 0. 

State the null hypothesis H0:  Husband and Wives watch the same (pi = .5)   State the alternative HA:  Husbands watch less (pi < .5)

 

Do we need to assume normality of the population to use this test? NO (sign test does not require normality) Test statistic ks = 2

 

P-value (be as specific as possible)  P-value = Pr(K <= ks) = 1/1024 + 10/1024 + 45/1025 = 56/1024 =  binomcdf(10,.5,2) = .055  ( >  alpha)  Conclusion of test   Retain the Null   

Detailed conclusion (be specific in context of the problem): There is insufficient evidence (P-value = .055) to conclude that husbands watch less TV.  

 

8.  TV-viewing again.  Repeat problem 9 but use the t test instead. 

Ten married couples are randomly selected.  Their weekly viewing times, in hours, are as follows

 

    Couple

1

2

3

4

5

6

7

8

9

10

Mean

SD

Husband

23

56

34

30

41

35

26

38

27

30

34.0

9.52

Wife

26

55

52

34

32

38

38

45

29

41

39.0

9.49

    Difference

+3

-1

+18

+4

-9

+3

+12

+7

+2

+11

5.0

7.51

At the 5% level of significance, use the t test to decide whether married men watch less TV, on average, than their wives. 

 

State the null hypothesis H0:  Husband and Wives watch the same (mu_{Diff} = 0) State the alternative HA:  Wives watch more (mu_{Diff} > 0)

 

Do we need to assume normality of the population to use this test? YES Degrees of Freedom df = n – 1 = 9 Test statistic ts = (5 – 0)/(7.51/sqrt[10]) = 2.105

 

P-value (be as specific as possible) .03 < P-value < .04 Conclusion of test   Reject the Null   

Detailed conclusion (be specific in context of the problem):  There is sufficient evidence (.03 < P-value < .04) to conclude that wives watch more TV.

 

9.  It is known that 68% of the registered voters in Fairfax County, Virginia are registered as Democrats.  To test a new telephone sampling method, we call 500 Fairfax County voters and ask their party.  We do this 5 times.  The results are  59.2%,  58.9%,  60.5%,  58.4%  and  61.3%  Democratic.  The sampling method appears to have 

 

high bias and low chance error;

 

Here x-bar = 59.66% and s = 1.201%.                    BIAS (approx)= x-bar – mu = 59.66%  – 68% =  – 8.34%                             Chance Error (approx)= s = 1.201%

 

10.  Using the same data, you compute a 95% confidence interval and a 99% confidence interval.

 

the 99% confidence interval is wider.

 

11.  New laser instruments were used to measure the speed of light; 225 measurements were taken.  They averaged out to 299,775 km per sec, with a standard deviation of 45 km/sec.  Assume the Gauss (this should have read NORMAL) model with no bias.  Fill in the blanks in part [a] and answer the rest True of False.  Here we use z_{.975} = 1.96 (approx)= 2

 

a.  The true speed of light is estimated as 299,775 km/sec This estimate is likely to be off by 45/sqrt[225] = 3 km/sec or so. 

 

False  b.  299,775  +  6 km/sec is a 95% confidence interval for the average of the 225 readings. (We know with 100% certainty that the average of the 225 readings is 299,775 km/sec)

 

True c.  299,775  +  6 km/sec is a 95% confidence interval for the true speed of light. (Yes that’s what confidence intervals say)

 

False d.  There is about a 95% probability that the next reading will be in the range 299,775  +  6 km/sec. (It should be 299,775 +/- 90 km/sec)

 

False  e.  About 95% of the 225 readings were in the range 299,775  +  6 km/sec. (It should be 299,775 +/- 90 km/sec)

 

True f.  If another 225 readings are made, there is about a 95% probability that their average will be in the range 299,775  +  6 km/sec. (Law of averages for sample mean)

 

12.  Two drugs, zidovudine and didanosine, were tested for the effectiveness in preventing progression of HIV disease in children.  In a double-blind clinical trial, 276 children with HIV were given zidovudine, and 281 were given didanosine.  The following table shows the survival data for the two groups.

 

 

Died

Survived

Row Total

Zidovudine

15

259

274

Didanosine

  5

274

279

Column Total

20

533

553

 

Use the binomial exact test at the α = 10% level of significance to decide whether Didanosine is associated with a lower death rate than Zidovudine.

 

Null Hypothesis Same death rate for both Alternative Hypothesis Didanosine has lower death rate

 

Here ks = 5, n = 20, pi = 279/553  (pi = .5 would be acceptable here).

 

P-value (be as specific as possible) P-value = Pr(K <= ks) = Pr(K <= 5) = binomcdf(20,279/535, 5) = .0188

 

Circle the correct decision:   Reject the Null  Detailed conclusion: There is sufficient evidence (P-value = .0188) to conclude that didanosine has a lower death rate than Zidovudine.

 

13.  The life of certain types of light bulbs follow the normal curve with a mean life span of 1000 hours and a standard deviation of 100 hours. 

 

[a]  Find the probability that one such component last more than 875 hours. Pr(X > 875) = Pr(Z > (875-1000/100)) = .9844

 

[b]  Find the probability that four such component all last more than 875 hours.(.9844)^4 = .6398

 

[c]  Find the probability that at least one of four such components last more than 875 hours. 1 – (.1056)^4 = .99988

 

[d]  Find the probability that four such components have a mean life exceeding 875 hours. Pr(X-bar > 875) = Pr(Z > (875-1000)/(100/sqrt[4])) = Pr(z > -2.5) = .99379

 

14.  Batteries are tested by keeping flashlights on until the light deteriorates by 50%.  When 25 such batteries were tested, the results followed a normal curve with a mean of 20 hours and a standard deviation of 3 hours.  Here we use z_{.975} = 1.96 (approx)= 2

 

[a]  About 95% of the batteries lasted between 14 and 26 hours.

 

[b]  A two-sided 95% confidence interval for the mean life of batteries of this type is μ = 20 +/- (1.96)(3/sqrt[25]) = 20 +/- 1.2 or 18.8 < mu < 21.2

 

[c]  A one-sided 95% confidence interval for the mean life of batteries of this type is μ <  20 + 1.6448(3/sqrt[25]) = 20 +.9869 (approx)= 21

 

[d]  Test the hypothesis that the mean life of batteries of this type is at least 22 hours. Use α = .05.  The one sided 95% confidence interval does not contain 22.  So we retain the hypothesis that the mean life of batteries of this type is not as much as 22 hours. 

 

15.  Los Angeles has about four times as many registered voters as San Diego.  A simple random sample of registered voters is taken in each city to estimate the percentage who will vote for school bonds.  Other things being equal, a sample of  4,000  taken in Los Angeles will be about

 

  twice as accurate            

 

as a sample of  1,000  taken in San Diego.

Choose one option, and say why.  Accuracy is proportional to the square root of the sample size.  So a sample 4 times as large will give results sqrt[4] = 2 as accurate.  

 

16.  To test a new drug, a group of 10 matched pairs of volunteers is used.  In each pair, one is treated with the drug while the other is treated with a placebo as a control.  At the end of the experiment, a physician examines each pair and declares which of the two is healthier.  Let  X  be the number of pairs in which the treated patient was declared healthier than the control.  The null hypothesis is that the drug is absolutely ineffective (neither good nor bad).  It has been decided to reject H0 whenever  X > 8 and to retain H0 otherwise.  Here Reject H0 means {X > 8}  and   Retain H0 means {X <= 8}.

 

[a]  Find a, the probability of falsely rejecting H0._____________

 

alpha = Pr(Rejecting H0), when H0 is actually true (pi = .5).

alpha = Pr(X > 8) = Pr(X = 9) + Pr(X = 10)  = 10/1024 + 1/1024 = 11/1024 = .0107

 

[b]  Suppose that the drug is so effective that for each matched pair, the probability is 90% that the treated patient will be declared healthier.  Find b, the probability of falsely accepting H0.___________

 

beta = Pr(Retaining H0), when H0 is false (in this case when pi = .90).

beta = Pr(X <= 8) = 1 – Pr(X > 8} = 1 – {Pr(X = 9) + Pr(X = 10)} =  1 – {10(.9)^9(1-.9)^1 + 1(.9)^10(1-.9)^0} = 1 - .7361 = .2639