BA 578-Fall2012: Midterm Test

Total 250 points

True/False (3 points each)

1. When determining the sample size n, if the value found for n is 79.2, we would choose to sample 79 observations. (Ch8)

F We always round up

2. If we examine selected items from the population, we are not conducting a census of the population. (Ch1)

T We are conducting sample

3. For a continuous distribution, Probability of (X greater than or equal to 10) is greater than the probability of (X greater than 10) (Ch6)

F They are equal for continuous distribution.

4. When the population is normally distributed and the population standard deviation s is unknown, then for any sample size n, it is appropriate to build the confidence interval of the sample mean X-bar based on the t distribution. (Ch8)

T This is true although for very large samples we "may" use Z distribution as approximation.

5. If the population is normally distributed with known variance then the sample mean may not be normally distributed for a very small sample size. (Ch7)

F In this case the sample mean is Normally distributed for any sample size.

6. We do not need to perform the continuity correction even if the population is 20 times or more than the sample size. (Ch6)

F The size of population is relevant only for "Finite Population Correction". For continuity correction, the sample size matters.

7. If the random variable of X is normally distributed, 99.73% of all possible observed values of X will be within three standard deviations of the mean. (Ch3)

T The "Empirical Rule"

8. If the Union (that is, the Probability using the conjunction "or") of two events equals 1, they are called complements. (Ch. 4)

1

F They are called "Exhaustive". To be complements they need additional property of being mutually exclusive.

9. When establishing the classes for a frequency table it is general rule that the more classes you use the better your frequency table will be. (Ch2)

F Having too many classes is not good: it beats the purpose of grouping. That is why we have the 2^k rule.

10. The reason sample variance has a divisor of n-1 rather than n is that it makes the standard deviation an unbiased estimate of the population standard deviation. (Chs. 3 and 7)

F The sample variance is an unbiased estimator but the sample standard deviation is not.

11. For a binomial probability experiment, with n=150 and p=.1, we can use the normal approximation to the binomial distribution even without continuity correction. (Ch6)

T Here np(1-p) = 13.5 > 10

12. When the level of confidence and sample proportion  p remain the same, a confidence interval for a population proportion p based on a sample of n=100 will be wider than a confidence interval for p based on a sample of n=150. (Ch8)

T The larger the sample size the smaller is the standard error. Hence the result.

13. If the population is normal and its standard deviation is known, then the t- distribution is appropriate for a sample size of 20. (Ch8)

F The Z distribution is appropriate if the standard deviation is known.

14. The sampling distribution of the sample mean is always normally distributed according to the Central Limit Theorem (Chaps. 6 and 7)

F It is always normally distributed if the parent population is normal, but is approximately normal for sufficiently large samples only if the parent distribution is not normal.

15. For a given sample size the variance of the sample proportion will be larger if p = 0.5 than if p = 0.6 (Ch7)T This is obvious from the formula for the variance of p which involves p(1-p).

16. If a population is known to be normally distributed, then it follows that the sample mean must equal the population mean.(Ch7)

F The mean of the sample mean is equal to the population mean, but the sample mean for any sample may be greater than or less than the population mean.

2

Multiple Choices (7 points each)

1. A fair die with six faces is rolled 10 times. What is the probability that an even number (2, 4 or 6) will occur between 2 and 4 times (inclusive)? (Ch5)

A. 0.6123B. 0.1709C. 0.1611D. 0.3662 correctE. 0.3223

For discrete distribution, P(2<= X <= 4) = P(X <=4) - P(X<= 1) = 0.37695 -0.01074 = 0.36621 using Binomdist function. You can also use the binomial table to add probabilities P(X= 2) + P(X =3) + P(X = 4) to get the same result for n= 10 and p = 0.5.

2. A person's telephone area code is an example of a(n) _____________ variable. (Ch1)

A. Nominal correct This number is only for identification.B. OrdinalC. IntervalD. RatioE. Continuous

3. In a statistic class, 10 scores were randomly selected with the following results were obtained: 74, 73, 77, 77, 71, 68, 65, 77, 67, and 66.  What is the median? (Ch3)

A. 71.5B. 72.0correctC. 77.0D. 71.0E. 73.0

Ordering the given numbers we have 65, 66, 67, 68, 71, 73, 74, 77, 77, and 77. The average of the two middle values (71 and 73) is 72.

4. In a rating of the satisfaction with their instructor, 13 students gave the following scores from a scale of 1 to 5 (mean=3): 3, 2, 1, 1, 5, 5, 4, 3, 3, 2, 4, 3, and 3. What is the standard deviation? (Ch3)

A. 3B. 4C. 2.779

3

D. 1.667E. 1.291 correct

Using MegaStat

Descriptive statistics

  # 1 count 13 mean 3.00 sample variance 1.67 sample standard deviation 1.29 minimum 1 maximum 5 range 4 sum 39.00

sum of squares137.0

0 deviation sum of squares (SSX) 20.00

5. The width of a confidence interval will be: (Ch8)

A. Narrower for 99% confidence than 95% confidenceB. Narrower for 90% confidence than 95% confidence correctC. Wider for a sample size of 100 than for a sample size of 50D. Wider when the sample standard deviation (s) is small than when s is large.

The Z value for 90% confidence interval is 1.645 and for 95% is 1.96. The confidence interval is narrower for lower Z value.

6. In a study conducted by UCLA, it was found that 25% of college freshmen support increased military spending. If 6 college freshmen are randomly selected, find the probability that: Exactly 3 support increased military spending. (Ch5)

A. 0.1318 correct Use Binomdist(3, 6, .25, 0)B. 0.0330C. 0.7844D. 0.9624

7. In a manufacturing process a random sample of 9 bolts is taken which gives a mean length of 3 inches with a variance of .09. What is the 90% confidence interval for the true mean length of the bolt? (Ch8)

A. 2.8355 to 3.1645B. 2.5065 to 3.4935

4

C. 2.4420 to 3.5580D. 2.8140 to 3.1860 correctE. 2.9442 to 3.0558

Remember that the variance is 0.09 which gives standard deviation of 0.3.

Confidence interval - mean

90% confidence level3 mean

0.3 std. dev.9 n

1.860 t (df = 8)

0.186 half-width

3.186

upper confidence limit

2.814

lower confidence limit

8. In a manufacturing process, we are interested in measuring the average length of a certain type of bolt. Past data indicates that the standard deviation is .25 inches. How many bolts should be sampled in order to make us 95% confident that the sample mean bolt length is within .02 inches of the true mean bolt length? (Ch8)

A. 25B. 601 correctC. 423D. 49E. 1225

Sample size - mean

0.02E, error tolerance

0.25standard deviation

95% confidence level1.960 z

600.228 sample size601 rounded up

9. A salesperson goes door-to-door in a residential area to demonstrate the use of a new Household appliance to potential customers. She has found from her years of experience that

5

after demonstration, the probability of purchase (long run average) is 0.40. To perform satisfactory on the job, the salesperson needs More than Four orders this week. If she wants to be at least 95 % confident of getting satisfactory evaluation, what is the minimum number of demonstrations she must make? (Ch5)

A. 15B. 17C. 19D. 20E. 21 correct

Binomial distribution21 n

0.4 p    cumulative

X P(X) probability0 0.00002 0.000021 0.00031 0.000332 0.00205 0.002383 0.00864 0.011024 0.02593 0.036965 0.05878 0.095746 0.10451 0.200257 0.14929 0.349548 0.17418 0.523729 0.16773 0.69144

10 0.13418 0.8256211 0.08945 0.9150812 0.04970 0.9647713 0.02294 0.9877114 0.00874 0.9964515 0.00272 0.9991716 0.00068 0.9998417 0.00013 0.9999818 0.00002 1.0000019 0.00000 1.0000020 0.00000 1.0000021 0.00000 1.00000

Binomial distribution20 n

0.4 p

    cumulativeX P(X) probability0 0.00004 0.00004

6

1 0.00049 0.000522 0.00309 0.003613 0.01235 0.015964 0.03499 0.050955 0.07465 0.125606 0.12441 0.250017 0.16588 0.415898 0.17971 0.595609 0.15974 0.75534

10 0.11714 0.8724811 0.07099 0.9434712 0.03550 0.9789713 0.01456 0.9935314 0.00485 0.9983915 0.00129 0.9996816 0.00027 0.9999517 0.00004 0.9999918 0.00000 1.0000019 0.00000 1.0000020 0.00000 1.00000

10. Consider a sampling distribution formed based on a small sample of 5 items only. The standard deviation of the sampling distribution of all sample means (sigma sub X-bar) is ______________ ______________ than the standard deviation of the population of individual measurements. (Ch7)

A. Always, Less correctB. Sometimes, LessC. Always MoreD. sometimes, MoreE. Can be equal to or less than or greater

This should be obvious from the formula for the standard deviation of the sample mean.

11. If the sampled population has a mean 48 and standard deviation 16, then the mean and the standard deviation for the sampling distribution of the sample mean X-bar  for n=64 are: (Ch7)

A. 4 and 4B. 12 and 4C. 48 and 2 correct D. 48 and 1/4E. 48 and 16

The standard deviation of the sample mean is the standard deviation of population divided by the square root of the sample size.

7

12. According to a hospital administrator, historical records over the past 10 years have shown that 20% of the major surgery patients are dissatisfied with after-surgery care in the hospital. A scientific poll based on 400 hospital patients has just been conducted. What is the probability that at least 70 patients will not be satisfied with the after-surgery care? (Ch6)

A. 89.44% correctB. 39.44%C. 10.56%D. 78.88%E. 84.49%

Since n π (1- π) = 400*0.2*0.8 = 64 > 10 we do not need continuity correction for normal

approximation. The standard error of p = √ [0.20*0.80/400] = 0.02. Here p = 70/400 = 0.175. Therefore, Z score = (0.175- 0.2)/0.02 = - 1.25. Note the phrase "at least", which means more than or equal to.

We need P(Z ≥ -1.25) 1 - P(Z ≤ -1.25) = 1 - 0.1056 = 0.8944

13. If the wages of workers for a given company are normally distributed with a mean of $15 per hour, then the proportion of the workers earning more than $13 per hour: (Ch 6)

A. Is greater than the proportion earning less than $18 per hourB. Is less than 50%C. Is less than the proportion earning more than the mean wageD. Is greater than the proportion earning less than $13 per hour correctE. Cannot tell without knowing the standard deviation

The value $13 is to the left of the mean ($15). Now, look at the Normal curve. It should be obvious that the area to the right of $13 is larger than the area to the left.

14. The fill weight of a certain brand of adult cereal is normally distributed with a mean of 910 grams and a standard deviation of 5 grams. If we select one box of cereal at random from this population, what is the probability that it will weigh more than 904 grams? (Ch6)

A. 0.3849B. 0.8849 correctC. 0.1151D. 0.7698E. 0.2302

The Z score is (904 - 910)/5 = -1.2. We need P(Z ≥ -1.2) = 1- 0.1151

8

15. If a population distribution is known to be normal, then it follows that: (Ch 7)

A. The sample mean must equal the population meanB. The sample mean is skewed for small samples but becomes more and more normal as sample size increasesC. The sample standard deviation must equal the population standard deviationD. The Sample proportion must equal the population proportionE. None of the above correct

Only the mean of the sample mean is equal to the population mean. Hence A is incorrect. If population is normal the sample mean is normal (not skewed) even for small samples. Hence B is incorrect. The sample standard deviation is rarely equal to the population standard deviation: C incorrect. Only the mean of the sample proportion in equal to the population proportion : D incorrect.

16. Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay (LOS) of their patients. In 1996, the average LOS for non-heart patient was 4.6 days. A random sample of 20 hospitals in one state had a mean LOS for non-heart patients in 2000 of 3.8 days and a standard deviation of 1.2 days. Calculate a 95% confidence interval for the population mean LOS for non-heart patients in the state's hospitals in 2000. (Ch8)

A. 3.24 to 4.36 correctB. 3.67 to 3.93C. 3.34 to 4.26D. 3.38 to 4.22E. 3.27 to 4.33

Confidence interval - mean

95% confidence level3.8 mean1.2 std. dev.20 n

2.093 t (df = 19)

0.562 half-width

4.362

upper confidence limit

3.238

lower confidence limit

9

Essay Type (10 points each)

1. The following frequency table summarizes the distances in miles of 70 patients from a regional hospital.Distance     Frequency 

0-2              25

2-4              30

4-6              10

6-8              5

Calculate the sample standard deviation for this data (since it is a case of grouped data: use group or class midpoints in the formula in place of X values). (Ch3)

Distance Frequencymid-point (X) fX (X-X ) (X-X )² f(X-X )²

0-2 25 1 25 -1.86 3.45 86.22

2-4 30 3 90 0.14 0.02 0.61

4-6 10 5 50 2.14 4.59 45.92

6-8 5 7 35 4.14 17.16 85.82

Total 70 200 218.57 ΣFX/ΣF 2.86

ΣFX/n 2.86

ΣF(X-Ẋ)²/n-1 3.17

√{ΣF(X-Ẋ)²/n-1} 1.78

2. Container 1 has 10 items, 3 of which are defective. Container 2 has 6 items, 2 of which are defective. If one item is drawn independently from each container, find the probability distribution for X defined as the number of defective items drawn (Hint: You have to use both multiplicative and additive rules to find P(X=1), where as P(X=0) and P(X=2) can be found only by multiplicative rules). (Ch4)

P(X = 0) = 0.7*(4/6) = 0.467

10

P(X = 1) = 0.7*(2/6) + 0.3*(4/6) = 0.433

P(X = 2) = 0.3*(2/6) = 0.1

Total of probabilities = 1.0

3. The 300 employees of a local university have been classified according to gender and job type.

Job                                                       Male (M)     Female (F)

Faculty(FA)                                           110            10

Salaried Staff (SS)                                  35              45

Hourly Staff (HS)                                   55              45

Are Gender and type of job independent? Explain with probabilities. (you have to show how you got the answer using probabilities). (Ch4)

Contigency Table

Job Male (M) Female (F) TotalFaculty (FA) 110 10 120Salaried Staff (SS) 35 45 80Hourly Staff (HS) 55 45 100Total 200 100 300

We can use any conditional or joint probability to prove this. For example P(FA) = 120/300 = 0.4 but P(FA/M) = (110/300)/ (200/300) = 0.55. These two are not equal showing that type of job and Gender are not independent.

4. The probability that an appliance is in repair is .6. If an apartment complex has 100 such appliances, what is the probability that at least 70 will be in repair? Use the normal approximation to the binomial and show the Z values and the steps used in the calculation. (Ch6)

nπ (1- π) = 100*0.6*0.4 = 24 ˃ 10 : we can do normal approximation without continuity correction.

σp = √π (1- π)/100 = 0.04899

Z-score of 70/100 is (0.7 - 0.6)/0.04899 = 2.04

P(Z ≥ 2.04) = 1- 0.9793 = 0.0207

11

5. The flying time of a drone airplane has a normal distribution with mean 4.76 hours and standard deviation of .04 hours. What is the probability that the drone will fly between 4.72 and 4.82 hours? (Ch6)

The Z-scores for the two given values are -1.0 and 1.5. Therefore, the required probability is = 0.9332 – 0.1587 = 0.7745 = 77.45%

6. It has been reported that the average time to download the home page from a government website was 0.9 seconds. Suppose that the download times were normally distributed with a standard deviation of 0.3 seconds. If random samples of 36 download times are selected what two values symmetrically distributed around the population mean have 80% probability of containing the sample mean? (Ch8)

0.3/√36 = 0.05

Given that X N (0.9, 0.3), we have X ~ N (0.9, 0.05). Standardizing X into Z

we need to find two values L and U such that P (X < L) = 0.10 and P (X > U) = 0.10.

We find P (Z ≤ -1.282) = 0.10 and P (Z ≥ 1.282) = 0.10

Thus we have, Xlower = 0.9 - 0.05*1.282 = 0.836 and Xupper = 0.9 + 0.05*1.282 = 0.964

7. Suppose that 40 percent of the voters in a particular region support a candidate. Find the probability that a sample of 900 voters would yield a sample proportion in favor of the candidate within 3 percentage points of the actual proportion. (Ch7)

It can be easily checked that the conditions for normal approximation without continuity correction are fulfilled. Let p be the proportion of voters in the region support the candidate. By

12

Central Limit Theorem p is normal with mean π = 0.40 and standard error σp = √(0.4*0.6/900) = 0.0163299.

We need P (0.40 – 0.03 ≤ p ≤ 0.40 + 0.03) or P (0.37 ≤ p ≤ 0.43)

Standardizing the variable p we have,

P (0.37 ≤ p ≤ 0.43) = P (-1.83712 ≤ Z ≤ 1.83712) = 0.9338

8. An insurance analyst working for a car insurance company would like to determine the proportion of accident claims covered the company. A random sample of 200 claims shows that the insurance company covered 80 accident claims while 120 claims were not covered. Construct a 95% confidence interval estimate of the true proportion of claims covered by the insurance company.(Ch8)

Here p = 80/200 = 0.40. Standard error = √(0.4*0.6/200) = 0.035 Lower boundary of confidence interval = 0.40 - 1.96*0.035 = 0.3314 and upper boundary = 0.40 + 1.96*0.035 = 0.4686

9. The weight of a product is normally distributed with a standard deviation of .5 ounces. What should the average weight be if the production manager wants no more than 10% of the products to weigh more than 5.8 ounces? (Ch6)

The corresponding Z-score is 1.282 because P(Z ≥ 1.282) = 0.10. This gives

µ = 5.8 - σ*Z = 5.8 - 0.5*1.282 = 5.16 (rounded)

13