Fall 2005Midterm 2Prof. Alexeev
S370
1.Consider the following statements:
I. When the sample is small and the population is normal, sample mean is distributed normally;
II. When the sample is large, sample mean is approximately normally distributed;
III. When the sample is small and the population standard deviation is unknown, sample mean is distributed according to a t-distribution.
Of these statements,
A. I and II are always trueB. II and III are always true
C. I and III are always trueD. All are always true
E.None of the above.
Answer: A.
2. Let X be a normally distributed random variable with a mean of X and standard deviation of 6. Define where random variable is the mean of a sample of size 8 taken from the population of X. The mean and variance of Y are:
E(Y) = ______; Y2 = ______
Answer: E(Y)=(1/3)E(X)=(1/3)(X*X)= X2/3; Y2 =(X2x-bar2/32)=(X2(36/8)/9)=X2/2.
3. A random sample of four test scores is drawnfrom a test in alarge class. Assuming that the test scores are distributed normally with a mean of 70 and standard deviation of 8, what is the probability that the sum of the test scores is below300?
A. 0.3944B. 0.8944
C. 0.1056D. 0.9999
E. None of the above.
Answer: B. Define random variable Y=X1+X2+X3+X4. E(Y)=280 and Y=16. Therefore, we need to find P(z<(300-280)/16=1.25) = .5+0.3944=0.8944.
4. The proportion of voters who favor an increase in sales tax is 0.8. If 400 registered voters are surveyed what is the probability that the proportion of voters who favor the tax increase is between .77 and .82?
A. 0.3413 B. 0.4332C. 0.7036D. 0.7745
E. None of the above.
Answer: D.
5. When sampling from a population with a standard deviation of 14, using a sample size of 144, what is the probability that the sample mean will be at least 3 units away from the population mean?
A. 0.0051 B. 0.0257 C. 0.2157
D. 0.0102 E. 0.0514 F. None of the above
Answer: D
6. A survey of 49 people revealed that the mean number of minutes a person talks on his or her wireless phone in one day in a particular locality is 26, with a standard deviation of 2.3. Give a 99% confidence interval for the average daily usage.
A. [25.12, 26.88]B. [25.14, 26.86] C. [25.16, 26.84]
D. [25.17, 26.83] E. [25.08, 26.92] F. None of the above
Answer: C (actually, it should be [25.15,26.85]).
7. It is known that hourly wages of a group of workers are normally distributed. A 98% confidence interval for the workers' hourly wages is computed based on a random sample of 7 observations and under the assumption that the population standard deviation of the workers' hourly wages is $1/hour. If the population standard deviation was not known in advance, but was estimated as $1/hour based on the above sample data, the 95% confidence interval would be:
A.Narrower and involve a lower risk of not containing the true μ.
B.Narrower and involve a greater risk of not containing the true μ.
C.Wider and involve a lower risk of not containing the true μ.
D.Wider and involve a greater risk of not containing the true μ.
E. Cannot determine from the information provided.
Answer: D. Clearly, the risk of not containing the true is greater for the 95% confidence interval. Also, the half length of the 98% CI with known STD of X is Z0.01*1/70.5=2.326*1/70.5, while the half length of the 95% CI with unknown STD of X is t0.025,6*1/70.5=2.447*1/70.5. Given that 2.3262.447, the 95% CI with unknown STD of X is wider.
8. A random sample of 36 SAT scores of students from a large university was taken. The sample standard deviation of the scores was 120. The average score, in the sample, was 1250. The 97% confidence interval for the true mean examination score is
A. 1250NORMINV(0.015,1250,120) 1250+NORMINV(0.015,1250,120)
B. 1250NORMINV(0.03,1250,20) 1250+NORMINV(0.03,1250,20)
C. 1250+NORMSINV(0.015)*20 1250NORMSINV(0.015)*20
D. 1250+NORMSINV(0.03)*20 1250NORMSINV(0.03)*20
E. None of the above
Answer: C. X-bar~N(,20); z0.015=NORMSINV(0.015)<0 C.
9. Given that the sample is taken out of a normal population and sample mean is 10, sample standard deviation is 4, the sample size is 16, and the width of the confidence interval is 3.5, what is the level of confidence that was used to construct the interval?
A. 90%B. 92%C. 95%D. 99%
E. Not enough information providedF. None of the above
Answer: A. Because we have a small sample and unknown population standard deviation, we use t-distribution. We know that t,15*4/4=3.5/2=1.75. Therefore, t/2,15=1.75. From the t table, this implies that /2=0.05 (check the row of the t-table corresponding to 15 degrees of freedom) or 1=0.9 or 90%.
10. In a department store, it is found that out of a randomly selected 64 customers, 21 buy cards. Construct a 95% confidence interval for the proportion of customers buying cards.
A. [0.3177, 0.3445] B. [0.2316, 0.42471]C. [0.3214, 0.3349]
D. [0.2484, 0.5922] E. [0.2131, 0.4432] F. None of the above
Answer: E
11. Mike is a frequent traveler. His time from leaving home to getting to the gate of his flight at the airport is normally distributed with a mean of 60 minutes and standard deviation of 12 minutes. How much in advance does Mike need to leave if he wants to be at the gate on time or earlier at least in 99% of the cases?
A. NORMDIST(.99,60,12,1)B. NORMINV(.99,60,12)
C. 60+12*NORMSINV(.01)D. 60+12*(1NORSMINV(.01))
E. None of the above
Answer: B.
12. The mean bill of 16 randomly selected customers in a restaurant is found to be $32, with a standard deviation of $3.50. Construct a 99% confidence interval for the mean bill of a customer.
A. [28.570, 35.570] B. [29.421, 34.579]C. [29,740, 34.260]
D. [29.747, 34.253]E. [31.355, 32.645] F. None of the above
Answer: B.
13. If the sample size is cut to 1/9 of its present size while the confidence level and all relevant sample statistics remain the same, the confidence interval for a population mean based on the smaller sample will always be exactly:
A. Three times as wideB. One third as wide
C. Nine times as wideD. One ninth as wide
E. Cannot determine from the information given.
Answer: E.
14. A survey of 500 students at a large university found that 50% favor eliminating statistics from the curriculum. The variance of the population proportion of students who would like to banish statistics is
A. 0.25=(1-), because population proportion is a binomial variable
B. 0, because population proportion is not a random variable
C. 0.0005=(1-)/n, because this is the variance of sample proportion
D. 125=(1-)n, because population proportion is a binomial variable
D. None of the above
Answer: B. Population proportion is not a random variable. It’s a fixed number. Therefore, its variance is 0.
15. A sample of 9 randomly selected customers in a restaurant is taken. From the sample, the mean bill and standard deviation are found to be, respectively, $32 and $3.30. The 94% confidence interval for the mean bill of the restaurant’s customer is:
A. [32TINV(.06,8)*1.1, 32+TINV(0.06,8)*1.1]
B. [32+TINV(.06,8)*1.1, 32TINV(0.06,8)*1.1]
C. [32TINV(.03,8)*1.1, 32+TINV(0.03,8)*1.1]
D. [32TINV(.06,8)*3.3, 32+TINV(0.06,8)*3.3]
E. [32TINV(.03,9)*1.1, 32+TINV(0.03,9)*1.1]
F. None of the above
Answer: A. The 94% CI is given by 32 t0.03,8*3.3/3=32TINV(0.06,8)*1.1. Recall also that TINV(p,d.f.) returns the number T such that P(|t|>T)=p.
16. The management of a local restaurant wants to estimate the average amount their customers spend at the restaurant to within $0.50, with a 95% confidence. What is the minimum sample size required, if the standard deviation is assumed to be $3.50?
A. 272 B. 189 C. 325
D. 196 E. none of the above
Answer: B.
17. A pollster would like to determine voters’ preferences with respect to two candidates, A and B, in an election. If she wants to determine the proportion of voters who prefer candidate A to within 0.06 with 95% confidence, what sample size should she use?
A. 137B. 136C. 266 D. 267 E. Not enough information provided
F. None of the above
Answer: D. n=z0.025^2*(.5*.5)/.06^2=266.8n=267.
18. The “no-show” rate at a popular resort hotel is 8%. If a random sample of 100 reservations is taken, what is the probability that the number of “no shows” will exceed 10?
A. 0.14B. 0.18C. 0.23
D. 0.77E. 0.86F. None of the above
Answer: B. You should treat this problem as a normal approximation of the binomial distribution and use the continuity correction factor. Then the problem is to determine the probability of a normal random variable X with a mean of 8 and standard deviation of (0.08*0.92*100)^.5=2.713 to be above 10.5. Standardizing, (10.5-8)/2.7130.92. From the table, the appropriate probability is 0.3212. Therefore, the answer is 0.5-0.32120.18. However, if you use the approach based on looking at the sample proportion, you would get C as the answer. (This approach would actually be equivalent to using normal approximation of the binomial without employing the continuity correction factor.) Since it was not quite clear from the question which approach to use, I gave full 10 points for either answer.