Fall 2008Midterm 2Prof. Alexeev
S370(The test has 18 problems)
1. As the random sample size increases,
A. The sample approaches the normal distribution.
B. The population approaches the distribution of a normal random variable.
C. The sampling distribution of a sample mean approaches that of the population.
D. The sampling distribution of a sample mean approaches normal.
E. None of the above is correct
Answer: D. (See Central Limit Theorem.)
2. X is normally distributed with a mean of X and standard deviation of X. Define where random variable is the mean of a sample of size 9 taken from the population of X. The mean and variance of Y are:
Answer: E(Y) = µx·x/(2µx)=x/2 ; Y2 = x2·x2/(4·9·µx2)=x4/(36µx2)
3. A random sample of 9 people from a large city is taken and their heights are measured. The sample mean turns out to be equal to 6 feet. Suppose that it is known that the true variance of the sample mean of people’s heights for samples of 9 is equal to 0.18. What is the population variance of people’s heights?
A. 1.62B. 0.54C. 0.18
D. 0.06E. 0.02
F. Cannot be determined unless we assume that people’s heights are normally distributed
Answer: A. Variance of the sample mean for samples of size 9 is x2/9. Given that it equals 0.18, we find that x=9*0.18=1.62.
4. Mike knows that hiscell phone battery lasts H number of hours, where H is approximately normally distributed with a mean of 11 days and standard deviation of 1. What is the probability that the battery will discharge in 12.5 days or less?
A. .0668B. 0.5332C. 0.8668D. 0.9332
E. None of the above.
Answer: D.P(H≤12.5)=P(z≤(12.5-11)/1)=P(z≤1.5)=0.5+0.4332=0.9332.
5. Mike knows that his cell phone battery lasts H number of hours, where H is approximately normally distributed with a mean of 10 days and standard deviation of 2. Mike does not like to recharge the battery frequently, but he wants to make 90% sure that it doesn’t run out on him before he recharges it. What should be his schedule for recharging the battery?
A. Every 7.07 daysB. Every 7.44 daysC. Every 8.19 days
D. Every 12.56 daysE. None of the above
Answer: B. Mike needs to find C such that P(H>C)=P((H-10)/2>(C-10)/2)=P(z>(C-10)/2)=0.9. Therefore, (C-10)/2=-1.28 and C=-2*1.28+10=7.44.
6. Mike knows that his cell phone battery lasts H number of hours, where H is approximately normally distributed with standard deviation of 2. He also knows that the probability that the battery would run out of juice in 11 days or earlier is 0.85. What is the mean of H?
A. 8.92B. 9.53C. 9.96D. 13.08
E. None of the above.
Answer: A.It is given that P((H-µH)/2≤(11-µH)/2)=0.85. Therefore, (11-µH)/2=1.04 and
µH=11-2*1.04=8.92.
7. On average, 40% of the population recycle their garbage. If a sample of 40 people is taken, what is the probability that either fewer than 22 or more than 33 of them recycle their garbage?
A. 1 NORMDIST(33.5,16,9.6,1) – NORMDIST(21.5,16,9.6,1)
B. 1 – NORMDIST(32.5,16,3.1,1) – NORMDIST(21.5,16,3.1,1)
C. 1 – NORMDIST(33.5,16,3.1,1) + NORMDIST(21.5,16,3.1,1)
D. 1 – NORMDIST(33.5,16,3.1,1) + NORMDIST(22.5,16,3.1,1)
E. None of the above
Answer: C. The number of people in the sample who recycle is a binomial random variable, B, with a mean of 16 and standard deviation of approximately 3.1. Let X be a normal random variable with the same mean and standard deviation. Then, P(B>33)P(X>33.5)=1 P(X<33.5) and P(B<22)P(X<21.5), implying that P(B>33)+P(B<22)1 P(X<33.5)+P(X<21.5).
8. Length of life of a certain brand of tires is approximately normally distributed with standard deviation of 1000 miles. A random sample of 25 tires is tested. What is the probability that the sample mean is within 500 miles of the true population mean life of the tires?
A. 0.3850 B. 0.4938C. 0.9876D. More than 0.9990
E. None of the above
F. Cannot determine from the information provided
Answer:C. P(|X-bar µ|≤500)=P(-500/(1000/5)≤Z≤500/(1000/5))=
P(-2.5≤Z≤2.5)=2*.4938=.9876.
9. It is known that 70% of newly organized small businesses experience cash flow problems during their first year of operation. A consultant for the Small Business Administration takes a random sample of 81 small businesses that have been in operation for one year. The probability that more than 90% of the sample has experienced cash flow problems is:
A. NORMDIST(0.9,0.9,0.051,1)B. 1NORMDIST (0.9,0.7,0.051,1)
C. 1NORMDIST(0.9,0.7,0.0026,1)D. NORMINV(0.9,0.7,0.026)
E. None of the above
Answer: B.Sample proportion has a mean of 0.7 and standard deviation of (.7*.3/81)^0.50.051. P(X>0.9)=1 P(X<0.9) where X~N(0.7,0.0512).
10. The sample size is 100. Given that the sample mean is 80 and sample standard deviation is 10, the 85% confidence interval for the population mean is:
A. [77.55, 82.45]B. [78.96, 81.04]C. [79.86, 80.14]
D. [78.85, 81.15]E. None of the above
Answer: E.85% CI is X-bar ± z0.075*10/10 80 ± 1.44 or [78.56,80.44].
11.An analyst, using only information from a random sample of 25 observations, obtained the following 95% confidence interval for mean weekly family income of households in a city: $1000≤ μ≤$1200. Assuming that household incomes are normally distributed,we can infer that sample standard deviation was:
A. 242.25B. 255.10C. 292.23D. 484.50
E. None of the above.
Answer: A.Half length of the confidence interval here is 100 and it is also equal to t0.025,24*s/5. Therefore, s=500/t0.025,24=500/2.064242.25.
12. From a random sample of 120 employees of a business it is found that 72 of them are males. Construct a 97% confidence interval for the proportion of males in this business.
A. [0.596,0.604]B. [0.516,0.684]C.[0.503,0.697]D. [0.512,0.688]
E. None of the above.
Answer: C.Sample proportion is 72/120=0.6. Therefore, an estimate of its standard deviation is (0.6*0.4/120)^0.50.0447. 97% CI is 0.6 ± z0.015*0.0447=0.6 ± 2.17*0.0447 or [0.503,0.697]
13. Compare distributions of t for 11 degrees of freedom and z. Denote the probability that 0 t k by Pt, and the probability that 0 z k by Pz, where k is a given positive constant. Also, let f(z) and g(t) be the probability density functions for z and t, respectively. Then:
A. Pt > Pz, always
B. Pt < Pz, always
C. Pt > Pz, only for k such that g(k)>f(k)
D. Pt < Pz, only for k such that g(k)<f(k)
E. Cannot determine without additional information
Answer: B. Obviously, for k such that g(k)<f(k) (i.e., k close to zero) the area under f(z) and between 0 and k is greater than the corresponding area under g(t). For k such that g(k)>f(k) (i.e., for k in the right tail) the area under g(t) and to the right of k is greater than the area under f(z) and to the right of k. (This is because t-distribution has fatter tails.) But this means that for these k, the area under f(z) and between 0 and k is again greater than the corresponding area under g(t) because these areas are equal to 0.5 (the area in the tail of the corresponding distribution).
14. The mean time for a direct flight from New York to Boston over a random sample of 25 flights is found to be 55 minutes with a sample standard deviation of 9 minutes. Assuming that the flight times are distributed normally, a 99% confidence interval for the mean flying time is:
A. [55TINV(.005,24)*1.8, 55+TINV(0.005,24)*1.8]
B. [55TINV(.005,24)*9, 55+TINV(0.005,24)*9]
C. [55TINV(.01,24)*1.8, 55+TINV(0.01,24)*1.8]
D. [55TINV(.01,24)*9, 55+TINV(0.01,24)*9]
E. None of the above
Answer: C. 99% confidence interval is 55 ± t0.005,24*9/5. Note, however, that TINV(p, d.f.) returns the t0 value such that P(|t|≥t0)=p (i.e., the t value corresponding to the two-tailed probability). Therefore, we need to use TINV(0.01,24).
15. Analyst A took a random sample of 81 observations from a normal population and constructed a 95% confidence interval for population mean based on sample information. Analyst B took a sample of 16 observations and constructed a 90% confidence interval. Given that sample standard deviations happened to be the same in both samples, the ratio of the length of confidence interval constructed by A to the length of confidence interval constructed by B is:
A. 1.118B. 0.221C. 0.650D. 0.497
E. None of the above F. Cannot determine from the information provided.
Answer: D. Analyst A’s 95% CI has a half-length of 1.96*s/9. Analyst B’s 90% CI is t0.05,15*s/4 = 1.753*s/4. Therefore, the ratio is (1.96/9)/(1.753/4)=0.497.
16. A cash management analyst wishes to estimate the average elapsed time between the issuing of a billing statement by a large retailer and the receipt of payment. From experience the analyst knows that the elapsed time tends to be normally distributed, with a standard deviation of 6.19 days. Find the appropriate sample size if the analyst wishes to estimate the mean elapsed time within 1 day with 95% confidence.
A. 104B. 105C. 147D. 148
Answer: D. Half-length of 95% CI is 1.96*6.2/n0.5=1. Therefore, n≥(1.96*6.19)2 =147.2, implying that n=148.
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 0.03 with 99% of confidence, what sample size should she use?
A. 716B. 1844
C. 1523D. 1785
E. None of the above.
Answer: B. Since no estimate of the population proportion is provided, the pollster should be conservative and use the assumption that p=0.5. Therefore, half-length of the 99% CI is z0.005*0.5/n0.5 = 2.576*0.5/n0.5 = 0.03, implying that n ≥(2.576*0.5/0.03)2 = 1843.27. Therefore, n=1844.
18. A random sample of 81 people is drawn from a population of students and their IQ’s are measured. Sample mean and standard deviation are, respectively 107 and 10. The 90% confidence interval for the true mean IQ of the student population is
A. NORMINV(0.05,107,1.11) NORMINV(0.95,107,1.11)
B. NORMINV(0.05,107,10) NORMINV(0.95,107,10)
C. 107NORMSINV(0.05)*1.11 107+NORMSINV(0.05)*1.11
D. 107+NORMSINV(0.05)*10 107NORMSINV(0.05)*10
E. None of the above
Answer: A. Sample mean has standard deviation of 10/91.11. Answer C is not correct, because NORMSINV(0.05) is a negative number.
Formulas
NORMDIST(x,mean,standard_dev,cumulative) – returns the normal cumulative distribution for the specified mean and standard deviation;
NORMINV(probability,mean,standard_dev) – returns the inverse of the cumulative normal distribution for the specified mean and standard deviation;
NORMSDIST(x) = NORMDIST(x,0,1,1); NORMSINV(prob-ty) = NORMINV(prob-ty,0,1);
TDIST(x, deg_freedom, tails) – returns the Student’s t distribution; x is the number such that the area in the tail to the right of x is TDIST (if tails=2, TDIST =area in both tails)
TINV(probability, deg_freedom): returns the inverse of the t-distribution; “probability” is the two-tailed probability.
, with n1 degrees of freedom, where s is sample standard deviation of X.
mean of a binomial distribution: = n*p; variance of a binomial distribution: 2 = n*p*(1-p)
Standard deviation of sample proportion,
Suppose sample statistic Y is an unbiased estimator of population parameter (i.e. expected value of Y is ). If Y ~N(,2) then the confidence interval for based on Y is Y Z/2Y, where (1-) is the confidence level. Note that you need to use the mean and standard deviation appropriate for the random variable Y that you are dealing with. For example, if Y is a sample mean and the sample is taken from a population with standard deviation X, then Y=X/n0.5.
Below, X and Y are random variables; α and β are constant numbers.
E(α + βX) = α + βE(X); Var(α + βX) = β2E(X)
E(αX + βY) = αE(X) + βE(Y)Var(αX + βY) = α2Var(X) + β2Var(Y) + 2αβCov(X,Y)