1. An African hospital is treating patients who have Lhassa Fever. Forty percent (0.40) of patients with this disease survive and each patient lives or dies independently of other patients. Answer the following:
a. There are currently 7 Lhassa Fever patients in the hospital. Use binomial calculations (Table C attached) to determine the exact probability that at least 6 survive. (6 points)Let X = number of patients surviving out of 7. Then X ~ B(7, .4) We are interested here in finding
P(X greater than or equal 6) = P(X = 6) + P(X = 7)
We get these values from Table C = .0172 + .0016 = .0188
See example 5.5 in text.b. Over the next six months the hospital expects 50 patients with Lhassa Fever. Approximate the probability that more than 70% of these patients die. (6 points)
There are several ways to do this problem. Here is one way. More than 70% of the patients dying is the same as less than or equal 30% survive. Here we approximate this probability using the normal approximation for a sample proportion. The mean p hat here is .4 and the standard deviation of p hat is sqrt[(.4)(1-.4)/50] = .0693. The approximation is then
P(p hat <= .3) is approximately equal to P(Z <= (.3 - .4)/.0693)
= P(Z <= -1.44) = .0749 See page the box on page 383.
c. How do you know whether your approximation in part b will work well or not?2. In a study of plant safety, it was found that the time it took for machine operators to react to a warning light was normally distributed with mean 1 second and standard deviation 0.3 second. Ten (10) operators are independently tested for their reaction times. What is the probability that the average reaction time of these 10 exceeds 1.2 seconds? (8 points)
(5 points)The normal approximation used is appropriate when np >= 10 and n(1-p) >= 10. Here n is 50 and p is .4 so np = 20 and n(1-p) = 30. Both are greater than or equal to 10 so the approximation will work well. See the lecture notes and the bottom of the box on page 383 of our text for this rule.
3. Define the following events A = student is a business major and B = student owns a computer. Suppose it is known that at UAF the probability that a randomly selected student is a business major is P(A) = 0.35, and similarly P(B) = 0.40, and P(A and B) = 0.10 Determine the following (5 points each):
a. P(A or B) = P(A) + P(B) - P(A and B) = .35 + .40 - .10 = .65b. Find the probability that a randomly selected student is a business major given that they own a computer.
This question askes you to find P(A|B) = P(A and B) / P(B) = .10 / .4 = .25
c. Are the events A and B independent? The answer is NO.Explanation: If A and B were independent events, then P(A and B) = P(A)P(B). Here P(A and B) = .10 but P(A)P(B) = .35 x .40 = .14 Since .10 is not equal to .14 these events are not independent. Your answer must show numerically that the events are not independent, you can not argue this based upon the events described.
d. Find P(A|Bc) = P(A and Bc) / P(Bc) = [P(A) - P(A and B)] / [1 - P(B)]4. Suppose you decide to enter a local lottery. You spend $5 for a lottery ticket. If you win the drawing, you will receive a prize worth $1005. Only 500 tickets are sold for the lottery and the winner is to be selected by a random drawing of one ticket. What is your expected (average) gain or loss. (6 points)
= [.35 - .10] / [1 - .40] = .4166
Here the P(winning) = 1/500 and P(loosing) = 499/500. If you
win your gain is $1005 - $5 = $1000. If you loose, the loss is -$
5. Therefore
Expected gain or loss = $1000 ( 1/500) + (-$5) ( 499/500) = $2 -
$4.99 = - $2.99
5. True or False (Circle one) The standard deviation of the difference of two random variables is the sum of the standard deviations of the two variables. (4 points)
This statement is false. It is true that the variance
of the difference of two random variables is the sum of the variances,
but it is not true for standard deviations. See Rules for variances
in section 4.4 page 337.
6. Many opinion surveys ask the respondent their reaction to a statement
like "the legislature should use the Alaska Permanent Fund for this years
budget". The respondent is asked to circle one of the possible responses
(1) Strongly agree (2) Agree (3) Neutral (4) Disagree
and (5) Strongly disagree. The random variable X records which answer
a respondent gives to the question. So the possible values of X are
1, 2, 3, 4, and 5. Answer the following:
a. The random variable X is a (record the letter of the correct statement in the answer space)(4 points) Answer = ii) discrete random variable that is not binomial
b. Suppose you are told the distribution of X is
| X | 1 | 2 | 3 | 4 | 5 |
| P(X) | -0.2 | 0.4 | 0.3 | 0.2 | 0.3 |
7. You want to compute a 98% confidence interval for a population mean. Assume that the population variance is 100. What sample size is required so that the margin of error in estimating the mean is 3. (6 points)
This question is taken from section 6.1 of the text. The correct formula is given on page 443 and example 6.5 illustrates the method. Here z = 2.33, sigma = 10, and m = 3. The required sample size is then n = [(z)(sigma)/m]2 = [2.33 (10)]/3]2 = 60.32 so use n = 61
8. Give three (3) factors that influence the width of a confidence interval
for the population mean.
(6 points)
Answer = i) sample size ii) confidence level iii) sigma;
the natural variation in the population. See section 6.1 in text.
9. Suppose that cellulose content of hay is normally distributed with standard deviation 10 mg/g. A sample of 25 plots of hay yielded a mean content of = 145 mg/g. You want to test H0: µ = 140 versus Ha: µ > 140
a. Calculate the value of the test statistic for this problem. (6 points)10. Suppose you observe a value of z = 1.83 when testing H0: µ = 12 versus Ha: µ not = 12. Determine the P-value of your test. (5 points)Z = (x bar - mu)/[sigma/sqrt(n)] = [145 - 140] / [10/sqrt(25)] = 5/2 = 2.5
See section 6.2 in text.b. Describe in terms of this problem what a type II error would be. (5 points)
A type II error would be concluding that the true average cellulose content of the hay was 140 when in fact it greater than 140. See section 6.4 in text.
c. For the purpose of conducting this hypothesis test, is it important that the cellulose content of hay is normally distributed? Why or why not? (5 points)
Hard to tell here. See the discussion in the second to the last bullet on page 444. The same rules that apply to confidence intervals apply to hypothesis tests. Since n = 25, the test is reasonably valid without normality as long as the distribution is not skewed and there are no outliers. There really is not enough information given here to answer this question well.
11. A significance test gives a P-value of 0.04. From this
we can (record the letter of the correct statement in the answer space).
(4 points) Answer = b
See section 6.2 in text.
a. reject H0 using a significance level of
= 0.01
b. reject H0 using a significance level of
= 0.05
c. say that the probability that H0 is false
is 0.04
d. say that the probability that H0 is true
is 0.04
e. none of the above