Practice Midterm Solution

STAT C141 Spring 2005

Practice Midterm Solution

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2. From the question, we know that

P(Jack Phones) = P(Ed Phones) = P(Steve Phones) = 1/3
P(Blank message | Jack Phones) = 0.2
P(Blank message | Ed Phones) = 0.3
P(Blank message | Steve Phones) = 0.1

So, P(Jack Phones | Blank message)

Similarly,

P(Ed Phones | Blank message)

= ,

and

P(Steve Phones | Blank message)

= .

3. Let be the winnings on ith bet on red (i = 1, 2, …, 50), then

, and

(a) So mean of gambler’s capital after 50 plays

The variance of gambler’s capital after 50 plays

(b) Let . We need to calculate P(Y>0).

As Y is asymptotically normal distributed, . So

, where is standard normal N(0,1) distributed. Then by looking up the normal table, we have

4. Joint distribution table for (X,Y) is

Y / Distribution of X
(row sums)
0 / 1
X / 0 / 147/927 / 454/927 / 601/927=0.6483
1 / 77/927 / 249/927 / 326/927=0.3517
Distribution of Y
(column sums) / 224/927=0.242 / 703/927=0.758 / 1 (total sum)

The conditional distribution of X given Y=0:

The conditional distribution of X given Y=1:

As the information on Y does not help much in guessing X, X and Y are not associated and are almost independent.

We can calculate the correlation coefficient between X and Y to further confirm the independent statement on X and Y:

So X and Y are almost completely uncorrelated.

5. To discuss the minimum value that the walk ever reaches before eventually reaching 1, we install an artificial stopping boundary at the value y, where . Then the probability that the unrestricted walk finishes at the artificial boundary y rather than at the value is

Thus if is the minimum height achieved by the walk, since the probability that the walk ever achieves -Lis equivalent to the probability that the minimum height achieved by the walk.

6. Each time when she rides the elevator, the probability to be trapped is .

1)The probability that she will never be trapped =

2)The probability that she will be trapped at least twice =

7. Assume that Urn 1 contains 1 white ball and 2 black balls; Urn 2 contains 2 white balls and 1 black ball.

1)H0: Urn 1 was chosen;

H1: Urn 2 was chosen

We will reject H0 if there are too many white balls: . The type I error is

We can use c=6 as the cutoff. The type I error at c=6 is .

2)If we observe w=7, we will reject the null hypothesis based on above test. And the chance to be wrong = the type I error = .

8. Average height = 68; SD = 3; r=0.95; football-shaped.

1)Without any further information, we should use the average height to guess: 68.

2)We are using the average height to estimate the height of any one of the twins, thus the r.m.s error should be SD = 3.

3)If we have known the height of one of the twins, we can use regression method to guess the height of the other one. The regress line is: . Since 6 feet 6 is 78, the height of other one is predicted as . The slope here is r because we are studying the heights of twins thus the SDs for x and y should be same (SD=3).

4)The r.m.s. error for regression method is =0.94

Note: If r=1, you should guess the height of the second twin by the height of the first twin. In this problem, the correlation is 0.95, which is less than 1, so we should regress the second twin backward the mean a little bit. Also there is quite a large reduction in r.m.s. error when you use the regression line.

a)Expected cell counts

Category /
Total
1 / 2 / 3 / 4 / 5
50 / 50 / 250 / 50 / 100 / 500

b)Alternative hypothesis: the category probabilities are different from p1 =.1, p2 =.1, p3 =.5, p4 =.1, and p5 =.2.

c), with degree of freedom = 5-1 = 4. Further . So we reject the null hypothesis.

10.

a)Anova table

Source / SS / df / MS / F
Between treatments / 23.1667 / 2 / 11.58 / 5.26
Within treatments / 19.75 / 9 / 2.2
Total / 42.9167 / 2, 9

b)The p-value for F=5.26 is . So the difference is significant.

11.

So .

12. , .

1)The least square estimates, of a and b are the values which minimize the sum of squares where .

By solving and , we can find that

and .

, and

Note that

3).