45-733 PROBABILITY AND STATISTICS I Practice Final

Probability and Statistics
Name__________________________

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

H

The decision rule for this problem is:

If_{Ù}(p - por_{0})/[p_{0}(1 - p_{0})/n]^{1/2}> z_{a/2}_{Ù}(p - pthen Reject_{0})/[p_{0}(1 - p_{0})/n]^{1/2}< -z_{a/2}H_{0}:If_{Ù}-zthen Do Not Reject_{a/2}< (p - p_{0})/[p_{0}(1 - p_{0})/n]^{1/2}< z_{a/2}H_{0}:_{Ù}-zTest Statistic = (.525 - .5)/[(.5*.5)/400]_{.10}= 1.28, p = 210/400 = .525^{1/2}= 1.0

Probability and Statistics
Name__________________________

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

H

The decision rule for this problem is:

If

Where

Test Statistic: (n-1)s

So Do Not Reject H

The P-Value of the Test = 2*.1712 = .3424

In EViews, the command:

SCALAR PVALUE=@CHISQ(6.4,4)

gives us the tail area above 6.4 which equals .1712.

Probability and Statistics
Name__________________________

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

3. Suppose we have the discrete bivariate probability distribution:æ c(x^{3}+ y + 3) x = 0,1,2 f(x,y) = ç y = 0,1,2 è 0 otherwise

- Find
**c**.

**y 0 1 2 ------------ 0 | 3 4 5 | 12 | | x 1 | 4 5 6 | 15 | | 2 |11 12 13 | 36 | | --------------- 18 21 24 | 63 Hence c = 1/63** - Find
**g**._{1}(x | y=2)

æ 5/24 x = 0 ç ç 6/24 x = 1 g_{1}(x | y=2) = f(x,2)/f_{2}(2) = ç ç 13/24 x = 2 ç è 0 otherwise

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

4. We wish to estimate the average length of bolts in a shipment to our factory. Suppose the population standard deviation is known to be 3.5cm. We take a random sample of 49 bolts. What is the probability that the absolute difference between the sample mean and the true mean will not exceed .2cm._ _ P[|X_{n}- m| £ .2] = P[-.2 £ X_{n}- m £ .2] = _ P[-.2/(3.5/7) £ (X_{n}- m)/(3.5/7) £ .2/(3.5/7)] = P[-.4 £ Z £ .4] = F(.4) - F(-.4) = .6554 - .3446 = .3108

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

VAR[X

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

- (5 Points) Find the probability that a randomly chosen person from
this population will contract at least one disease.
**P(I È II) = P(I) + P(II) - P(I Ç II) = .25 + .23 - .09 = .39** - (5 Points) Find the probability that a randomly chosen person from
this population will contract both diseases, given that he/she has contracted
at least one disease.
**P(I Ç II | I È II) = P(I Ç II)/ P(I È II) = .09/.39 = .2308**

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

Confidence Limits are:_{Ù}_{Ù}_{Ù}p ± z_{a/2}[p(1 - p)/n]^{1/2}_{Ù}p = .64, z_{.005}= 2.58 .64 ± 2.58*[(.64*.36)/400]^{1/2}= .64 ± .06192 so the 99% limits are: (.578, .702)

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

Assume that the cleaning scale values for both brands of soap are normally distributed with equal variances. Test the hypothesis that the two soaps are equally good at cleaning sweatshirts against the alternative thatClean Better ------------------ 8 7 8 5 7 10 9 8 6 6 9 6 7 8 -----------------

H

The decision rule for this problem is: _ _ If (X_{n}- Y_{m})/[s(1/n + 1/m)^{1/2}] > t_{a}then Reject H_{0}: _ _ If (X_{n}- Y_{m})/[s(1/n + 1/m)^{1/2}] < t_{a}then Do Not Reject H_{0}: where s^{2}= [(n - 1)s_{1}^{2}+ (m - 1)s_{2}^{2}]/(n + m - 2) _ _ X_{n}= 7.714, Y_{m}= 7.143, s_{1}^{2}= 1.238, s_{2}^{2}= 2.810 s^{2}= (6*1.238 + 6*2.810)/12 = 2.024 Test Statistic = (7.714 - 7.143)/[2.024*(1/7 + 1/7)]^{1/2}= .751 Using the EViews Command: SCALAR PVAL=@TDIST(.751,12) Produces a two-tailed P-Value of .4671 which we divide by 2 to get: P-Value = .234

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

105 91 109 141 122 99 159 138 88and the measurements for Cleveland in centimeters are:

135 121 99 167 90 89 199Assume that the depth of potholes in both cities is normally distributed. Find 98 percent confidence limits for the difference between the true depths of the populations of Pittsburgh and Cleveland potholes.

_ _ _ _ P{X_{n}- Y_{m}- t_{a/2}s(1/n + 1/m)^{1/2}< m_{x}- m_{y}< X_{n}- Y_{m}+ t_{a/2}s(1/n + 1/m)^{1/2}} = 1 - a where s^{2}= [(n - 1)(s_{x})^{2}+ (m - 1)(s_{y})^{2}]/(n + m - 2) _ _ X_{n}= 116.889, Y_{m}= 128.571, s_{x}^{2}= 606.861, s_{y}^{2}= 1743.952 t_{.01,14}= 2.624, and s^{2}= (8*606.861 + 6*1743.952)/14 = 1094.186 Confidence Limits are: 116.889 - 128.571 ± 2.624[1094.186(1/9 + 1/7)]^{1/2}= -11.682 ± 43.742, or (-55.424, 32.060)

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

m = 1.1 and s

Hence: P[(å

P{[(å

P(Z < .393) = F(.393) = .6528

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(5 Points)

P(1 £ l £ 3) = F(3) - F(0) = .433 - .018 = .415

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

P[(n-1)s

90% Confidence Limits:

C

(n-1)s

(n-1)s

98% Confidence Limits:

C

(n-1)s

(n-1)s

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

95 191 119 81 102 111 100 108 98and the measurements for Cleveland in centimeters are:

115 171 99 117 130 99 179 107Assume that the depth of potholes in both cities is normally distributed. Test the null hypothesis that the variances of the two populations are the same against the alternative hypothesis that the variances are not the same (

H

The decision rule for this problem is:

If s

Where P(F

K

K

s

Test Statistic = 999.5/980.125 = 1.02 So Do Not Reject H

To get the P-Value, use the EViews command:

SCALAR PVAL=@FDIST(1.02,8,7)

which produces a value of .496158. Multiply this by 2 to get the result:

P-Value = .992316

For the One Tail Test set K

Test Statistic = 1.02 we still Do Not Reject H

Note that the P-Value = .496158. This is why that I said in class that I prefer this version of the F-Test. But I will accept either version on the final.

Spring 1999 Flex-Mode and Flex-Time 45-733

Practice Final

Keith Poole

(10 Points)

14. SupposeFindæ [4(2 - xy)]/7 0 < x < 1 f(x,y) = ç 0 < y < 1 è 0 otherwise

E(XY) = ò

(4/7) ò

E(X) = ò

(4/7) ò

E(Y) = ò

(4/7) ò

COV(X,Y) = 2/9 - (10/21)