STAT 203  Answers to Sample Questions for Midterm

 

1a:  .21 + .18 = .39

1b:  Pr(Early | Contaminated) = Pr(Early and Contaminated)/Pr(Contaminated) = .21/.39 = 7/13

 

2a:  Expect mu = nπ = (20)(.10) = 2, give or take sigma = sqrt[nπ(1-π)] = sqrt[1.8] = 1.34, or so.   

2b:  20C1(.1)^1(.9)^19 = .27017                                              

2c:  1 – Pr(none has iron deficiency) = 1 – (.9)^(20) = .8784

2d:  Pr( 1 <= K <= 3) = binomcdf(20,.10,3) – binomcdf(20,.10,0) = .7455

 

3.  One year.  Use the 68% rule and 95% rule from page 37.  If the answer were ONE MONTH it would mean that 95% of all freshmen are born within two months of each other.  This cannot be true because freshmen have birthdays throughout the entire year.  If the answer were FIVE YEARS it would mean 68% of all freshmen are between 13 and 23 years old.  This cannot be true because most of the freshmen are close to age 18. 

 

4.  Then answer to 4c is (.53)(.75) = .3975 because the events are independent. 

The answer to 4a is 53 + .75 - .3975 = .8825. 

The answer to 4d is the opposite of 4a, 1 - .8825 = .1175. 

Alternatively, the answer to 4d is (1 - .53)(1 - .75) = .1175 or . 

Also, the answer to 4a is the opposite of 4d, or 1 - .1175 = .8825. 

The answer to 4b is .8825 - .3975 = .4850. 

 

5.  Of the patients who died, let π be the proportion of patients given Didanosine. 

H0:  π = .50, 

HA:  π < .50,

alpha = .10, 

Test statistic is ks = 5. 

P-value = Pr(K <= ks) = F(5) = .0207.  Look in Table 2 on page 347.  Since P-value < alpha, Reject H0.  In English:  There is sufficient evidence (P-value = .0207) to decide that didanosine is associated with a lower death rate than zidovudine.

 

6.  Here mu = n π  = 3.5, sigma = sqrt[70 π (1 - π )] = 1.8235.   Pr(K >= 3) =  1 – binomcdf(70,.05,2)  = .6863

 

7a.  mu = 4,                     

7b:  sigma = sqrt[1/3] = .57735,          

7c:  Pr(1 < X < 3.5) = .25.

 

8.  f(x) = 0 for x < 0, f(x) = 3x2 for 0<x<1, and f(x) = 0 for x > 1. 

mu = .75,  sigma = sqrt[3/80] = .19365, Pr(.5<X<2) = F(2) – F(.5) = 1 – (.5)^3 = 7/8.

 

9.  π = 1/48, n = 12, Pr(K >= 1) = 1 – Pr(K = 0) = 1 – (47/48)^12 = .22325.

 

10a.  f(0) = (9/12)(8/11) = 12/22, f(2) = (3/12)(2/11) = 1/22, f(1) = 1 – 12/22 – 1/22 = 9/22, f(x) = 0 elsewhere. 

10b:  F(x) = 0 for x < 0, F(x) = 12/22 for 0 <= x < 1, F(x) = 21/22 for 1 <= x < 2, F(x) = 1 for x >= 2.

10c:  mu = n π = 2(3/12) = ½. 

10d:  Var = E(X2) – E2(X) = {(0^2)(12/22) + (1^2)(9/22) + (2^2)(1/22)} – (1/2)^2 = {13/22} – (1/4) = 15/44. 

 

11.  Let π  be the population proportion who prefer the new formulation.  

H0:  π = .5,

HA:  π > .5.

 alpha = .05, 

Here n = 95 since 5 of the 100 were undecided.  Test statistic ks = 38. 

P-value = Pr(K <= 38) = binomcdf(95, .5, 38) = .0321.  We can’t use Table 2 because it only goes up to n = 20.  Since P-value < alpha, Reject the null. 

Detailed conclusion in English:  There is sufficient evidence (P-value = .0321) to decide that the new formulation is preferred. 

 

PART 2

 

1a: .060224 = 6.0224% 

1b:  .004058 = .4048%

 

2a:

k

f(k) = (5 take k)(11 take 2 – k)/ (16 take 2)

0

11/24

1

11/24

2

2/24

                                 Total

24/24

 

2b:   1 – 11/24 = 13/24  

2c:  mu = npi = 2(5/16) = 5/8   sigma = sqrt[(16-2)/16-1)]sqrt[n(pi)(1-pi)] = sqrt[.401 = .63328

 

3a:  mu = 3, sigma = 1.5969 

3b  (.85)^20 = .0388 

3c:  .1368,

3d:  1 – Pr(none) = 1 - .0388 = .9612

3e:  1 – Pr(K <= 1)  =  1 - {.03988 + .1368}

 

4: Pr(K < 120) = binomcdf(1000,.15,119) = .00277

 

5:  F(x) = 0 for x < 0

     F(x) = integral from 0 to x  of the function 3 t^2/8 dt  = (x^3)/8   for 0 < x < 2

     F(x) = 1 for x > 2.

 

5b:  Pr(X > 1) = 1 = F(1) = 1 – 1^3/8 = 7/8 

5c:  E(X) = mu = integral from –infinity to + infinity of xf(x) = 0 + 3/2 + 0 = 3/2

5d:  sigma^2 = VAR  = X(X^2) – mu^2 = 12/5 – 9/4 = 3/20, sigma = sqrt[3.20] = .3873

 

6:  Of the ones that developed cancer, let pi be the probability of someone being in the low-cal group.

H0:  low-cal diet has no effect on cancer (pi = ½ ), 

HA:  low-cal diet reduced incidence of cancer (pi < ½),

alpha = .05, 

test statistic = ts = 2,

P-value = Pr(K <= 2) = binomcdf(9,.50,2) = .0898, Retain H0. 

There is insufficient evidence (P-value = .0898) to conclude that the low-cal diet is associated with lower rates of cancer.

 

7.  pi = population proportion of pulse decreases.

H0:  pi = .5,

HA: pi > .5,

alpha = .10,

test statistic ks = 0

P-value = (1/2)^4 = .0625.  Decision is to Reject H0.

There is sufficient evidence (P-value = .0625) to conclude that physical fitness program lowers at-rest pulse.