Download 1 - Hinsdale Township High School District 86

AP STATISTICS:
INFERENCE ANSWER KEY
1. For this problem, I will use a 1-sample z test. (I am assuming the population standard deviation is 12 grams) I
am interested in the parameter , the mean weight of an apple.
H0: = 120 The true mean weight of apples from the company is 120 grams.
Ha:   120 The true mean weight of apples from the company is not 120 grams.
Conditions: Assume a SRS and the sample size (49) is larger than 30, so a z-test is appropriate.
x   122.5  120

 1.4583
/ n
12 / 49
P( z  1.4583 or z  1.4583)  .1447
z
Use 2*normalcdf(.1.4583,E99) to get the probability using the TI83.
To find the critical values, use the formula:
116.64
123.36
x    z *  / n  120  1.960(12) / 49  (116.64,123.36)
You can also find the critical values on the calculator.
The critical region is shaded and the
values are the critical values.
This is not statistically significant, so I would not reject H0. Hence, the evidence suggests the mean weight is around
120 grams.
To verify these calculations using the TI-83, see below.
IF WE DO NOT ASSUME THAT THE 12 IS THE POPULATION STANDARD
DEVIATION, then you should use a one-sample t-test. The null and alternate hypotheses are the same, p-value
would be slightly different, but the conclusion is the same.
x   122.5  120

 1.4583
s/ n
12 / 49
df  48
P(t  1.4583 or t  1.4583)  .1513
t
AP STATISTICS:
INFERENCE ANSWER KEY
2. For this problem, I will use a 1-sample t test. I am interested in the parameter , the mean number of hours per
week of part-time work of high school seniors.
H0: = 10.6 The true mean number of hours per week of part-time work of high school seniors is 10.6.
Ha: 10.6 The true mean number of hours per week of part-time work of high school seniors is not 10.6.
Conditions: We are given the data is a SRS and the sample size (50) is larger than 40, so a t-test is appropriate.
x   12.5  10.6

 10.33
s/ n
1.3/ 50
df  49
P(t  10.33 or t  10.33)  6.6738  10 14
t
Use 2*tcdf(.10.33,E99,49) to get the probability using the TI83.
To find t* = 2.0096, you have to use df = 49 and Table C or the calculator.
10.231
10.969
The critical region is shaded and the
values are the critical values.
To find the critical values, use the formula.
x    t * s / n  10.6  2.0096(1.3) / 50  (10.231,10.969)
You can also find the critical values on the calculator.
This is statistically significant, so I would reject H0. Hence, the evidence suggests the mean number of hours per
week of part-time work of high school seniors is not 10.6.
To verify these calculations using the TI-83, see below.
AP STATISTICS:
INFERENCE ANSWER KEY
3. For this problem, I will use a 1-sample proportions test. I am interested in the parameter , the population
proportion of defective items.
H0: = 0.03 The true proportion of defective items is 0.03.
Ha: > 0.03 The true proportion of defective items is greater than 0.03.
Conditions: We are given the data is a SRS, N > 10n, np 0>10 n(1-p0)>10. In checking these conditions, I will
assume that the population is ten times as large as the sample size. Second, in checking 250(.03) > 10 7.5> 10 is
NOT satisfied, but 250(.7) > 10  242.5 > 10. Since one of the conditions does not held true, I will proceed with
caution.
pˆ  p0
8 / 250  .03

 .1853
p0 (1  p0 )
.09(1  .03)
n
250
P( z  .1853)  .42646
z
Use normalcdf(.1853,E99) to get the probability using the TI83.
This is not statistically significant, so I would not reject H0. Hence, the evidence suggests the true proportion of
defective items is around 0.03.
To verify these calculations using the TI-83, see below.
AP STATISTICS:
INFERENCE ANSWER KEY
4. For this problem, I will use a two-sample t test. I am interested in determining if there is a difference between the
mean study times between the two schools are different.
H0: u1 = u2 The mean study time at the two school are equal.
Ha: u1 > u2 The mean study time Driscoll is more than the mean study time at Fenton.
Conditions: We are given the data is a SRS from two different populations. Also, n 1 + n2 = 140 > 40.
t
x1  x2
2
1
2
2
s
s

n1 n 2
df  64

18.56  17.95
4.352 4.872

65
75
 .7827
P(t  .7827)  .21834
To get the probability, tcdf(.7827,E99,64) on the TI-83.
This is not statistically significant, so I would not reject H0. Hence, the evidence suggests mean study time at both
schools is about the same.
To verify these calculations using the TI-83, see below. NOTE: THE CALCULATOR USED THE ADVANCED
DEGREES OF FREEDOM TO CALCULATE THE P-VALUE. THIS IS NOT THE SAME AS ABOVE.
AP STATISTICS:
INFERENCE ANSWER KEY
5. For this problem, I will use a matched pairs t-test. I am interested in determining if the SAT test preparation
course will improve a students Math SAT score at Fenton by 50 points.
H0: = 50 The SAT test preparation course increased a students Math SAT score by fifty points.
Ha: > 50 The SAT test preparation course increased a students Math SAT score by more than fifty points.
Conditions: We are given the data is a SRS. The sample size (10) is less than 15,but the normal quantile plot show
the data is not non-normal. However, there is an outlier, so I will proceed with caution and be skeptical about my
results.
Normal Prob. Plot
Box and Whiskers
I calculated the difference between scores (After - Before) to be 25,28,20,65,10,43,8135, -12,38. The mean of this
set of data is 33.3 with a standard deviation of 26.39.
x
33.3  50

 2.001
s / n 26.39 / 10
df  9
P (t  1.977)  .9618
t
To get the probability, tcdf(-1.977,E99,9) on the TI-83.
This is not statistically significant, so I would reject H0. Hence, the evidence suggests SAT test preparation course
does not improve a Fenton high school students score by more than fifty points.
To verify these calculations using the TI-83, Enter the data into L1 and L2, calculate L3=L2 - L1, and run the test
below.
AP STATISTICS:
INFERENCE ANSWER KEY
6. For this problem, I will use a 2-sample proportions test. I am interested in determining if the population
proportions are the same.
H0: 1 = 2 The two proportions of students who intend to pursue post-secondary education are equal.
Ha: 1 2 The two proportions of students who intend to pursue post-secondary education are not equal.
Conditions: We are given the data consist of two SRS. I will assume that each population is ten times as large as
the sample size (N > 10n1 and N > 10n2). I also need to check the conditions below (see p. 684) with the pooled phat.
pˆ 
p1  p2 36  31 67


 .60909
n1  n2 60  50 110
n1 pˆ  5  60(.60909)  36.54  5
n1 (1  pˆ )  5  60(.3909)  23.45  5
n2 pˆ  5  50(.60909)  30.45  5
n2 (1  pˆ )  5  50(.3909)  19.54  5
All of the conditions are met.
z
pˆ1  pˆ 2

pˆ (1  pˆ ) pˆ (1  pˆ )

n1
n2
36 / 60  31/ 50
 .21405
.609(.3909) .609(.3909)

60
50
P( z  .21405 or z  .21405)  .8305
Use 2*normalcdf(.21405,E99) to get the probability using the TI83.
This is not statistically significant, so I would not reject H0. Hence, the evidence suggests the true proportion of
students that plan to pursue post-secondary education are the same in the urban and suburban high schools.
To verify these calculations using the TI-83, see below.
AP STATISTICS:
INFERENCE ANSWER KEY
7. For this problem, I will use a 1-sample z test. I am interested in the parameter , the mean number of days their
room deodorizers last.
H0: = 17.3 The mean number of days their room deodorizers last is 17.3.
Ha:  17.3 The mean number of days their room deodorizers last is not 17.3.
Conditions: Assume a SRS and the sample size (20) is not larger than 40, so a t-test should be used with if there are
not any outlier or strong skewness in the data (see p. 606).
x   15.2  17.3

 4.472
/ n
2.1/ 20
P( z  4.472 or z  4.472)  7.751  10 6
z
Use 2*normalcdf(-4.472,E99) to get the probability using the TI83.
To find the critical values, use the formula:
16.38
18.22
x    z *  / n  17.3  1.960(2.1) / 20  (16.37,18.22)
You can also find the critical values on the calculator.
The critical region is shaded and the
values are the critical values.
This is statistically significant, so I would reject H0. Hence, the evidence suggests the average number of days a
room deodorizer lasts is not 17.3 days as the company claims.
To verify these calculations using the TI-83, see below.
IF WE DO NOT ASSUME THAT THE 2.1 IS THE POPULATION STANDARD
DEVIATION, then you should use a one-sample t-test. The null and alternate hypotheses are the same, p-value
would be slightly different, but the conclusion is the same.
x   15.2  17.3

 4.472
s/ n
2.1/ 20
df  19
P(t  4.472 or t  4.472)  2.6127  104
t
AP STATISTICS:
INFERENCE ANSWER KEY
8. For this problem, I will use a two-sample t test. I am interested in determining if the Fowler motor lasts longer
than the Malloy motor.
H0: u1 = u2 The Fowler and Malloy motors mean life are equal.
Ha: u1 > u2 The mean life of a Fowler motor is larger than the mean life of a Malloy motor.
Conditions: We are given the data is a SRS from two different populations. Also, n 1 + n2 = 95 > 40.
t
x1  x2
2
1
2
2
s
s

n1 n 2
df  44

28.5  29
2.12 2.12

45
50
 1.1587
P(t  1.1587)  ..87359
To get the probability, tcdf(.-1.1587,E99,44) on the TI-83.
This is not statistically significant, so I would not reject H0. Hence, the evidence suggests the mean life of the
motors are the same.
To verify these calculations using the TI-83, see below. NOTE: THE CALCULATOR USED THE ADVANCED
DEGREES OF FREEDOM TO CALCULATE THE P-VALUE. THIS IS NOT THE SAME AS ABOVE.
Problem
9
Parameter of Interest
Mu
Choice of Test
Two-sample t-test
10
P
One-sample proportions
test
11
Mu
Two sample t-test
12
P
Two-sample proportions
test
13
Mu
Matched pairs
Null Hypothesis in words
The new halogen bulb
lasts twenty five more
hours than the
incandescent bulb
The population proportion
of students who prefer the
Mehta bookbag is .55
The average score on the
state test is the same for
urban districts and
suburban districts
The proportion of males
and females going to postsecondary education after
graduation are equal.
The difference between
the pre-test and post-test
scores of third graders is
zero.
AP STATISTICS:
INFERENCE ANSWER KEY
14. For this problem, I will use a Chi-square, goodness of fit test. I am interested in determining if the proportion of
prizes claimed by the manufacturer is the same as the observed proportions.
H0: The proportions of getting various prizes claimed by the manufacturer are the same as the observed proportions.
Ha: At least on of the proportions of getting various prizes claimed by the manufacturer are different than the
observed proportions.
Conditions: I can use this test if all expected at least 80% of the expected counts are larger than five and all are at
least one. The expected counts are in L2, thus the condition is satisfied.
 
2
df  4
observed  exp ected 
2
exp ected
186  204 

204
2
 327  306

306
2
 ...  9.693
P(  2  9.693)  .0459
To get the probability, use X2cdf(9.693,E99,4)
This is statistically not significant at the 5% level. Thus, there is no reason not to believe that the observed
proportions are equal to the population proportions.
AP STATISTICS:
INFERENCE ANSWER KEY
15. For this problem, I will use a 2-sample proportions test. I am interested in determining the proportion of
families that watch the TV show in rural Indiana is the same as the proportion of families that watch the TV show in
urban Indiana.
H0: 1 = 2 The two proportions of urban and rural families that watch the TV show are equal.
Ha: 1 > 2 The proportion of urban families that watch the TV show is greater than the proportion of rural families
that watch the show.
Conditions: We are given the data consist of two SRS. I will assume that each population is ten times as large as
the sample size (N > 10n1 and N > 10n2). I also need to check the conditions below (see p. 684) with the pooled phat.
pˆ 
p1  p2 40  22
62


 .41333
n1  n2 100  50 150
n1 pˆ  5  100(.41333)  41.3333  5
n1 (1  pˆ )  5  100(.5866)  58.666  5
n2 pˆ  5  50(.41333)  20.666  5
n2 (1  pˆ )  5  50(.5866)  29.3333  5
All of the conditions are met.
z
pˆ1  pˆ 2

pˆ (1  pˆ ) pˆ (1  pˆ )

n1
n2
.44  .40
 .46897
.41333(.58666) .41333(.58666)

50
100
P( z  .46897)  .31954
Use normalcdf(.46897,E99) to get the probability using the TI83.
This is not statistically significant, so I would not reject H0. Hence, the evidence suggests the two proportions f
urban and rural families that watch the TV show are equal.
To verify these calculations using the TI-83, see below.
These problems do not include any inference related to section
13.2, the chi-square tests for independence or homogeneity.
You should review these types of problems.