Download 25-7 AP Statistics Quiz A – Chapter 25 Name Before you took this

AP Statistics Quiz A – Chapter 25
Name
Before you took this course, you probably heard many stories about Statistics courses. Oftentimes
parents of students have had bad experiences with Statistics courses and pass on their anxieties to
their children. To test whether actually taking AP Statistics decreases students’ anxieties about
Statistics, an AP Statistics instructor gave a test to rate student anxiety at the beginning and end of
his course. Anxiety levels were measured on a scale of 0-10. Here are the data for 16 randomly
chosen students from a class of 180 students:
Pre-course anxiety level 7 6 9 5 6 7 5 7 6 4 3 2 1 3 4 2
Post-course anxiety level 4 3 7 3 4 5 4 6 5 3 2 2 1 3 4 3
Difference (Post – Pre)
-3 -3 -2 -2 -2 -2 -1 -1 -1 -1 -1 0 0 0 0 1
1. Do the data indicate that anxiety levels about Statistics decreases after students take AP
Statistics? Test an appropriate hypothesis and state your conclusion.
2. Create and interpret a 90% confidence interval.
25-7
AP Statistics Quiz A – Chapter 25 – Key
1. Let d = Post-course anxiety level – Pre-course anxiety level.
H 0 : µ d = 0 The mean difference in the anxiety levels is zero.
H A : µ d < 0 The mean difference in the anxiety levels is less than zero.
* Paired data: The data are paired because they are measurements on the same individuals both
before and after the intro stats course.
* Independence: The anxiety level of any student is
5
independent of the anxiety level of any other student, so the
4
differences are independent.
3
* Randomization: We are told this is a random sample from
2
the class.
1
* 10% condition: Our sample of 16 students is less than 10%
of the students who take AP Statistics.
-3
-1
1
* Nearly Normal condition: The histogram of the differences
Difference (Post-Pre)
is unimodal and roughly symmetric.
Under these conditions the sampling distribution of the differences can be modeled by a
Student’s t-model with (n – 1) = 15 degrees of freedom, and we will use a paired t-test.
We find from the data: n = 16 , d = −1.125 , and sd = 1.1475 .
We estimate the standard deviation of d using: SE ( d ) =
sd
n
=
1.15
16
= 0.287 .
d − 0 −1.125 − 0
=
= −3.92 , so our P-value is P(t15 < −3.82) = 0.0007
SE ( d )
0.287
With a P-value this small, we can reject the null hypothesis. We have strong evidence that
taking AP Statistics class reduces the anxiety level of students.
t15 =
2. Under these conditions the sampling distribution of the differences can be modeled by a
Student’s t-model with (n – 1) = 15 degrees of freedom, and we will use a paired t-interval.
The 90% critical value for t15 is 1.753 (from table).
The margin of error: ME = t15* × SE ( d ) = 1.753(0.287 ) = 0.503
So the 90% confidence interval is −1.125 ± 0.503 , or an interval of (-1.63, -0.62).
We are 90% confident that taking AP Statistics will decrease a student’s anxiety between 0.6
and 1.6 points (on our scale).
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AP Statistics Quiz B – Chapter 25
Name
Most people are definitely dominant on one side of their body – either right or left. For some
sports being able to use both sides is an advantage, such as batting in baseball or softball. In order
to determine if there is a difference in strength between the dominant and non-dominant sides, a
few switch-hitting members of some school baseball and softball teams were asked to hit from
both sides of the plate during batting practice. The longest hit (in feet) from each side was recorded
for each player. The data are shown in the table at the right. Does this sample indicate that there is
a difference in the distance a ball is hit by batters who are switch-hitters?
1. Test an appropriate hypothesis and state your conclusion.
Dominant Non-dominant
Side
Side
142
144
153
148
146
149
138
145
153
160
163
170
169
151
152
167
164
165
163
119
118
126
119
121
125
116
120
124
138
135
144
142
128
131
141
140
140
138
2. Create and interpret a 95% confidence interval.
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AP Statistics Quiz B – Chapter 25 – Key
1. Let d = difference in the hit lengths between the dominant side and the non-dominant side of
the plate.
H 0 : µ d = 0 The mean difference in the hit lengths (in feet) is zero.
H A : µ d > 0 The mean difference in the hit lengths (in feet) is greater then zero.
* Paired data: The data are paired because they are measurements on the same individual from
the dominant side and the non-dominant side of the plate.
* Independence: The hit length by any individual is
independent of the hit length of other individuals, so the
differences are independent.
* Randomization: We assume these switch-hitters, though
not randomly selected, are representative of all switchhitters.
* 10% condition: Our inference is about hit lengths from
dominant and non-dominant sides, not about
individuals, so we do not need to check this condition.
* Nearly Normal condition: The histogram of the
differences is unimodal and symmetric.
Under these conditions the sampling distribution of the differences can be modeled by a
Student’s t-model with (n – 1) = 18 degrees of freedom. We will use a paired t-test.
We find from the data: n = 19 pairs, d = 25.1 feet, and sd = 2.31 feet.
We estimate the standard deviation of d using: SE ( d ) =
sd
n
=
2.31
19
= 0.53 .
d − 0 25.1 − 0
=
= 47.4
P = P(t18 > 47.4) < 0.0001
SE ( d )
0.53
With a P-value this small, we can reject the null hypothesis. We have strong evidence that hits
tend to be longer from the dominant side of the plate.
t18 =
2. Create and interpret a 95% confidence interval.
Under these conditions the sampling distribution of the differences can be modeled by a
Student’s t-model with (n – 1) = 18 degrees of freedom. We will use a paired t-interval.
We find from the data: n = 19 pairs, d = 25.1 feet, and sd = 2.31 feet.
We estimate the standard deviation of d using: SE ( d ) =
The 95% critical value for t18 is 2.101 (from table).
sd
n
=
2.31
19
= 0.53 .
The margin of error: ME = t18* × SE ( d ) = 2.101(0.53) = 1.11
So the 95% confidence interval is 25.1 ± 1.11 , or an interval of (23.99, 26.22) feet.
We are 95% confident that hits from the dominant side will average between 24.0 and 26.2 feet
longer than hits from the non-dominant side of the plate.
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