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#4 pg 502
Hypothesis Test for Effects of Marijuana Use on College Student: In a study of the
effects of marijuana use , light and heavy users of marijuana in college were tested for
memory recall with the results given below. Use a 0.05 significance level to test the claim
that the population of heavy marijuana users has a standard deviation from that of light
users.
Items sorted correctly by light marijuana users:
Items sorted correctly by heavy marijuana users:
n = 64, “x bar” = 53.3, s = 3.6
n = 65, “x bar” = 51.3, s = 4.5
Solution:
Null Hypothesis:
The population of heavy marijuana users has no standard deviation from
that of light users.
x1  x 2
Alternative Hypothesis:
The population of heavy marijuana users has a standard deviation from
that of light users.
x1  x 2
Critical Value:
Since the sample size greater we can use the Z-test statistics. The critical
value from the Z-table at the 0.025 level of significance is 1.96 (since the test is two
sided).
Test Statistics:
Z-test statistics is given by,
Z=
x1  x 2
s12 s22

n1 n2
By plugging the values in the formula we have,
Z=
Z=
53.3  51.3
3 .6 2 4 . 5 2

64
65
2
= 2.7895
0.7169
Decision Rule:
If the calculated value of Z is less than the table value then we can accept
the null hypothesis. Otherwise reject the null and accept the alternative hypothesis.
Conclusion:
Since the calculated value of Z 2.7895 is greater than the critical value
1.96 we have to reject the null hypothesis and accept the alternative hypothesis that the
population of heavy marijuana has a standard deviation from that of light marijuana.
#12 pg 481 Cigaretts Filters and Nicotine Refer to the sample result listed in the
measured nicotine contents of randomly selected filtered and nonfiltered king size
cigarettes. All measurements are in milligrams, and the data are from the Federal Trade
Commission.
a. Use a 0.05 significance level to test the claim that king size cigarettes with
filters have a lower mean amount of nicotine than the mean amount of
nicotine in nonfiltered king size cigarettes.
b. Construct a 90% confidence interval estimate of the difference between the
two population means
c. Do cigarette filters appear to be effective in reducing nicotine?
Filter
Kings
Nicotine (mg)
Non Filter
Kings
n1 = 21
x(bar) =0.94
s1 = 0.31
n2 = 8
x2(bar) = 1.65
s2 =0.16
Solution:
a. Test:
Null Hypothesis:
The king size cigarettes with filters have a higher or equal mean amount of
nicotine than the mean amount of nicotine in non-filtered king size cigarettes.
x1  x 2
Alternative Hypothesis:
The king size cigarettes with filters have a lower amount of nicotine than
the mean amount of nicotine in non-filtered king size cigarettes.
x1  x 2
Critical Value:
Since we are about to use t-test statistic we have the critical value from the
t-table at the 0.05 level of significance and at 27 degrees of freedom as 1.703.
Test Statistics:
The t-test statistics is given by the formula,
t=
x1  x 2
s12 s22

n1 n2
By substituting the values in the formula we get,
0.94  1.65
t=
t=
0.312 0.16 2

21
8
 0.71
= -8.051
0.0881
|t| = 8.051
Decision Rule:
If the calculated value of Z is less than the table value then we can accept
the null hypothesis. Otherwise reject the null and accept the alternative hypothesis.
Conclusion:
Since the critical value 1.703 is less than the calculated value 8.051 we
have to reject the null hypothesis and accept the claim that there is lower mean amount of
nicotine in filtered cigarettes than the non-filtered cigarettes.
b.
The confidence interval is given by the formula,
t0.05 =
( x1  x 2 )  ( 1   2 )
s12 s22

n1 n2
 0.71  ( 1   2 )
1.703 =
0.0881
(1  2 )  0.71  0.15
The confidence interval for the difference between the means at 90% confidence interval
and at 27 degrees of freedom is,
 0.86  (1  2 )  0.56
C. It is obvious from the test that the cigarette which has filter has the lower amount of
nicotine than the non-filtered cigarette.
#6. pg.464
Calculations for Testing Claims: In exercise 6 assume that you plan to use a significance
level of 0.05 to test the claim that P1 = P2. Use the given sample sizes and numbers of
successes to find (a) the pooled estimate “p bar” (b) the z test statistic (c) the critical z
values and (d) the P-value
Low Activity High Activity
N1=10239 N2=9877
X1=101
X2=56
Solution:
Null Hypothesis:
Two population proportions are equal.
P1=P2
Alternative Hypothesis:
Two population proportions are not equal.
P1  P2
Critical value:
Since the sample size is larger we can use the Z test statistics. We have the
critical value from the Z table at the 0.025 level of significance (since the test is two
sided) as 1.96.
Test Statistics:
Since we are advised to use the pooled estimate for the proportion we
have,
X  X2
101  56
Pˆ  1
=
= 0.0078
N1  N 2 10239  9877
Qˆ  1  Pˆ  1  0.0078  0.9922
P1 
X1
101

 0.00986
N1 10239
X2
56

 0.00567
N 2 9877
The test statistics is given by,
P2 
Z=
( P1  P2 )  0
1 1
Pˆ Qˆ (  )
n1 n2
By substituting all the values in the formula we get,
Z=
Z=
(0.00986  0.00567)
1
1
0.0078 * 0.9922(

)
10239 9877
0.00419
 3.3763
0.001241
P-Value for Z-calculated value is (1-0.9996) 0.0004.
Conclusion Based on critical value:
Since the calculated value 3.3763 is greater than the critical value 1.96
from the Z table at the 0.025 level of significance we reject the null hypothesis that two
population proportions are equal.
Conclusion Based on P-Value:
Since the P-Value for the calculated value 0.0004 is less than the given
level of significance 0.025 we have to reject the null hypothesis and accept the
alternative.
#4pg 547
Stocks and Super Bowl Data Set 25 includes pairs of data for the Dow-Jones Industrial
Average (DJIA) high value and total number of points scored in the Super Bowl for 21
different years. Excel was used to find that the value of the linear correlation coefficient
is r = -0.133 and the regression equation is y (dependent variable) = 53.3 – 0.000442x,
where x is the high value of the DJIA. Also the mean number of Super Bowl points is
51.4. What is the best predicted value for the total number of Super Bowl points scored in
a year with a DJIA high of 1200?
Solution:
The best predicted value for the total number of super bowl points scored in a
year with DJIA high of 1200 is found by plugging the 1200 in the given linear regression
equation.
Y = 53.3 – 0.000442X
Y = 53.3 – 0.000442*1200
Y = 52.7696