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Means: Con fid ence Intervals a nd Hypothesis Testin g
What are C onfiden ce Intervals?
Definitions: unbiased estimator
confidence interval
point estimate
maximum error of estimate or
margin of error (E)
Know:
* Be able to find and to interpret confidence intervals for given
means
* Be able to determine the minimum sample size needed to
estimate a mean for a specific error value.
* Be able to compare and contrast the standard normal distribution
to the Student-t distribution.
Notes:
* Confidence Intervals (CI) are used to answer the question,
How good is x at estimating µ ?
* Be able to: Create a confidence interval, x ± E for means with degree of
confidence, 1 " ! (change to percentage) and the interpretation of such an
interval; i.e., X% confident that the interval contains the true value.
* The interval is constructed under the assumption that the sample is random.
* Confidence interval about the mean, approximately normally distributed,
!
standard deviation known (! known): E = z"
OR
2
n
s
standard deviation unknown: E = t!
,
2
n
with degrees of freedom being n-1.
E is the maximum error of estimate for the confidence interval, x ± E .
* For the number needed in a sample, solve for n in the above equation:
& z( '
n=$ 2
$ E
%
#
!
!
"
2
Example Problems:
1. When 16 seventeen-year-old women randomly were selected and
tested with an aerobics test, the mean of their scores was found to be
37.9 with a sample standard deviation of 7.3. Construct both a 90%
confidence interval and a 99% confidence interval for the mean. Write
out the meaning of your findings.
2. Verbal SAT scores of a random sample of 250 high school students
have a mean of 470.3 and a population standard deviation of 138.
Construct a 95% and a 99% confidence interval and explain the reason
for the difference in these intervals.
3. Suppose that you take a random sample of 36 athletes who try out
for the Mexican Olympic men's basketball team. Your sample averages 75
inches tall with a standard deviation of 3 inches. Find a 95% confidence
interval and interpret your findings.
ANS.
1. This is a small sample since n = 16. The Student-t distribution is needed. The 90%
confidence interval is from 34.7 to 41.1. The 99% confidence interval is from 32.5 to
43.3. Interpretation: You can be 90% (or 99%) confident that the true mean of the
population is inside the interval 34.7 to 41.1 (or 32.5 to 43.3). OR There is a .90 (or
.99) probability that the true mean is in the given interval.
2. 95% interval is 453.2 to 487.4, 99% interval is 447.8 to 492.8. The more confident
(higher percentage), the wider the interval of values.
3. E = 1.0. There is a 95% probability that the true mean of the athletes who try out
for the Mexican Olympic men's basketball team is between 74 inches and 76 inches.
What is Hypothesis Testing?
HYPOTHESIS TESTING is a decision making process by which we analyze a
sample in an attempt to distinguish between results that can easily occur and
results that are highly unlikely. It is the use of sample statistics to determine if
the sample is statistically significantly different from the claim. It is used to
determine if there is a significant difference or a significant relationship.
Hypothesis - a claim or statement that something is true
Null hypothesis - the claim or statement of a zero or null difference (or no
relationship or association); the null hypothesis is the hypothesis or claim that is
directly statistically tested. The null hypothesis always includes "equals."
The process of hypothesis testing begins with a claim or statement that
something is believed to be true or something needs to be tested. This belief or
need is the reason for the study. This claim or belief may or may not be a null
hypothesis.
The assumptions for the selected hypothesis testing process for means MUST
be met. . .
1. Random sample
2. No outliers
3. Normally distributed
4. Interval or ratio data
5. For two or more means (one or both n < 30), homogeneity of variances
Hypothesis Testing Procedure:
1. Null Hypothesis, Ho, statement that there is no difference between a parameter
and a value or no difference between/among parameters or no relationship. Ho
includes the “=”, the claim being “tested” statistically; may or may not include the
original claim or hypothesis (which is the reason for collecting the data). The
assumptions of the test MUST be met or the results are meaningless!
2. Test Statistic is the value obtained from information from the sample; it is a
value based on sample data and used in making the decision to reject or to fail to
reject the null hypothesis. For means, one sample:
x"µ
x!µ
, ! known t =
, ! unknown, the usual case
z=
!
s
n
n
3. Critical value is a number, based on significance level ( ! ), separating the set
of values that is the rejection region from the set of values that is the nonrejection region. It is a value from a probability distribution such as the standard
normal distribution or t-distribution. This value is dependent on whether or not
the test is a two-tailed (=) or one-tailed test (", !) .
Note: The only reason to use a one-tailed test is if there has been a pilot
study or other evidence to indicate such. The original claim (the reason for
collecting the data) determines whether the test is one-tailed or two-tailed.
In addition, a p-value can be used instead of finding the critical value. Pvalues are available on computer printouts and in the computer program. A pvalue is the probability of getting the results using your sample data given that
the null is true.
4. Conclusion: ONLY TWO possibilities:
* Reject the null hypothesis (test statistic is in the rejection region) or
* Fail to reject the null hypothesis (test statistic is not in the rejection
region).
Remember
* Statistics cannot PROVE anything
* Never "accept" a null hypothesis
5. Inference: Interpret for the particular situation, summarize results.
Two types of errors in hypothesis testing:
! is the probability of rejecting the null hypothesis when the null is TRUE
(Type I error).
! is the probability of failing to reject the null hypothesis when the null is
FALSE (Type II erro)r; 1 " ! is the power of the test, the probability of rejecting a
false null hypothesis.
True state:
Action:
Reject the null hypothesis
Null true
Null False
!
Correct
Fail to reject the null hypothesis
Correct
!
Level of significance - the ! (alpha level) at which you are testing the null
hypothesis.
Maximum Type I error is given as ! (alpha). It is a measure of error so the
smaller the alpha level stated, the stronger the statement when one rejects the
null hypothesis. In rejecting the null hypothesis, use the smallest alpha possible
where the null hypothesis is still rejected.
Without the variance and mean of the population, one cannot compute ! (beta).
Beta is a theoretical concept, giving the probability of the second type error one
can have in testing the null hypothesis.
One-tailed tests: All the rejection region is in one tail of the distribution.
Two-tailed tests: Half of the alpha value is in each tail and the rejection region is
split, half in the right tail and half in the left tail.
NOTE: Some fields of endeavor state null hypotheses using only "=" and not !
or ! statements although they might state alternate hypotheses with an
inequality. In these instances one should look at the alternate hypothesis (when
given) to determine if the test is one-tailed or two-tailed.
Know:
* Classical hypothesis testing procedure: null hypothesis, test
statistic, Critical Value (CV), conclusion, interpretation.
* Be able to show the hypothesis testing procedure to compare a
mean of a sample to a given value, and to compare two means, both
independent and dependent.
* For small samples be able to check homogeneity of variances.
* Be able to relate the hypothesis testing procedure to the related
confidence interval.
Important Facts:
* Never "accept" a null hypothesis! The two possible conclusions
in the hypothesis testing procedure are to reject the null hypothesis or to
fail to reject the null hypothesis.
* If the null hypothesis can be rejected at an alpha of .05, then
reject the null. If the null hypothesis is rejected, make the strongest
statement possible by either showing the smallest value of alpha at which
the null hypothesis can be rejected or determining the p-value of the test
statistic.
* Statistics can NEVER PROVE anything! Statistics can NOT
SHOW CAUSE! Be careful in wording your inference statement.
* Statistics is used to determine if there is a significant difference
or if there is an association or relationship.
Hypothesis Testing, One Samp le Mean s:
Test statistics:
x"µ
, ! known
z=
!
n
t=
x!µ
, ! unknown, the usual case
s
n
Example Problems (one sample, mean):
1. Organic chemists often purify organic compounds by fractional
crystallization. A laboratory technician desired to prepare and purify
several samples, each of 4.85 grams aniline. He claims that his procedure
theoretically will produce 3.43 grams of acetanilide. In a sample of 16,
the mean was found to be 3.50 with a standard deviation of .55. Test
the claim made by the chemist.
2. The final exam in Math 107 usually produces a mean score of 78
(percent). Twenty students in an above-average class take this final
exam and produce a mean score of 82 and a standard deviation of 6.2
(with no outliers). The professor claims that the above-average class
scored significantly higher than the previously accepted average of 78.
Evaluate this situation with an appropriate hypothesis testing procedure.
3. In a random sample of 18 high school juniors, the mean of scores on
the verbal portion of the SAT scores was found to be 510 with a standard
deviation for the sample of 140 points.
(a) Construct a 98% confidence interval for the mean. Interpret your
findings. (b) The mean for these scores is believed to be 500. Test the
claim that the sample mean is not significantly different from the value
500. State the null hypothesis, test statistic, critical value for alpha of
.02, conclusion, interpretation. (c) Explain how the results of the above
two parts are related.
4. The U. S. Department of the Treasury claims that the mean weight of
quarters minted is 5.670 grams with a standard deviation of .072, for the
population of quarters. A random sample of 50 quarters is gathered at a
local bank. The mean weight is found to be 5.638. (1) Is the random
sample significantly different from the weight claimed by the U.S.
Department of the Treasury? (2) Construct a 99% confidence interval
for the mean weight of quarters found by the bank. (3) Explain how the
results of the above two answers are consistent. (4) Explain how the U.
S. Treasury is or is not correct in the claim that the weight of quarters
minted has a mean of 5.670 grams.
5. A Ford engineer claims that a new fuel injection design increases the
mean mileage on the Taurus above its current 30 miles per gallon level.
Twenty of the new designs were checked and the mean recorded as
32.00 miles per gallon with a standard deviation of 3.87 miles per gallon.
Evaluate this claim.
6. Weights of cereal in 16-ounce boxes are normally distributed with a
mean of 16 ounces and a population standard deviation of .12 ounce.
Respond to the following: (a) What is the probability that a cereal box
selected at random will have at least 15.95 ounces? (b) What is the
probability that the mean of a sample of 16 boxes will be at least 15.95
ounces? (c) In a production of 10,000 boxes, how many would you
expect to be below 15.95 ounces? (d) The production manager was
concerned that his boxes of cereal had significantly lower weight than the
expected value of 16 ounces. A sample of 50 boxes was obtained and
the mean weight found to be 15.95. Assume ! = 0.12. Evaluate this
situation using appropriate hypothesis testing.
7. When 16 seventeen-year-old women randomly were selected and
tested with an aerobics test, the mean of their scores was found to be
37.9 with a sample standard deviation of 7.3. (a) Construct both a 90%
and a 99% confidence interval for the mean. Interpret the 99% interval.
(b) Explain why one interval is larger than the other. (c) How many
people would need to be interviewed so that one could be 95% confident
that the mean is in error by no more than 1.5 units? (d) If you could be
guaranteed that a minimum of 95% of the data would be available, how
many would you need in your sample? (e) It is believed that this sample
of seventeen-year-olds are not as aerobic as the national average of
those participating in aerobics. The national average on this test is 40.
Use appropriate hypothesis testing techniques to check the claim. Use an
alpha of .05.
Answers:
1. Ho: µ = 3.43, t = .51 (test statistic). There is insufficient evidence to reject the null
hypothesis; i.e., fail to reject. The laboratory technician seems to be correct.
2. Ho: µ ≤ 78. The test statistic is t = 2.885. The critical value is t(19) = 2.861 for
alpha of .005. Reject the null hypothesis at alpha of .005. The professor's claim seems
to be true, the above average class scored significantly higher than the usual average on
the final exam.
3. (a) 425 < µ < 595. 98% confident that the population mean of the verbal SAT is in
the interval. (b) Ho: µ = 500, t = .303 (test statistic), t(17) = ±2.567 (C.V.). Fail to
reject the null hypothesis. The sample has a mean that is not significantly different from
500. (c) The value 500 (part b) is inside the interval 425 < µ < 595 (part a).
4. (1) Ho: µ = 5.670, test statistic of z = -3.14, critical value at alpha of .01 is
±2.576, p-value is .0016, reject Ho. There is a significant difference between the
weight of the quarters gathered by a local bank and the claim by the U.S. Department of
the Treasury. (2) 5.638 ± .026 gives 5.612 < µ < 5.664. (3) The claimed value of
5.670 grams is NOT in the confidence interval of 5.664 to 5.612 and the null hypothesis
that the mean equals 5.670 was rejected. These are consistent results. (4) The U.S.
Treasury could be correct in that originally quarters weighed an average of 5.670. From
use in circulation, the coins could have been worn down and could have weighed less.
The U. S. Treasury is incorrect in assuming that quarters weigh 5.670 g each. More
information is needed.
5. Ho: µ ≤ 30, test statistic of t = 2.311, t(19) = 2.093 for alpha of .025, reject Ho.
The new fuel injection engine gives significantly better gas mileage than the old fuel
injection design.
6. (a) P(x > 15.95) = P(z > -.42) = .6628; (b) This is a Central Limit Theorem
problem. P( z > -1.67) = .9525; (c) P(x < 15.95) = P(z < -.42) = .3372. And the
product {number of boxes times the probability of any given box}: (10,000)(.3372) =
3372 boxes. (d) Ho: µ ! 16 ; z = -2.95; p-value is .0016; Reject the null hypothesis;
The evidence suggests that the average weight is significantly lower than 16 oz.
7. (a) 90%:
34.7 < µ < 41.1 and 99%: 32.5 < µ < 43.3 . I am 99% confident that the
population mean is in the interval of 32.5 to 43.3. (b) The more confident you want to
be, the larger the interval. (c) 91 (d) 96 (e) H 0 : µ ! 40; t = -1.15 is the test
statistic; t = -1.753 is the critical value; fail to reject the null hypothesis. There is
insufficient evidence to conclude that this group is not as aerobic as the national
population.
Hypothesis Testing, Two Samp les Mean s:
Know:
* Know the assumptions for each statistical test.
* For dependent samples, be able to show appropriate testing
procedure for testing the mean of the difference is zero (Ho: µd = 0).
* Be able to show the hypothesis testing procedure for the
difference in means for both large and small independent samples.
* Check homogeneity of variance (Ho: ∂12 = ∂22) prior to testing
difference of means for small independent samples. If the variances are
homogeneous, use pooled variance in the test statistic. If the variances
are significantly different (at alpha of .05), a nonparametric test
(Wilcoxon) should be used to compare the two samples.
* Be able to calculate a confidence interval for the difference of
two means. Relate these to the results of hypothesis testing.
* Before working with any data set, check for outliers and for
normality (a quick test is Pearson's index of skewness).
Additional Notes:
(1) Two Means, independent large (both n ! 30) samples, from a normal distribution:
test statistic:
z=
( x1 ! x2 ) ! ( µ1 ! µ 2 )
s12 s22
+
n1 n2
Confidence Interval: ( x1 ! x2 ) ± E where E = z!
2
s12 s22
+
n1 n2
(2) Two means, independent small (one or both samples n < 30) samples from a normal
distribution, ! unknown.
* Check homogeneity of variances first:
s12 2
H o : !1 = ! 2 with F = 2 , s1 > s22 compare to F with n1 -1 df in numerator and n2 -1 df in
s2
denominator, ! area in right tail for the Rejection Region.
2
When the variances are not significantly different, use pooled variance.
Otherwise, variances are not pooled and many statisticians feel that a t-test is not
appropriate; therefore, we will use the Wilcoxon test for independent samples, a
nonparametric measure included later.
* test statistic:
s 2p =
( x ' x2 ) ' ( µ1 ' µ 2 )
(n1 ! 1) s12 + (n2 ! 1) s22
for t = 1
with n1 + n2 - 2 df
n1 + n2 ! 2
&
#
1
1
s 2p $$ + !!
% n1 n2 "
We will not work confidence interval for these small samples.
(3) Small dependent matched pairs, test the mean of the difference:
* Check for outliers in the distribution of D and check the distribution of D for
normality.
* test statistic:
t=
D ! µ D with n-1 df AND Confidence Interval: E = t sD for D ± E
!
2
sD
n
n
Important Facts:
* If the data sets do not meet the assumptions for the statistical
model of choice, find another method for comparing the two samples.
Nonparametric tests may be found later in this notebook.
* If a 95% confidence interval for the difference of two means (or
of two proportions) contains zero, then the related null hypothesis should
not be rejected.
Example Problems:
1. In a study to determine strength of a basic back kick and a reverse high
punch in karate, data were collected from 10 black belt subjects. The values
given in the table represent a measure of strength for each strike. It is believed
that there is no significant difference in the two as far as strength when
executed by an expert. Use hypothesis testing techniques to determine if this
belief seems to be true.
Su bj ect
A
B
C
D
E
F
G
H
I
J
pu nch
4.6 6 5.7 0 5.3 7 3.3 4 3.7 7 7.4 3 4.1 5 6.2 1 5.9 0 5.7 7
kick
4.6 3 6.3 4 5.7 2 3.2 3 3.6 0 6.6 9 3.6 6 5.8 1 5.6 1 5.3 3
2. It is believed that Seney Hall has poor heat regulation. Two thermostats still
work. Records are taken from these two thermostats, one on the second floor
and one on the fourth floor. The temperature is recorded at noon for 15
working days. The second floor had a mean of 72.3 with a variance of 7.2
degrees and the fourth floor had a mean of 74.5 with a variance of 5.8 degrees.
(a) Test the claim that the two samples have equal variances. Give the null
hypothesis, the test statistic, the critical value for alpha of .05. and your
conclusion. (b) If appropriate, test the claim that the fourth floor is hotter than
the second floor. State the null hypothesis, test statistic, critical value, and
conclusion. Interpret your findings.
3. Newton County and Fulton County use two different procedures for selecting
jurors. The mean waiting time for the 40 randomly selected subjects from
Newton County is 183.0 minutes with a standard deviation of 21 minutes. The
mean waiting time for the 50 randomly selected subjects from Fulton County is
253.1 minutes with a standard deviation of 29.2 minutes. (Use the z test
statistic.) (a) Test the claim that the mean waiting time of prospective jurors is
the same for both counties. (b) Construct a 98% confidence interval estimate
of the difference between the two mean waiting times. Is your conclusion in
part a consistent with the confidence interval found in part b?
4. It is believed that freshmen have higher averages in Math 111 than do the
sophomores in Math 111. The results of past classes reveal an average for the
freshmen (n = 57) of 83.63 with a standard deviation of 6.3 and an average for
the sophomores (n=19) of 79.3 with a standard deviation of 8.6. Is the claim
realistic in light of the past two years?
5. Before beginning a four-week exercise program, the weights of the students
were taken.
Stu den t
A
B
C
D
E
F
G
H
I
pre- we ig ht
118 126 134 162 145 188 173 125 137
post -we ight
123 131 132 159 138 182 170 128 137
Test the claim that the difference in the pre-weight and the post-weight is zero.
6. Two different cereal companies make a raisin bran cereal. Twelve different
18-ounce boxes are randomly selected for each brand and the contents
weighed. Brand A has a mean of 18.05 oz and a standard deviation of .20 oz.
Brand B has a mean of 17.98 oz and a standard deviation of .24 oz. Is there a
significant difference in the variances or in the weights of the two brands?
Interpret your results.
7. A study of obsessive-compulsive people included a measurement of the
volume of the brain by x-ray tomography. Samples of 10 healthy persons and of
10 obsessive-compulsive persons were obtained. Obsessive-compulsive: mean
volume = 1390.03, s = 156.84. Healthy: mean volume = 1268.41, s = 137.97.
Is there a significant difference in brain size between the two groups?
8. Given the following data set of weights (in pounds) of newborn boys, taken
from a middle class neighborhood hospital:
7.8
8.2
9.0
8.8
6.5
2.3
7.4
8.4
8.3
9.4
10.3 6.8
8.1
7.3
6.2
8.0
8.7
7.9
7.8
8.6
(a) Is there an outlier? If so, discard. Explain your decision. (b) It is assumed
that the distribution of weights of newborn boys is normally distributed. Check
to make sure that the distribution is not significantly skewed. (c) The national
average of weights of newborn boys is 7.8 pounds. Is the above sample
significantly different from the national average?
(d) A second sample was taken from a low socioeconomic area:
7.4
6.2
8.5
7.6
6.8
5.6
6.2
8.6
9.0
5.7
4.8
5.8
7.7
8.4
7.8
7.4
6.6
6.3
6.9
7.2
Check this sample for outliers and discard if needed. Is this distribution
significantly skewed? Is this sample significantly different from the national
average? (e) For the two samples, are the variances homogeneous? (f) It is
believed that the birth weight of newborns in lower socioeconomic homes is
lower than the birth weights of those in middle or upper socioeconomic homes.
With these samples, test this belief.
9. A study of reaction times was conducted with 12 volunteers. The reaction
time of each individual (in milliseconds) was measured before and after the
consumption of one alcoholic beverage:
Volunteer
1
2
3
4
5
6
7
8
9
10
11 12
Before:
After:
469
697
563
814
693
850
737
933
706
821
595
788
634
818
511
761
620 522
792 1235
496
763
Use hypothesis testing techniques to test the claim that reaction time is
different after consumption of this beverage. Check for all assumptions.
577
787
10. A regional sales manages selects two similar offices to study the
effectiveness of a new training program aimed at increasing sales. One office
institutes the program and the other does not. The office that does not, Office
1, has 47 sales people. For this office, the mean sales per person over the next
month is $3,193 with a standard deviation of $102. Office 2, the office that
does institute the training program, has 51 sales people. For this office, the
mean sales per person over the next month is $3,229 with a standard deviation
of $107. (Use the z test statistic for these.) (a) At a 5% significance level,
does the training program appear to increase sales? (Show all steps clearly.)
(b) Construct a 90% confidence interval for the difference in mean sales of the
two offices. Interpret your findings. (c) Are the results of parts (a) and (b)
consistent? Clearly explain your reasoning.
11. An instructor wants to compare the number of hours a student studies for
a calculus exam to the number of hours a student studies for a statistics exam.
The instructor takes a random sample resulting in the following data (in number
of hours studied):
Calculus
Statistics
2.7 6.8 5.4 3.0 4.8
2.1 2.2 3.1 2.5 4.0 2.4 6.0
2.5 4.0 1.5 5.0 7.0
2.9 3.0 2.1 4.7 5.2 3.0 3.6
Determine, at a 5% significance level, whether the two courses have different
mean study times for an exam. Clearly check all appropriate assumptions.
Answers:
1. Ho: µd = 0. The mean of the differences is .168 and the standard deviation of the
differences is .411. You need to be able to find these. The test statistic is t = 1.29.
The test criterion is t(9) = 2.262 for apha of .05. Fail to reject the null hypotheses.
There seems to be no difference in strength between the basic back kick and the reverse
high punch.
2. (a) Ho: ∂ 1 = ∂ 2. Test statistic is 1.241 with critical value of F(14, 14) needed, use
F(12, 14) = 3.05. Fail to reject. There is not a significant difference between the
variances. The assumption of homogeneity of variance is met.
(b) Ho: µ4 ≤ µ2. Test statistic is t = 2.363. For alpha = .05, t(28) = 1.701, onetailed test. Note: sp 2 = 6.5. Reject the null hypothesis at alpha of .05. The fourth
floor has a significantly higher temperature than the second floor.
3. (a) Ho: µ1 = µ2, test statistic is z = -13.229, highly significant for a value of z on the
standard normal table; p-value is p < .0001. Reject the null hypothesis. There is a
significant difference in mean waiting time for the two counties.
(b) -82.4 < µ1 - µ2 < -57.8. The interval does not include zero and the null hypothesis
in part a was rejected. These are consistent results.
4. Check to see if the variances are not significantly different. Ho: ∂12 = ∂ 22 gives
test statistic F = 1.863. Critical value at alpha of .05 is (two-tailed) F(18,56) <
F(15,40) = 2.18. Fail to reject Ho. The assumption of homogeneity of variances has
been met. Test the claim of µ1 > µ2 with Ho: µ1 ≤ µ2 or use the notation, Ho: µ1 - µ2 ≤
0 where µ1 represents freshmen. One of the samples is less than 30, so small samples
test statistic, t = 2.359. Critical value, t(74) < t(70) = 1.994 at alpha of .025. (Note:
df is19 + 57 - 2 = 74). Reject Ho, the freshmen have a significantly higher average than
the sophomores in Math 111.
5. Ho: µd = 0. The mean of the difference is .8889 and the standard deviation of the
difference is 4.4566. Test statistic, t = .5984. For alpha of .05, t(8) = 2.306. Fail to
reject Ho. The exercise program did not seem to make a significant difference in the
weight of the subjects.
6. The variances are not significantly different: Ho: ∂12 = ∂ 22 gave test statistic of F
= 1.44. The critical value, F(11,11) is approximated by F(10,11) = 3.53 (for alpha of
.025 in one tail). Fail to reject Ho. Variances are not significantly different. The
assumption of homogeneity of variances is met. The weights of the two brands are not
significantly different: Ho: µ1 - µ2 = 0 gives test statistic of t = .7762. (NOTE: sp2 =
.0488) Critical value: t(22) = 2.074 at alpha of .05. Fail to reject Ho. The weights are
not significantly different between brands of raisin bran cereal.
7. The variances are not significantly different: Testing Ho: ∂ 12 = ∂ 22 gave a test
statistic of F = 1.2922. The critical value for alpha of .05 is F(9,9) = 4.03 (two-tailed
test). Fail to reject Ho, the assumption of homogeneity of variances is met. There is
not a significant difference in the brain size between the two groups: Ho: µ1 = µ2 for
small samples. Test statistic is t = 1.84. Critical value for alpha of .05 is t(18) =
2.101. Fail to reject Ho.
8. (a) 2.3 is an outlier by both definitions; (b) I = .06, not skewed; (c)
H 0 : µ = 7.8 ;
t = 1.228 is the test statistic; t(18) = 2.101 is the critical value; Fail to reject the null
hypothesis. The sample is not significantly different from the national average birth
weight of boys. (d) No outliers; I = -.066, not significantly skewed; H 0 : µ = 7.8 ;
t = -3.051 is the test statistic; t(19) = 2.093 is the critical value; reject the null
hypothesis. This sample is significantly different from the national average. (e)
H 0 : !1 = ! 2 ; F = 1.306 is the test statistic; F(19,18) < F(15,18) = 2.67 is the critical
value for alpha of .05; The samples are homogeneous; (f)
H 0 : µ 2 ! µ1 ; t = -3.077, p =
.002; t(37) < t(36) = 1.688, alpha of .05; Reject the null hypothesis. The birth weight
of boys from lower socioeconomic homes is significant lower than the birth weight of
boys from the middle socioeconomic homes.
9. The value -713 (or 713 depending on which you subtracted) is an outlier, so remove
it.
I = -.403, so not significantly skewed. Data set: mean = -202.1, s = 45.4, median = 196, n = 11. The claim is that the difference is not zero. H 0 : µD = 0 ; test statistic, t =
-14.76; critical value for alpha of .05, t(10) = ± 2.228; Reject the null hypothesis.
There is sufficient evidence to support the claim that reaction time is different after
consumption of the beverage.
!
10. (a) Claim:
µ 2 > µ1 and H 0 : µ 2 " µ1 ! 0 . Test statistic, z = 1.705; Critical value
at alpha of .05, z = 1.645; p-value is .0436 (area in right tail when using the test
statistic value as the cut-off). Reject the null hypothesis. The training program seems
to produce a significant increase in sales. (b) E = 34.7, and the interval is (1.3, 70.7).
We are 90% confident that the difference between the means is in the interval given.
(c) Yes. The null hypothesis was rejected and the value 0 is not in the above interval
since all points in the interval are positive. The results are consistent. This problem
used a one-tailed test because it was reasonable to assume that the training would
increase sales.
11. Assumptions: No outliers; for skewness: I = -.21 for the calculus group and I =
.836 for the statistics group, therefore neither are significantly skewed. Test for
2
2
homogeneity of variances: H 0 : !1 = ! 2 . F = 2.307 and the critical value, F(9,13) =
3.31. Variances are homogeneous.
2
Test the means: H 0 : µ1 ! µ 2 . (Note: s p = 2.284 ) Test statistic t = 1.486; at alpha of
.05, the critical value is t(22) = ± 2.074 . Fail to reject the null hypothesis. There does
not seem to be a significant difference in the mean study time for the students in the
two courses.