Case 1:
1 For each variable measured in this study, success would mean a reduction over
time. Excel is used to measure the difference scores between the before and after
measures, and since it performs this calculation as BEFORE – AFTER, a positive number
indicates a reduction. This means that in general, to decide if a variable is reduced
significantly, the following hypotheses are tested:
H 0 μ D = 0 program makes no diff
H 1μ >D0 program is successful
Rejection region: Degrees of freedom are ν = n D1 = 33 – 1 = 32. Assume a 5%
significance level. This is a one tailed test.
t > tα ,ν t0.05 ,321.6939
The test statistic for matched pairs is
t = xD−μ D
s
D
nD
Excel is used to calculate the results:
Weight ~
tTest: Paired Two Sample for
Means
Weight 1 Weight 2
Mean 79.8879 78.6212
Variance 255.5367 251.0892
Observations 33 33
Pearson Correlation 0.9876
Hypothesized Mean Difference 0
df 32
t Stat 2.8969
P(T<=t) onetail 0.0034
t Critical onetail 1.6939
P(T<=t) twotail 0.0067
t Critical twotail 2.0369
t=2.8969 exceeds the critical value and so the null hypothesis is rejected. This means that
there is sufficient evidence to infer that the program leads to positive reductions – in
other words people lost weight. ((repeat for Chl lvl, fat, …)) all get rejected – significant reductions in each variable..
overall the program is successful.
2. To determine if gender affects reduction in each of the variables measured in the
study, the hypotheses to be tested are:
H 0 μ F μ M = 0 Gender makes no diff to reduction
H 1 μ F μ M ≠ 0 Gender makes a diff to reduction
To perform this test properly, it must first be determined whether or not equal variances
can be assumed. This is determined by testing (for each variable), the following
hypotheses:
2
H : σ 1 =1 Equal variances can be assumed
0 σ 2
2
2
σ 1
H 1 2 ≠1 Equal variances cannot be assumed
σ 2
(where population 1 represents females and population 2 males)
Rejection Region: Assume a 5% significance level. Degrees of freedom are
v1= (n 11) = 16 – 1 = 15 v2= (n 21) = 17 – 1 = 16
This is a two tailed test and so the null hypothesis is rejected if
1 1
F < F = F = 0.3497 or F > F α / 2 ,ν 1,ν0.025 ,15 ,1679
α / 2,ν 2,ν 10.025 ,16 ,15
The test statistic is calculated in Excel using the formula:
2
s1
F = 2
s2
Results:
Weight ~
FTest TwoSample for
Variances
Variable 1 Variable 2
Mean 1.31875 1.217647059
Variance 1.542958333 11.16654412
Observations 16 17
df 15 16
F 0.13817689 F=0.1382 falls within the rejection region, and so the null hypothesis is rejected. This
means that equal variances cannot be assumed for this variable.
Cholesterol Level~
FTest TwoSample for
Variances
Variable 1 Variable 2
Mean 0.481875 0.717058824
Variance 0.48061625 0.551822059
Observations 16 17
df 15 16
F 0.870962373
F=0.8710 does not fall within the rejection region, and so the null hypothesis is not
rejected. This means that equal variances can be assumed for this variable.
Fat Intake~
FTest TwoSample for
Variances
Variable 1 Variable 2
Mean 14.5 24.87647059
Variance 229.9146667 906.7619118
Observations 16 17
df 15 16
F 0.253555717
F=0.2536 falls within the rejection region, and so the null hypothesis is rejected. This
means that equal variances cannot be assumed for this variable.
Cholesterol Intake~
FTest TwoSample for
Variances
Variable 1 Variable 2
Mean 38.4563 90.5588
Variance 5683.4920 46983.5176
Observations 16 17
df 15 16
F 0.1210
F=0.1210 falls within the rejection region, and so the null hypothesis is rejected. This
means that equal variances cannot be assumed for this variable.
Percent Daily Calories from Fat~
FTest TwoSample for
Variances Variable 1 Variable 2
Mean 6.875 4.623529412
Variance 18.694 34.28691176
Observations 16 17
df 15 16
F 0.545222624
F=0.5452 does not fall within the rejection region, and so the null hypothesis is not
rejected. This means that equal variances can be assumed for this variable.
Equal variance can be assumed for both Cholesterol Level and Percent Daily Calories
from Fat. The two variables will be dealt with first.
H 0 μ −Fμ M = 0 Gender makes no diff to reduction
H 1 μ −Fμ M ≠ 0 Gender makes a diff to reduction
Rejection Region: Assume a 5% significance level. Degrees of freedom are
v = n1+ n 2 2 = 16 + 17 – 2 = 31
This is a two tailed test so the rejection region is
t < −t α / 2 ,ν−t0.025 ,312.040 or t > α / 2,ν t0.025 ,312.040
The test statistic is calculated in Excel as
(x1− x 2 − (μ 1 μ ) 2
t = (n −1)s + (n −1)s 2
2 1 1 sp= 1 1 2 2
sp + n1+ n 2 2
n1 n 2
Cholesterol Level~
tTest: TwoSample Assuming Equal
Variances
Variable 1 Variable 2
Mean 0.481875 0.717058824
Variance 0.48061625 0.551822059
Observations 16 17
Pooled Variance 0.517367635
Hypothesized Mean Difference 0
df 31

t Stat 0.938718334
P(T<=t) twotail 0.355132326
t Critical twotail 2.039514584 t = 0.9387 does not fall within the rejection region so the null is not rejected. There is not
enough evidence to infer that gender affect cholesterol level reduction.
Percent Daily Calories from Fat~
tTest: TwoSample Assuming Equal
Variances
Variable 1 Variable 2
Mean 6.875 4.623529412
Variance 18.694 34.28691176
Observations 16 17
Pooled Variance 26.74195446
Hypothesized Mean Difference 0
df 31
t Stat 1.249963086
P(T<=t) twotail 0.220667338
t Critical twotail 2.039514584
t = 1.250 does not fall within the rejection region so the null is not rejected. There is not
enough evidence to infer that gender affect percent daily calories from fat reduction.
With the other variables, unequal variances must be assumed. The degrees of freedom
will be different in each case since the formula is dependent on more than just the sample
size
2
s2 s2
1 + 2
n1 n 2
v= 2 2
s1 s2
n n
1 2
+
n1−1 n2−1
It is left to Excel to calculate the degrees of freedom for each case and as a result, the
critical values also need to be determined by Excel.
Weight~
tTest: Two
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