The data collected from the 12 men & the calculations asking in the project shown in the table below: Male Time in Number of Favorite Days Workout Second(X) Workout M1 83 6889 Arms Chest M2 65 4225 6 1849 Arms Back M3 43 3 М4 34 1156 2 M5 92 8464 1. Arms M6 87 7569 Legs Legs M7 177 31329 M8 182 33124 Arms M9 45 2025 Chest M10 177 31329 1 Chest M11 181 32761 3 Arms Legs Mode= M12 24 576 4 ΣΧ. Ex? n= 12 Median= %3D Mean (M) = N SS = The standard deviation: S = n-1 12-1 The Standard error of the mean: SM =

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t Value (From the chart) = ±1.796
90% confidence interval for the men = M ± CV X SM
The data collected from the 12 men & the calculations asking in the project are
The data collected from the 8 women & the calculations asking in the project
shown in the table below:
are shown in the table below:
国
Male
Time in
X²
Number of
Favorite
Female
Time in
Number of Days
Favorite
Second(X)
Workout
second(X)
Workout
Workout
Days
Workout
FM1
149
22201
2
Arms
M1
83
6889
1
Arms
FM2
93
8649
4
Arms
M2
65
4225
6
Chest
FM3
94
8836
6
Вack
M3
43
1849
3
Arms
FM4
112
12544
Chest
M4
34
1156
Back
FM5
142
20164
6
Chest
M5
92
8464
Arms
FM6
87
7569
Arms
M6
87
7569
6
Legs
FM7
70
4900
1
Вack
M7
177
31329
3
Legs
FM8
95
9025
1
Chest
M8
182
33124
1
Arms
n= 8
ΣΧ-
Ex? =
Med=
Mode
M9
45
2025
5
Chest
M10
177
31329
1
Chest
ΣΧ
Mean: M =
8
M11
181
32761
3
Arms
M12
24
576
4
Legs
SS =
n = 12
ΣΧ-
Ex? =
Median=
Mode=
The standard deviation: S =
ΣΧ
Mean (M) =
п-1
V8
-1
SS =
The standard error of the mean: SM =
Vn
Level a = 1-0.90 = 0.10.
The standard deviation: S =
=
n-1
12-1
df = n-1 = 8-1 =7
S
The Standard error of the mean: SM =
vn
%3D
V12
t value (From the chart) = F1.895
Level a = 0.10
90% confidence interval for the women= M ±CV (t-value) X SM
Transcribed Image Text:t Value (From the chart) = ±1.796 90% confidence interval for the men = M ± CV X SM The data collected from the 12 men & the calculations asking in the project are The data collected from the 8 women & the calculations asking in the project shown in the table below: are shown in the table below: 国 Male Time in X² Number of Favorite Female Time in Number of Days Favorite Second(X) Workout second(X) Workout Workout Days Workout FM1 149 22201 2 Arms M1 83 6889 1 Arms FM2 93 8649 4 Arms M2 65 4225 6 Chest FM3 94 8836 6 Вack M3 43 1849 3 Arms FM4 112 12544 Chest M4 34 1156 Back FM5 142 20164 6 Chest M5 92 8464 Arms FM6 87 7569 Arms M6 87 7569 6 Legs FM7 70 4900 1 Вack M7 177 31329 3 Legs FM8 95 9025 1 Chest M8 182 33124 1 Arms n= 8 ΣΧ- Ex? = Med= Mode M9 45 2025 5 Chest M10 177 31329 1 Chest ΣΧ Mean: M = 8 M11 181 32761 3 Arms M12 24 576 4 Legs SS = n = 12 ΣΧ- Ex? = Median= Mode= The standard deviation: S = ΣΧ Mean (M) = п-1 V8 -1 SS = The standard error of the mean: SM = Vn Level a = 1-0.90 = 0.10. The standard deviation: S = = n-1 12-1 df = n-1 = 8-1 =7 S The Standard error of the mean: SM = vn %3D V12 t value (From the chart) = F1.895 Level a = 0.10 90% confidence interval for the women= M ±CV (t-value) X SM
Hypothesis test:
(M1-M2)-(H1-H2) –
S(M1-M2)
12.5–15.75–0
-3.25
Male
Female
The t statistics: t =
= -2.48 (you will have
1.31
1.31
different answer)
n4= 12
n2= 8
M,= use mean from male group
M2 = use mean from female
group
The test statistics: Z, st = -2.48 (you will have different answer)
Ss,= use SS, from male group
SS2 = use SS2 from female group
df = n1 +n2 - 2
= 12 + 8 -2
E)
Decision:
df = 18
Level a = 0.05. use a from Original project!
A) The null hypothesis ( Ho ) : µ1-H2 = 0
B) The alternative hypothesis ( H1 ) :µ1 – Hz #0
C) The critical value :
t = + 2.101
-1
1
2
3
D) The test statistics : Detail calculations
Figure 2
Pooled variance: S?.
SS,+SS2 _70.22+78 _ 148.22
= 8.23
This is just a sample!
%3D
df1+df2
Make sure you level the CV and don't forget to shade the area of the
11+7
18
region!
s2,
+
n2
8.23
8,23
Estimated standard error: S(M2-M)=
+
8
12
V n1
=V0. 69 + 1.03 = V1.72 = 1.31(you will have different
answer)
F)
Conclusion about the claim: At
% significant level, the data_
provide sufficient evidence to
Transcribed Image Text:Hypothesis test: (M1-M2)-(H1-H2) – S(M1-M2) 12.5–15.75–0 -3.25 Male Female The t statistics: t = = -2.48 (you will have 1.31 1.31 different answer) n4= 12 n2= 8 M,= use mean from male group M2 = use mean from female group The test statistics: Z, st = -2.48 (you will have different answer) Ss,= use SS, from male group SS2 = use SS2 from female group df = n1 +n2 - 2 = 12 + 8 -2 E) Decision: df = 18 Level a = 0.05. use a from Original project! A) The null hypothesis ( Ho ) : µ1-H2 = 0 B) The alternative hypothesis ( H1 ) :µ1 – Hz #0 C) The critical value : t = + 2.101 -1 1 2 3 D) The test statistics : Detail calculations Figure 2 Pooled variance: S?. SS,+SS2 _70.22+78 _ 148.22 = 8.23 This is just a sample! %3D df1+df2 Make sure you level the CV and don't forget to shade the area of the 11+7 18 region! s2, + n2 8.23 8,23 Estimated standard error: S(M2-M)= + 8 12 V n1 =V0. 69 + 1.03 = V1.72 = 1.31(you will have different answer) F) Conclusion about the claim: At % significant level, the data_ provide sufficient evidence to
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