This research is a random sample of n-12 men and n=8 women. For a drawing of a random sample, we number each individual and then by using a random number table we select 12 men. In the same way we select 8 women and start counting from a number on a random table and then continue to the next 8 numbers. Therefore, since we are using a random table, these drawings are completely random, and the sample acquired is a random sample. After selecting the random samples, we ask the following questions below: How many days per week do they exercise? What is their favorite exercise? The Mean and Stand Deviation of the number of seconds the subject stayed balanced The Median number of days per week exercised The Mode of favorite exercise The 90% confidence Interval of the mean Construct a complete hypothesis test and determine if the two groups have significantly different balance using, u=0.05 The data collected from the 12 men & the calculations asking in the project are shown in the table below: Male Time in Second(X) Number of Days Workout Favorite Workout M1 83 6889 1 Arms M2 65 4225 6 Chest M3 43 1849 3 Arms M4 34 1156 2 Back M5 92 8464 1 Arms M6 87 7569 6 Legs M7 177 31329 3 Legs M8 182 33124 1 Arms M9 45 2025 5 Chest M10 177 31329 1 Chest M11 181 32761 3 Arms M12 24 576 4 Legs n = 12 1190 161296 Median= 3 Mode= Arms Mean (M)= 99.166666666667 SS= 43287.66667 For Median arrange the data in ascending order Sr. No. Data 1 1 2 1 3 1 4 1 5 2 6 3 7 3 8 3 9 4 10 5 11 6 12 6 Median= (n/2) observation + (n/2+1) observation 2 = (n/12) observation +(12/2+1) observation 2 =6th observation +7th observation 2 =3+3 3 =3 For Mode= Arms Mean: (M) =∑X/N=1190/12= 99.16667 SS= ∑X2−(∑X2) N =161296−11902 12 =43287.66667 The standard deviation: S = = = 62.7315 The Standard error of the mean: = = = 18.1090 Level α = 0.10 df = n-1 = 12-1= 11 t Value (From the chart) = ±1.796 90% confidence interval for the men = M ± CV X The data collected from the 8 women & the calculations asking in the project are shown in the table below: Female Time in second(X) Square the x value from left column for each row below Number of Days Workout Favorite Workout FM1 67 6 Arms FM2 71 6 Back FM3 31 7 Back FM4 184 7 Chest FM5 29 5 Chest FM6 49 4 Legs FM7 50 4 Chest FM8 57 4 Arms n = 8 ∑X = Med= Mode Mean: M = = SS = The standard deviation: S = = = The standard error of the mean: = = = Level α = 1-0.90 = 0.10. df = n-1 = 8-1 =7 t value (From the chart) = 1.895 90% confidence interval for the women= M ±CV (t-value) X Hypothesis test: Male Female = 12 = 8 = use mean from male group = use mean from female group = use from male group = use from female group df = + - 2 = 12 + 8 -2 df = 18 Level = 0.05. use from Original project! The null hypothesis ( ) : - = 0 The alternative hypothesis ( ) : ≠ 0 The critical value: t = ± 2.101 The test statistics: Detail calculations Pooled variance: = = = = 8.23 This is just a sample! Estimated standard error: = = = = = 1.31(you will have different answer) The t statistics: t = = = = -2.48 (you will have different answer) The test statistics: = -2.48 (you will have different answer) Decision: Figure 2 Make sure you level the CV and don’t forget to shade the area of the region! Conclusion about the claim: At _____% significant level, the data_______ provide sufficient evidence to ……………………………………………………………
This research is a random sample of n-12 men and n=8 women. For a drawing of a random sample, we number each individual and then by using a random number table we select 12 men. In the same way we select 8 women and start counting from a number on a random table and then continue to the next 8 numbers. Therefore, since we are using a random table, these drawings are completely random, and the sample acquired is a random sample.
After selecting the random samples, we ask the following questions below:
- How many days per week do they exercise?
- What is their favorite exercise?
- The Mean and Stand Deviation of the number of seconds the subject stayed balanced
- The
Median number of days per week exercised - The
Mode of favorite exercise - The 90% confidence Interval of the mean
- Construct a complete hypothesis test and determine if the two groups have significantly different balance using, u=0.05
The data collected from the 12 men & the calculations asking in the project are shown in the table below:
|
Male |
Time in Second(X) |
|
Number of Days Workout |
Favorite Workout |
|
M1 |
83 |
6889 |
1 |
Arms |
|
M2 |
65 |
4225 |
6 |
Chest |
|
M3 |
43 |
1849 |
3 |
Arms |
|
M4 |
34 |
1156 |
2 |
Back |
|
M5 |
92 |
8464 |
1 |
Arms |
|
M6 |
87 |
7569 |
6 |
Legs |
|
M7 |
177 |
31329 |
3 |
Legs |
|
M8 |
182 |
33124 |
1 |
Arms |
|
M9 |
45 |
2025 |
5 |
Chest |
|
M10 |
177 |
31329 |
1 |
Chest |
|
M11 |
181 |
32761 |
3 |
Arms |
|
M12 |
24 |
576 |
4 |
Legs |
n = 12 1190 161296 Median= 3 Mode= Arms
Mean (M)= 99.166666666667
SS= 43287.66667
For Median arrange the data in ascending order
|
Sr. No. |
Data |
|
1 |
1 |
|
2 |
1 |
|
3 |
1 |
|
4 |
1 |
|
5 |
2 |
|
6 |
3 |
|
7 |
3 |
|
8 |
3 |
|
9 |
4 |
|
10 |
5 |
|
11 |
6 |
|
12 |
6 |
Median= (n/2) observation + (n/2+1) observation
2
= (n/12) observation +(12/2+1) observation
2
=6th observation +7th observation
2
=3+3
3
=3
For Mode= Arms
Mean: (M) =∑X/N=1190/12= 99.16667
SS= ∑X2−(∑X2)
N
=161296−11902
12
=43287.66667
The standard deviation: S = = = 62.7315
The Standard error of the mean: = = = 18.1090
Level α = 0.10
df = n-1 = 12-1= 11
t Value (From the chart) = ±1.796
90% confidence interval for the men = M ± CV X
The data collected from the 8 women & the calculations asking in the project are shown in the table below:
|
Female |
Time in second(X) |
Square the x value from left column for each row below |
Number of Days Workout |
Favorite Workout |
|
FM1 |
67 |
6 |
Arms |
|
|
FM2 |
71 |
6 |
Back |
|
|
FM3 |
31 |
7 |
Back |
|
|
FM4 |
184 |
7 |
Chest |
|
|
FM5 |
29 |
5 |
Chest |
|
|
FM6 |
49 |
4 |
Legs |
|
|
FM7 |
50 |
4 |
Chest |
|
|
FM8 |
57 |
4 |
Arms |
n = 8 ∑X = Med= Mode
Mean: M = =
SS =
The standard deviation: S = = =
The standard error of the mean: = = =
Level α = 1-0.90 = 0.10.
df = n-1 = 8-1 =7
t value (From the chart) = 1.895
90% confidence interval for the women= M ±CV (t-value) X
Hypothesis test:
Male Female
= 12 = 8
= use mean from male group = use mean from female group
= use from male group = use from female group
df = + - 2
= 12 + 8 -2
df = 18
Level = 0.05. use from Original project!
- The null hypothesis ( ) : - = 0
- The alternative hypothesis ( ) : ≠ 0
- The critical value: t = ± 2.101
- The test statistics: Detail calculations
Pooled variance: = = = = 8.23 This is just a sample!
Estimated standard error: = =
= = = 1.31(you will have different answer)
The t statistics: t = = = = -2.48 (you will have different answer)
The test statistics: = -2.48 (you will have different answer)
- Decision:
Figure 2
Make sure you level the CV and don’t forget to shade the area of the region!
- Conclusion about the claim: At _____% significant level, the data_______ provide sufficient evidence to ………………………………………………………………..
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