T-test: Statistical Hypothesis Testing and Mean Age Essay

T-test
A t-test is a hypothesis test in which the test statistic follows a Student’s t-distribution under the null hypothesis - T-test: Statistical Hypothesis Testing and Mean Age Essay introduction. There are several different test statistics that fall into the category of a t-test. One-Sample t-test

x  0
s
n
df  n  1

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t

Independent Two-Sample t-test

t

x1  x2
2
sx1x2
n

1 2
2
sx1  sx2
2
df  2n  2
sx1x2 

Unequal Sample Size Two-Sample t-test

t

x1  x2
1 1
sx1x2

n1 n2

sx1x2 

 n1  1 sx2   n2  1 sx2
1

2

n1  n2  2

df  n1  n2  2
Unequal Sample Size and Unequal Variance Two-Sample t-test

t

x1  x2
sx1x2

sx1x2 

df 

2
s12 s2

n1 n2
2
 s12 s2 
 

 n1 n2 

2

  s 2 2  s 2 2 
2
 1 
  
  n1    n2  
  n  1  n  1 
2
 1




Dependent t-test for two samples
xD   D
sD
n
df  n  1

t

Some assumptions are made in order to use the t-test.
Assumptions for t-test
1. The data comes from normal distribution or the sample size is greater than 30. 2. The sample is a simple random sample
The t-test is a hypothesis test. There are seven steps for a hypothesis test. 1.
2.
3.
4.
5.
6.
7.

State the null hypothesis
State the alternative hypothesis
State the level of significance
State the test statistic
Calculate
Statistical Conclusion
Experimental Conclusion

Example One-Sample t-test
The age of the head of household residents of Phoenix, Arizona has increased in the recent years. Ten years ago the average age of head of household in Phoenix, Arizona was 33. A random sample of 40 head of household residents in Phoenix, Arizona was taken and the ages were recorded as follows.

25
46
63
37
72
20
47
41
59
64
32
37
80
59
63
31
38
35
45
36
45
22
36
44
48
69
31
36
34
37
70
33
65
29
31
76
51
25
27
21
27

We wish to use this data to test if the hypothesis that the average age of head of household residents in Phoenix, Arizona has increased.
This is a hypothesis test so we will need to go through the seven steps of a hypothesis testing.
Step 1: Null Hypothesis
Since in the past the mean Head of Household age was 33 that is the statement of no effect which is what the null hypothesis gives.

H 0 :   33
Step 2: Alternative Hypothesis
We want to know if the mean head of household age has increased so the alternative is

H A :   33
This is a one-tailed test as we are only looking at a one-sided alternative. Step 3: Level of Significance

  0.05
Step 4: Test Statistic
The test statistic needed is for a one-sample t-test. It is one-sample because we are only looking at one set of data values. The other requirements for a t-test have been met as the sample size is 40 > 30 and the sample is a simple random sample.

x  0
s
n
df  n  1

t

Step 5: Calculations
First the sample mean of the data should be calculated.
x

 x  1787  43.58537
n

40

Now find the sample standard deviation

s

 x  x 

2

 n  1

 16.53931

(For explanations of how to calculate the mean and the standard deviation see Measure of Center and Variation respectively)
Plug this information into the formula for the test statistic. x  0 43.58537  33 10.58537


s
16.53931
2.615095
n
40
 4.047795

tobs 
tobs

df  n  1  40  1  39

Find the critical value for a level of significance of 0.05 and 39 degrees of freedom. Use the student t-distribution table or a t-score calculator. (i.e. http://www.stattrek.com/online-calculator/t-distribution.aspx) t 0.05,df 39  1.685

Step 6: Statistical Conclusion
The null hypothesis is rejected if the observed test statistic is more extreme that the critical value. Since the alternative hypothesis is right tailed (greater than) then we reject if the observed value is greater than
the critical value.

Since tobs  4.047795  1.685  t 0.05,df 39 then we reject the null hypothesis at a level of significance of 0.05.
Step 7: Experimental Conclusion
There is sufficient evidence to indicate that the mean age of the Head of Household in Phoenix, Arizona has increased at a level of significance of 0.05. Example Dependent t-test for two samples
A research wishes to prove that a diet results in significant weight loss. A random sample of 40 people are weighed and then follow the diet for 6 weeks and weighed. The 40 participants were all in need of weight loss and of approximately the same health standings. The before and after weights are in the following table. Before

After
257 240
244 236

151
229
151
296
151
219
265
177
202
194
198
175
194
281
148
260
159
176
255
212
287
242
248
224
185
210
157
170
293
152
263
154
287
249
161
244
267
300
300

145
200
146
277
147
201
215
160
190
172
180
170
183
233
240
200
250
267
202
195
243
204
205
199
164
173
150
152
232
140
202
142
260
195
153
204
207
245
234

Step 1: Null Hypothesis
The null hypothesis is the statement of no effect, which in this case, means no weight loss, so the mean weight loss would be 0.

H 0 : D  0
Step 2: Alternative Hypothesis
The researcher believes that the diet causes weight loss, so the difference
between the weight before and after the diet should be greater than 0.

H A : D  0
Step 3: Level of Significance

  0.05
Step 4: Test statistic
We are comparing before and after results. This means we have paired data so the test statistic needed is the dependent test statistic for two samples. xD   D
sD
n
df  n  1

t

Step 5: Calculations
To start the calculation, we need to find the difference between the before and after weights for the 40 participants.
Before
After
257 240
244 236
151 145
229 200
151 146
296 277
151 147
219 201
265 215
177 160
202 190

Difference
17
8
6
29
5
19
4
18
50
17
12

194
198
175
194
281
148
260
159
176
255
212
287
242
248
224
185
210
157
170
293
152
263
154
287
249
161
244
267
300
300

172
180
170
183
233
140
200
150
167
202
195
243
204
205
199
164
173
150
152
232
140
202
142
260
195
153
204
207
245
234

22
18
5
11
48
8
60
9
9
53
17
44
38
43
25
21
37
7
18
61
12
61
12
27
54
8
40
60
55
66

Find the sample mean of the differences and the sample standard deviation.
(For explanations of how to calculate the mean and the standard deviation see Measure of Center and Variation Respectively)

x
s

 x  1134  27.65854
n

40

 x  x 
n 1

2

 19.85146

This information can be plugged into the test statistic.
xD   D 27.65854  0 27.65854


 8.81184
sD
19.85146
3.13879
40
n
df  n  1  40  1  39

tobs 

The critical value can be found using the t-table or a t-value calculator. (http://www.stattrek.com/online-calculator/t-distribution.aspx) t 0.05,df 39  1.685

Step 6: Statistical Conclusion
Reject the null hypothesis if the observed test statistic is more extreme than the critical value. Since the alternative hypothesis is right tailed (greater than) then we reject if the observed value is greater than the critical value.

Since tobs  8.81184  1.685  t 0.05,df 39 then we reject the null hypothesis. Step 7: Experimental Conclusion
There is significant evidence to indicate that at a level of significance of 0.05 the diet leads to statistically significant weight loss.
Example Unequal Sample Size Two-Sample t-test
An advertising company wishes to place billboards for their new client in cities where they will do the most good. They believe this means cities with the youngest population for the new product. They wish to compare Phoenix and Los Angeles. They took a random sample of adult residents from each city. The ages of 45 LA residents and 40 Phoenix residents are shown below. The researchers wish to prove that the mean age of the adult residents of Phoenix is less than the mean age of the adult residents of LA. Phoenix LA

85
50
64
46
24
32
87
24
18

88
37
54
85
84
73
25
37
23

85
31
21
35
29
58
69
31
27
89
34
26
57
50
26
28
23
47
50
86
65
55
30
46
19
28
19
80
81
21
31

59
56
78
26
72
22
57
31
68
88
77
90
90
89
82
68
90
30
19
22
35
42
46
64
78
58
83
56
30
63
54
70
78
79
82
50

Step 1: Null Hypothesis
The null hypothesis is the statement of no effect so that would be that the mean age of the adult residents of the two cities is the same.

H 0 : 1  2
Where population 1 is Phoenix and population 2 is LA.
Step 2: Alternative Hypothesis
The researchers’ theory is that the mean age of the adult residents in Phoenix (population 1) is less than the mean age of the adult residents in LA (population 2)

H A : 1  2
Step 3: Level of Significance

  0.05
Step 4: Test statistic
The test is comparing two data sets that have unequal sample size. n1  40 and n2  45 so we use the independent two sample t-test statistic for unequal sample size.

t

x1  x2
1 1
sx1x2

n1 n2

sx1x2 

 n1  1 sx2   n2  1 sx2
1

2

n1  n2  2

df  n1  n2  2
Step 5: Calculations
The calculations start with the sample mean for Phoenix and LA. x1 

 x  1807  45.175

x2 

 x  2688  59.733

n

40

n

45

Now find the standard deviation for each sample.

sx1 
sx2 

 x  x 

2

 22.9904

n 1

 x  x 
n 1

2

 23.0478

Plug this information into the test statistic.

 n1  1 sx2   n2  1 sx2

sx1x2 

1

2

n1  n2  2

 40  1 22.9904    45  1 23.0478  2

sx1x2 

2

40  45  2

 39  528.5585    44  531.2011

sx1x2 

83

sx1x2 

20613.7815  23372.8477
83

sx1x2 

43986.6292
83

sx1x2  529.9594
sx1x2  23.0208
tobs 

x1  x2
45.175  59.733

1 1
1
1
sx1x2

23.0208

n1 n2
40 45

14.558
14.558

23.0208 0.04722 5.0026
 2.9101

tobs 
tobs

df  n1  n2  2  40  45  2  83

Find the critical value by using either the t-distribution table or t-distribution calculator. t 0.05,df 83  1.663

Step 6: Statistical Conclusion
Reject the null hypothesis if the observed test statistic is more extreme than the critical value. Since we are interested in an alternative hypothesis that the mean of population 1 is less than the mean of population 2. The Rejection will occur if the observed test statistic is less than the critical value.

Since tobs  2.9101  1.663  t 0.05,df 83 then we reject the null hypothesis. Step 7: Experimental Conclusion
There is significant evidence to indicate that the mean age in Phoenix residents is less than the mean age of the adult residents in LA.

Example Independent Two-Sample t-test
An advertising company wishes to place billboards for their new client in cities where they will do the most good. They believe this means cities with the youngest population for the new product. They wish to compare Phoenix and Los Angeles residents. They took a random sample of adult residents from each city. The ages of 40 LA residents and 40 Phoenix residents are shown below. The researcher wishes to prove that the mean age of the adult residents of Phoenix is less than the mean age of the adult residents of LA.

Phoenix

LA

85

88

50

37

64

54

46

85

24

84

32

73

87

25

24

37

18

23

85

59

31

56

21

78

35

26

29

72

58

22

69

57

31

31

27

68

89

88

34

77

26

90

57

90

50

89

26

82

28

68

23

90

47

30

50

19

86

22

65

35

55

42

30

46

46

64

19

78

28

58

19

83

80

56

81

30

21

63

31

54

Step 1: Null Hypothesis
The null hypothesis is the statement of no effect which is that the mean age of population 1 (Phoenix) is the same as the mean age of population 2 (LA).

H 0 : 1  2
Step 2: Alternative Hypothesis
The alternative hypothesis of interest here is that the mean age of the population 1 (Phoenix) is less than the mean age of the population 2 (LA).

H A : 1  2
Step 3: Level of Significance

  0.05

Step 4: Test statistic
Independent two sample t-test statistic.

t

x1  x2
2
sx1x2
n

1 2
2
sx  sx2
2 1
df  2n  2
sx1x2 

Step 5: Calculations
To begin the calculations find the sample mean and variance of each sample. x1 

 x  1807  45.175

x2 

 x  2329  58.225

2

sx

n

40

n

40

x  x 

sx

 528.5583

n 1

1

2

2

x  x 

n 1

2

2

 561.0506

Use this to begin calculating the test statistic.

sx1x2 

1 2
2
sx1  sx2
2

sx1x2 

1
 528.5583  561.0506 
2

sx1x2 

1
1089.6089 
2

sx1x2  544.80445
sx1x2  23.341
x1  x2 45.175  58.225
13.05


 23.341 0.2236 
2
2
sx1x2
 23.341
n
40
13.05
tobs 
 2.5004
5.2192
df  2n  2  2  40   2  78
tobs 

Now find the critical value for   0.05 and degrees of freedom = 78 using either a tdistribution table or calculator.

t 0.05,df 78  1.665

Step 6: Statistical Conclusion
We reject the null hypothesis if the observed test statistic is more extreme than the critical value. We reject the null hypothesis if the observed test statistic is less than the critical value.
Since tobs  2.5004  1.665  t 0.05,df 78 then reject the null hypothesis. Step 7: Experimental Conclusion here is significant evidence to indicate that the mean age of Phoenix residents is less than the mean age of LA residents at a level of significance of 0.05. Example Unequal Sample Size and Unequal Variance Two-Sample t-test The publishers of a health magazine are going to publish an article on the healthiest cities. They believe LA and Phoenix are equal on all accounts except they need to compare the average weights of the adult residents. The variance of the two populations is known to be different. A random sample of 40 adult residents from each city was taken and their weights measured.

Phoenix LA
280
195
146
186
121
258
256
125
279
190
167
232
186
142
264
281
290
257
279
158
153
204
246
162
134
118
151
142
126
134
246
215
270
132
228
221
274
174

159
134
146
295
212
135
127
158
255
164
231
179
251
220
174
146
121
296
120
130
237

145
256
163
168
226
183
151
235
255
267
156
231
233
255
199
212
163
155
166
127
269

Step 1: Null Hypothesis
The null hypothesis is the statement of no effect. For this problem that means that the average weights of the adult residents of Phoenix and LA are the same. Let Phoenix be population 1 and LA be population 2.

H 0 : 1  2
Step 2: Alternative Hypothesis
The alternative hypothesis is would be that one city has an average weight less than the other. We will test the claim that the average weight of the adult residents of Phoenix (population 1) is less than the average weight of the adult residents of LA (population 2)

H A : 1  2
Step 3: Level of Significance

  0.05

Step 4: Test statistic
Since we are comparing two different populations with different variances the following test statistic is needed. This test statistic can be used even though we have equal sample size for each population.

t

x1  x2
sx1x2
2
s12 s2

n1 n2

sx1x2 

df 

2
 s12 s2 
  
 n1 n2 

2

  s 2 2  s 2 2 
2
 1 
  
  n1    n2  
  n  1  n  1 
2
 1




Step 5: Calculations
Find the mean and variance of each set of sample data.
x1 
x2 
2

sx

1

2

sx

2

x

1

n1

x

2

n2

 199.65
 193.525

x  x 

2

 3701.721

n 1

x  x 

n 1

2

 2254.358

Use this information to calculate the test statistic

sx1x2 

2
s12 s2
3701.721 2254.358



n1 n2
40
40

sx1x2  92.5401  56.3589  148.8991
sx1x2  12.2024
tobs 

x1  x2 199.65  193.525
6.125


sx1x2
12.2024
12.2024

tobs  0.5020
2
 s12 s2 
 

 n1 n2 

2

2

 3701.721 2254.358 



40
40


df 

2
2
2
2
2
  s2 
 s2     3701.721   2254.358  
1
 
 
 
   
40  
40
 
  n1 
 n2    


  n  1  n  1    40  1
 40  1 
2
 1
 



 


df 

148.9019 

2

 219.5952  81.4444

22171.7982
 73.6508  74
301.0396

Find the critical value for the t-distribution with alpha=0.05 and degrees of freedom of 74. Use either a table or a t-distribution calculator available on the web. t 0.05,df 74  1.666

Step 6: Statistical Conclusion
Reject the null hypothesis if the observed test statistic is more extreme than the critical value. Since we are looking at a left tailed alternative hypothesis that we reject the null hypothesis if the observed test statistic is less than the critical value. Since tobs  0.5020  1.666  t 0.05,df 74 then we fail to reject the null hypothesis. Step 7: Experimental Conclusion

There is not sufficient evidence to indicate that the mean weight of Phoenix is different than the mean weight of LA at a level of significance of 0.05.

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