# 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.