Question

MTH 438 A 4/27/2018 1. A simple random sample of 100 voters were asked which of three candidates they would vote for in an election. The number of voters supporting each candidate is shown in the table. Jones Greene Belton 41 32 27 Test the claim that the candidates are equally popular. Use alpha 0.01. Is there a difference in the leadership qualities of short people differ from those of tall people. Data collected are shown in the table. 2. Do-nothing-er Leader 12 32 Follower Short 14 Tall State and test a null hypothesis at alpha 0.05. Researcher asked mothers of autistic and non autistic children to say what time period they breastfed their children. The results are shown in the table. 3. 2 mo3 to 6 mo None 241 40 More than 6 mo Autistic 198 164 215 Non Autistic 25 27 is there a relationship between breast feeding and autism? Use alpha 0.05.

Answer 1

Answei Date: 10/05/2018 1) hypothesis is that the candidates are not equally popular at 1% level of e significance. The null and alternative hypothesis is, The null hypothesis is that the candidates are equally popular The alternative hypothesisis that the candidates are not equally popular The chi-square test statistic for goodness-of-fit test is By using MINITAB software, find chi-square with 1) Import the data. 2) Select Chi-Square Goodness-of-Fit Test (One Variable) from Tables in Stat menu 3) Select Obsered counts and peelfle proportions 4)Click Ok.

Chi-Square Goodness-of-Fit Test for Observed Counts in Variable: Observed Test Category Observed Proportion Expected 41 0.333333 33.3333 0.333333 33.3333 0.333333 33.3333 Contribution to Chi-Sq 1.76333 0.05333 1.20333 1 32 27 N DF Chi-Sq P-Value 3.02 100 2 0.221 From the MINITAB output, the chi-square test statistic for goodness-of-fit test is 3.02 The p-value is, From the MINITAB output, the chi-square test p-value is 0.221 . Thep-value is 0.221 Decision

The conclusion is that the significant p-value is greater than 0.01 which is 0.221, so the null hypothesis is not rejected at 1% level of significance. There is insufficient evidence to indicate that the candidates are not equally popular. The result is not statistically significant. 2) Test the claim that there is a difference in the leadership qualities of short people differ from those of tall people at 5% level of significance. The null and alternative hypothesis is, The null hypothesis is that there is no difference in the leadership qualities of short people differ from those of tall people. The altemative hypothesis is that there is a difference in the leadership qualities of short people differ from those of tall people. The chi-square test statistic of independence is, By using MINITAB software, find chi-square with help of following steps: 1) Import the data. 2) Select Chi-Square Test (Table in Worksheet) from Tables in Stat menu

3) Select Columns containing the table 4) Cick Ok Chi-Square Test: Leader, Follower, Do-nothing-er Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts Leader Follower Do-nothing-er Total 43 12 19.92 3.146 16.29 1.998 6.79 0.720 14 19.71 1.652 2 52 32 24.08 2.602 8.21 0.595 Total 36 15 95 Chi-Sq10.712, DF- 2, P-Value0.005 From the MINITAB output, the chi-square test statistic of independence is 10.712 The p-value is From the MINITAB output, the chi-square test p-value is 0.005

Thep-value is 0.005 Decision The conclusion is that the significant p-value is less than 0.05 which is 0.005, so the null hypothesis is rejected at 5% level of significance. There is sufficient evidence to indicate that there is a difference in the leadership qualities of short people differ from those of tall people. The result is statistically significant. 3) Test the claim that there is a relationship between breast feeding and autism at 5% level of significance. The null and alternative hypothesis is, The null hypothesis is that there is no relationship between breast feeding and autism The alternative hypothesis is that there is a relationship between breast feeding and autism

The chi-square test statistic of independence is, By using MINITAB software, find chi-square with help of following steps: 1)Import the data 2) Select Chi-Square Test (Table in Worksheet) from Tables in Stat menu 3) Select Columns containing the table 4)Click Ok Chi-Square Test: None, 2 mo, 3 to 6 mo, More than 6 mo Expected counts are printed below observed counts Chi-Square contributions are printed below expected counts More than one 2 mo 198 240.94 191.21 0.000 0.241 3 to 6 mo 164 163.77 0.000 6 mo Total 818 215 222.08 0.226 241 25 40.06 31.79 0.000 1.450 40 27 27.23 0.002 136 36.92 1.357 Total 281 223 191 259 954 Chi-Sq-3.276, DF- 3, P-Value0.351

From the MINITAB output, the chi-square test statistic ofindependence is 3.276 Thep-value is, From the MINITAB output, the chi-square testp-value is 0.351 The p-value is 0.351 Decisiorn The conclusion is that the significant p-value is higher than 0.05 which is 0.351, so the null hypothesis is not rejected at 5% level of significance. There is insufficient evidence to indicate that there is a relationship between breast feeding and autism. The result is not statistically significant