Question

please show all work and chi square graph please

3. In a study to determine whether there is any relationship between temperament and speed of professional advancement, a random sample of 420 men is taken and the following table is constructed: Speed of Advancement Slow Average Fast Very Fast Extreme Self-control 56 28 Average Self-control 47 21 13 26 1 39 19 Self-control Test, at the 0.05 level of significance, the hypothesis that there is no real relationship between temperament and speed of advancement.

Answer 1

(a) The step-by-step instructi ons to obtain Chi-square test for Asscoicati on output using Minitab 18 *Store the Temperament in the column C1 and Speed of Advancement in the rest of the columns of the Minitab18 worksheet *Choose the Stat Tables -Chi-square test for Assoicaition Choose the summarized data in a two-way table *Select all the columns (C2-C5) in the columns containing the variable Labels for the table: Rows: Temperam ent and Columns: Speed of Advancement *Click ok The Chi-square Test for Association output Minitab18

Chi-Square Test for Association: Temperament, ... sional advancement Rows: Temperament Columns: Professional advancement Slow Average Fast Very fast All Extreme self-comfort 42 31 56 28 157 26.54 51.96 53.08 25.42 Average self-control 16 82 47 21 166 28.06 54.94 56.12 26.88 Little Self-control 13 26 39 19 97 15.70 16.40 32.10 32,80 420 All 71 139 142 68 Cell Contents Count Expected count Chi-Square Test Chi-Square DF P-Value Pearson 42.895 6 0.000 Likelihood Ratio 42.366 0.000 6

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Step 1: To, test whether there is relati onship between the two qualitative variables temperament and speed of advancement at the 5% level of significance. We set up the null hypothesis that the temperament and speed of advancement independent. H The null hypothesis that there is no relationship between the two qualitative variables temperam ent and speed of advancement (In dependent) ver sus H The alternative hypothesis there is relati onship between the two qualitative vari ables temperament and speed of advancement (dependent). Step2: The level of significance is a 0.05 Step3: The test statistic is of the chi-square test association output of the Minitab18 is l4,-(48) Τ-Σ (4B 42.985 Step4: The formula for calculating the degrees of freedom is df (r-1) (c-) =(4-1) 3) =(3)(2) = 6 df = 6 Therefore, the degrees of freedom is df 6. Decision Rule: If 205612.592 We reject the null hypothesis Ho Otherwise, we do not reject the null hypothesis H 012.592 The critical value of y2 = 42.985 We reject the null hypothesis Ho Step5: The calculation of the P- value P- value P> 1-P() -1-P(z s 42.985) 1-1 0.00000 P-value 0.0000 <a0.05 We reject the null hypothesis H,. Step6: Conclusion Since, the test stati sti c is 42.985>056 12.592 We reject the null hypothesis Ho Since, the P-value 0.000 0.05 We reject the null hypothesis H There is sufficient evi dence to conclude that there is relationship between the two qualitative variables temperament and speed of advancement at the 5%level of significance The graph of the chi-square critical value, test statistic and the correspondingp-value:

X Distribution Plot Distribution Plot Chi-Square, df-6 0.14 0.12 0.10 0.08 12.592 0.06 Koos,6 0.04 42.985 0.02 P-value 1.2236E-07 0.00 0 X Density