Question

**************ALL OF THE PARTS ARE FROM ONE QUESTION.****************

**********ALL OF THE PARTS ARE FROM ONE QUESTION.***********

(a) The mathematical ability of a random sample of 306 students were as- sessed by means of a test. The test regards the mathematical ability of the student as either: low, medium or high. The handedness of each subject (right-handed of left-handed) was also recorded. The data is pre- sented in the table below. Low MediumHigh Total Left-handed19 393391 1 215 74 306 114 153 Right-handed60 Total 79 i. Compute the table of expected cell counts. ii. Is there evidence of association between handedness and mathemat- ical ability? (a-0.05) [8 marks] b) Two independent samples of size nı- 8 and n2 - 17 were used to compare two population variances. ơ and 0% The hypothesis being tested was: 2 2 2 S118.4 s2 4.7 The test statistic was - 3.91 with significance level a- 0.01. i. Find the critical value for this test. ii. State clearly the conclusion based on the critical value [8 marks] (c) A two-tailed large sample test for comparing two populations means gave i. Find the p-value and state whether the null hypothesis should be ii. On the basis of the decision made in (i), what error could you have the value of the test statistic z - 1.98 rejected at a significance level of 0.01. made? (d) A strong correlation has been found between two variables x and y (r 0.8). Does it mean that there exist a causal relation between x and y? [8 marks] (e) The following scatterplot shows bivariate data on the variables (x, y). A scuss briefly simple linear regression model has been fitted on these data, then the corresponding residual plot is produced. Scatterplot X and Y Residual plot i. List the assumptions of a linear regression model ii. Are all these assumptions fulfilled in the above situation? Discuss briefly

Answer 1

Total 91 215 306 Expected Values Row 1 Row 2 Total Column1 79x9123.493 79x215 55.507 79 Column 2 153x9145.5 153 215 107.5 153 Column 3 74x91 30622.007 306 74 215 51.993 306 306 74 Based on the observed and expected values, the squared distances can be computed according to the following formula: (E O)/E. The table with squared distances is shown below: Squared Distances Column1 Column 2 Column 3 (33-22.0075.492 (41-51.9932.324 3550.929 (19-23.4935 23.493 0.859 Row 1 114 1075 0.393 6055507) 0.364 Row 2 107

ii) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

H0: The two variables are independent

Ha: The two variables are dependent

This corresponds to a Chi-Square test of independence.

Rejection Region

Based on the information provided, the significance level is α=0.05 , the number of degrees of freedom is df=(2−1)×(3−1)=2, so then the rejection region for this test is R={χ^2 :χ^2 >5.991}.

Test Statistics The Chi-Squared statistic is computed as follows 72 XT(Oij Eijr-0.859 + 0.364 + 0.929 + 0.392+5.492 + 2.324 = 10.361

Decision about the null hypothesis

Since it is observed that χ^2=10.361>χc^2 =5.991, it is then concluded that the null hypothesis is rejected.

Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that there is association between two variables, at the 0.05 significance level.

b) i) Critical value for this test with df1= n1-1=8-1=7 and df2= n2-1=17-1=16 at alpha=0.01

F critical= 4.026

ii) Since F test statistic is greater than Fcritical therefore we DO NOT REJECT NULL HYPOTHESIS and may conclude that population variances are equal.

c) i) The P-Value is .047704.

The result is not significant because p > .01.

ii) Type 2 error

Fail to reject H0 when H1 is TRUE.

NOTE: I HAVE DONE THE FIRST FOUR PARTS OF THE QUESTION. PLEASE REPOST (d) and (e) Thank you.