A-Use a 0.01 significance level to test the claim that the proportion of men who

ID: 3324875 • Letter: A

Question

A-Use a 0.01 significance level to test the claim that the proportion of men who plan to vote in the next election is the same as the proportion of women who plan to vote. 300 men and 300 women were randomly selected and asked whether they planned to vote in the next election. The result are shown below: (3 marks) B- According to Berford’s law, a variety of different data sets includes numbers with leading (first) digits that follow the distribution shown in the nine row of Table 1. The bottom row lists the frequencies of leading digits of the populations of all 120 countries from New York and California combined. Test the claim that those 120 countries have populations with leading digits that follow Benford’s law. (3marks) Leading digit Benford’s law:Distibution of leading digit CA and NY country population 1 30.1% 33 2 17.6% 22 3 12.5% 10 4 9.7% 15 5 7.9% 10 6 6.7% 9 7 5.8% 5 8 5.1% 7 9 4.6% 9 (From Table A-4 the Critical Value for 2 at degree of freedom =8 is 15.507 and p-value=0.652) 3. A- One-way ANOVA table given below has 5 treatment and with the 15 observations per treatment. Find the values for the missing entries and compute F? (3marks) Source Df SS MS F Treatment 26.3 Error 25 4 Total 29 126.3 B- Evaluate the F test statistic for the following data, where n=4 for each sample (3marks) Group 1 Group 2 Group 3 Group 4 Sample Mean 6.6 3.4 3.0 1.2 Sample Variance 5.35 1.35 2.5 2.65

Explanation / Answer

Solution for the Question on incomplete ANOVA

The given figures are: DF: 25 for Error and 29 for Total. From this we conclude the DF is 4 [i.e., 29 - 25] for Treatment [This is also corroborated by the given fact that there are 5 treatments and hence DF = 5 – 1 = 4]

On SS, given SS = 26.3 for Treatment and 126.3 for Total, SS for Error = 126.3 – 26.3 = 100.

[This is also corroborated by the given fact that MS for Error is 4 and as already obtained above DF for Error is 25 and hence SS = 25 x 4 = 100.]

For Treatment, with DF = 4 and SS = 26.3, MS = 26.3/4 = 6.575.

So, F = MS (Treatment)/MS(Error) = 6.575/4 = 1.64375.

F has F-distribution with degrees of freedom 4 and 25.

Hence, p-value = P(F4,25 > 1.64375) = 0.194587

=> even at 19% level of significance, the null hypothesis is rejected.

ANOVA TABLE [given figures are in bold and other figures are derived]

Source

Fcal

p-value

Significance

Treatment

26.3

6.575

1.64375

0.194587

At 19%

Error

100.0

4.0

Total

126.3

DONE