Tutorial 5 - eview PDF

Preview

Summary

personal work...

Description

Tutorial 5 Exercise 5 a) Estimate the ratio of the two population variances with 95% confidence Sample variances are: s12 =13.621192 = 185.54 s22 = 10.039922 = 100.80 The confidence level is (1-α) × 100% = 95%. Fn1-1,n2-1,α/2 = F19,19,0.025 = 2.53 and Fn1-1,n2-1,1-α/2 = F19,19,0.975 =

1

1 =0.395 = F 19,19,0.025 2.53

The confidence interval estimate is

(

s21/s 22

s21 /s22

,

F n −1, n −1 ,α /2 F n −1,n −1,1−α /2 1

2

1

)=(

185.54/100.80 185.54 / 100.80 )=(0.728,4 .660) , 0.395 2.53

2

Therefore, with 95% confidence interval, the ratio of the variances of the two populations; purchasers and non- purchasers is between 0.728 and 4.66.

b) Can we conclude at the 5% significance level that the population variances differ? What do you conclude if we reduce the significance level to 1%? Do the calculations both manually and with EViews. The hypotheses are

H0:

α21 2

α2

=1, H A :

α 12 α 22

≠1

This is a two-tail test and can rely on the 95% c.i. The interval in (a) includes 1, thus the null hypothesis of equal variances cannot be rejected at the 5% significance level. The observed test statistic value is 2

F obs=

s1

s22

=

185.54 =1.841 100.80

The observed test statistic value is between the lower and upper critical values, we do not reject the null hypothesis.

The p-value of the F-test is 0.1927, so the null hypothesis of equal variances cannot be rejected at the 5% significance level. The three out of other four tests performed also not rejecting H0 so we can be quite confident in this conclusion. Therefore, we can conclude that at 5% significance level that the population variances are differ.

Exercise 6

In a public opinion survey, 60 out of a sample of 100 high-income voters and 40 out of a sample of 75 low-income voters supported the introduction of a new national security tax. Can we conclude at the 5% level of significance that there is a difference between the proportions of high- and lowincome voters favoring a new national security tax? Perform the required test manually. Let X1 be the population of the opinions of high-income voters and X2 the population of the opinions of low-income voters. The proportions of voters who are in favor of a new tax is p1 and p2. The hypotheses are;

H 0 : p 1− p 2=0, H A : p 1− p2 ≠ 0 The sample proportions are

^p1=

60 40 =0.6, ^p 2= =0.53 100 75

The estimates of the corresponding population proportions;

n ^p 1=100 ( 0.6) =60,n ( 1− ^p 1 )=40,n ^p2=75 ( 0.53 )=39.75, n( 1−^p2 )=35 Both of them are bigger than 5, so we can rely on the normal approximation and perform a Z-test The critical values; ±Zα /2 = ± Z0.025 = ±1.96 The null hypothesis the hypothesized difference between the two population proportions is D 0 = 0, so we can estimate the common population proportion from the pooled sample

^p=

f 1 + f 2 60+39.75 = =0.57 n1 + n2 100+75

The estimate of the standard error is

√

√

1 1 1 1 + )=0.0756 s ^p − ^p = ^p ^q ( + )= 0.57 ×0.43 ×( n1 n 2 100 75 1

2

Test statistic is

z obs=

^ 1− ^p 2 0.60−0.53 p = =0.9259 0.0756 s ^p − ^p 1

2

The calculated tests statistics is between ±1.96 the lower and upper critical values, thus we cannot reject the null hypothesis. Therefore, at 5% significance level there is no significant difference between the proportions of high income and low income voters....