ECON 3U03 ASSIGNMENT #2
The coefficient of EXPER indicates that on average a technical artist’s quality rating is increased
up by 0.076 for each extra year of experience.
t = t = 2.074, 95 % confidence interval for B is given by: b + 2.074 * 0.044 = (-0.015,
c (0.975, 22) 2 2 -
0.167). Therefore 95% confient that the procedure we have used for constructing a confidence
interval will yield an interval that include the true parameter B 2.
To test 02:0Hβ= against 12:0,Hβ≠ we use the test statistic t = b2/se(b2) = 0.076/0.044 = 1.727. The t
critical value for a two tail test with N − 2 = 22 degrees of freedom is 2.074. Since −2.074 < 1.727 < 2.074
we fail to reject the null hypothesis.
To test 02:0Hβ= against 12:0,Hβ> we use the t-value from part (c), namely 1.727t=, but the right-tail
critical value (0.95,22)1.717ctt==. Since 1.7271.717>, we reject 0H and conclude that 2β is positive.
Experience has a positive effect on quality rating.
The p-value of 0.0982 is given as the sum of the areas under the t-distribution to the left of −1.727 and to
the right of 1.727. We do not reject 0H because, for 0.05,α= p-value > 0.05. We can reject, or fail to
reject, the null hypothesis just based on an inspection of the p-value. Having the p-value > α is equivalent
to having 2.074ctt<=. 3.4
b1= t * se(b )1= 1.257 * 2.174 = 2.733
se(b 2=b /2 = 0.180/5.754 = 0.0313
p-value = 2 × (1 – P (t <1.257)) = 2 × (1 − 0.8926) = 0.2147
The estimated slope b 0.182 =dicates that a 1% increase in males 18 and older, who are high
school graduates, increases average income of those males by $180. The positive sign is as
expected; more education should lead to higher salaries.
Using t c t (0.995, 49).68, a 99% confidence interval for the slope is given by b ± t se(2 ) c 2
0.180 ± 2.68 × 0.0313 = (0.096, 0.264)
For testing H :0 =20.2,H : β1≠ 022 we calculate:
t=0.180 – 0.2/0.0313 = -0.639
The critical values for a two-tailed test with a 5% significance level and 49 degrees of freedom
are ± tc=± 2.01. Since t = −0.634 lies in the interval (−2.01, 2.01), we do not reject H . The nu0l
hypothesis suggests that a 1% increase in males 18 or older, who are high school graduates,
leads to an increase in average income for those males of $200. Non-rejection of H means 0
that this claim is compatible with the sample of data
A: H :β = 1 versus H : β ≠ 1. The economic relevance of this test is to test whether the return on
0 j 1 j
the firm’s stock is risky relative to the market portfolio. Each beta measures the volatility of the
stock relative to the market portfolio and volatility is often used to measure risk. A beta value of
one indicates that the stock’s volatility is the same as that of the market portfolio. The test
statistic given H0 is true, is
t = bj– 1/se(b) j t (118)
The rejection region is t < -1.980 and t> 1.980, where t (0.975, 118)980.
The results for each company are given in the following table:
Stock t-value Decision
Stock t-value Decision rule
Disney t = (0.9593 – 1)/ -1.98