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8 a) ttest hwage 30 One-sample t test Variable l Obs Mean Std. Err. Std. Dev. [95% Conf. Interval]

175 33.52778 1.73398 22.9384 30.10544 36.95012 mean mean(hwage) t 2.0345 Ho: mean-30 degrees of freedom174 Ha: mean<30 Ha: mean !-30 Ha: mean>30 Pr(T <t) 0.9783 Pr(ITI Itl) 0.0434 Pr(T>t) 0.0217

Mean Std. Err. (95% Conf. Interval] hwage 33.52778 1.73398 38.18544 36.95812 . summarize hwage Variable I Obs Mean Std. Dev Min Max hwage 175 33.52778 22.9384 2.051282 149.1966 . return list scalars: r(N) 175 r(sum w) -175 r(mean) 33.52778188831168 r(Var) 526.1781389182357 r(sd)22.93839878714886 r(min) 2.051282167434692 r(max) 149.1966400146484 r(sum) 5867.361676454544 b) display invttall(174,9925) 1.9736914 . generate avg r(mean) . generate se r(sd)8. Statistical Inference: Hypothesis Tests Continue with the same data as question7 Test at significance level 0.05 Ho population mean hourly wage $30 against Ha: it does not equal S30 (a) Use command ttest hwage = 30. Use the p-value approach. State clearly your conclusion (b) Repeat (a) but use the criticalt-value approach. State clearly your conclusion (c) Now calculate the same t test statistic manually (i.e. using , , Perform the test using the p-value approach

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Answer #1

a) H0 = 30

Ha != 30

significance level = 5%

n = 174

P value = 0.0434

since the p value < significance level

I.e .0434 < 0.05

we reject the null hypotheses that mean wage = 30

b )

H0 = 30

Ha != 30

tabulated t value from table = 2.0345

t value at 5% significance level with n-1= 174 -1 = 173 degrees pf freedom from the table = 1.96

since the calculated t value > t value from table

we reject the null hypotheses that mean wage = 30

c)

H0 = 30

Ha != 30

t = calculated mean - populated mean / standard error

33.52778 - 30/ 1.73398

= 2.0345(rounded off)

t value at 5% significance level with n-1= 174 -1 = 173 degrees pf freedom from the table = 1.96

since the calculated t value > t value from table

we reject the null hypotheses that mean wage = 30

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