Youarehiredbythegovernortostudywhetherataxonliquorhasdecreasedaverageliquorconsu

Question

Youarehiredbythegovernortostudywhetherataxonliquorhasdecreasedaverageliquorconsump- tion in your state. You are able to obtain, for a sample of individuals selected at random, the difference in liquor consumption (in ounces) for the years before and after the tax. For person i who is sampled randomly from the population, Yi denotes the change in liquor consumption.

(a) Write down the null hypothesis is that there was no change in average liquor consumption. State this formally in terms of .

(b) In terms of , state the alternative is that there was a change in liquor consumption. 1

(c) Now,supposeyoursamplesizeisn=900andyouobtaintheestimatesy =32.8ands=466.4. Calculate the t statistic for testing H0 against H1;Do you reject H0 at the 5% level? at the 1% level?

(d) Would you be able to do the above test if your sample size was only n = 25?

(e) Would you say that the estimated fall in consumption is large in magnitude? Comment on the practical versus statistical significance of this estimate. [Hint: Practical significance is whether the magnitude of the estimate is economically meaningful. Statistical significance is whether you reject that it is zero.]

(f) What has been implicitly assumed in your analysis about other determinants of liquor consump- tion over the two-year period in order to infer causality from the tax change to liquor consump- tion?

Explanation / Answer

(a)

The null hypothesis is

(b)

The alternative hypothesis is

(c)

The test statistics is

Degree of freedom: df=900-1=899

P-value of the test is: 0.0348

Since p-value is greater than 0.01 so we reject the null hypothesis at 1% level of signficance.

Since p-value is less than 0.05 so we fail to reject the null hypothesis at 5% level of signficance.

(d) Yes, otherwise we cannot apply these formulas. Result may be different, using CLT if sample size >30 we can assume that approximation is normal.

But the case where size< 30

Here n= 25

= -32.8/(466.4/25^1/2)

= -0.352

P(-ve value)= P(t<-0.352)

   = 0.3640

Since in both the cases at significance level 5% or at 1% . P value is greater than these two value. We don't have evidence to support any claims. So fails to reject Hu.

(e) No because p-value is not too small.

(f) Assuming that observations are paired.

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