A life insurance company wishes to examine the relationship between the amount o

Question

A life insurance company wishes to examine the relationship between the amount of life insurance held by a family and family income. From a random sample of 20 households, the company collected the data in the file insur.dat. The data are in units of thousands of dollars. (a) Estimate the linear regression with dependent variable INSURANCE and independent variable INCOME. Write down the fitted model and draw a sketch of the fitted function. Identify the estimated slope and intercept on the sketch. Locate the point of the means on the plot. (b) Discuss the relationship you estimated in (a). In particular, (i) What is your estimate of the resulting change in the amount of life insurance when income increases by $1,000? (ii) What is the standard error of the estimate in (i), and how do you use this standard error for interval estimation and hypothesis testing? (c) One member of the management board claims that for every $1,000 increase in income, the amount of life insurance held will go up by $5,000. Choose an alternative hypothesis and explain your choice. Does your estimated relationship support this claim? Use a 5% significance level. (d) Test the hypothesis that as income increases the amount of life insurance increases by the same amount. That is, test the hypothesis that the slope of the relationship is one.. regress insurance income. predict cr_pred (option xb assumed; fitted values) predict error, resid twoway(scatter insurance income) (connected cr_pred income)

Explanation / Answer

Part-b(i)

Assuming income measures in thousands, we have

when income increase by $1000, insurance INCREASE ON an average by 1*3.880=$3.88

Predicted insurance yhat= 6.854991+3.880186*1 =10.735177

Part-b(ii)

Standard error of estimate= SE(b1)*sqrt(1+1/n+(x0-xbar)2/((n-1)*SD(x)2))

Here put se(b1)=7.383473, n=20,x0=1,xbar=mean income, SD(x)=standard deviation of income from your data and get the standard error of prediction.

Part-c

We have to test the null hypothesis H0: =5 versus the alternative Ha: 5

Test statistic t=(betahat-5)/SE(betahat)=(3.880186-5)/0.1121246=-9.987228494

Degree of freedom =18

P-value=9.10828E-09 using excel function =TDIST(9.98723,18,2)

As p-value<0.05, we reject the null hypothesis and conclude that insurance increase up by amount different from $5000 for $ 1000 increase in income

Part-d

We have to test the null hypothesis H0: =1 versus the alternative Ha: 1

Test statistic t=(betahat-1)/SE(betahat)=(3.880186-1)/0.1121246=25.68736923

Degree of freedom =18

P-value=1.23041E-15using excel function =TDIST(25.69,18,2)

As p-value<0.05, we reject the null hypothesis and conclude that insurance increase up by amount different from the amount of life insurance by the same amouunt

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