Question

A car insurance company advertises that customers switching to their insurance save, on average, $435 on their yearly premiums. A market researcher at a competing insurance discounter is interested in showing that this value is an overestimate so he can provide evidence to government regulators that the company is falsely advertising their prices. He randomly samples 83 customers who recently switched to this insurance and finds an average savings of $399, with a standard deviation of $103.

(a) Are conditions for inference used in the lectures satisfied?

(b) What is your null hypothesis?

(c) What is your alternative hypothesis?

(d) Calculate the p-value for a one-sided hypothesis test.

(e) What is your conclusion if the test was conducted at the 95% level?

(f) Do you agree with the market researcher that the amount of savings advertised is an overestimate? Make sure you can explain your reasoning.

(g) Calculate a 90% confidence interval for the average amount of savings of all customers who switch their insurance.

(h) Do your results from the hypothesis test and the confidence interval agree? Make sure you can explain your reasoning.

Answer 1

a ) For performing hypothesis test , the sample must be a simple random sample from the population which it is ; also the data must be continuous and not discrete . In this case , that data on savings is continuous . Hence the conditions for performing a hypothesis test is satisfied here .

b ) Null hypothesis :

c ) Alternative hypothesis :

d ) To test :

Test statistic :

where ,

Degrees of freedom = 82

The value of the test statistics :  

e ) We reject H₀ if the

Now ,

Hence we reject the null hypothesis or H₀ .

f ) Thus we have sufficient evidence to support the claim that the amount of savings that the customers save on average when switching to their insurance is less than $ 435 on average . In other words ,  the amount of savings advertised is an overestimate

g ) The 90 % confidence interval of the population mean is :

where ,

Required confidence interval :  

h ) The results from both hypothesis testing and the confidence interval agree . The testing was done at 5 % level of significance . The p-value obtained is less than . However , even if the testing was done at 1 % level of significance , still then the p-value is less than , which results in the rejection of H₀ . Now , the 90 % confidence interval is   which does not contain the value of $ 435 , thus resulting in the rejection of H₀ . Hence results from the hypothesis test and the confidence interval do agree .