A recent study compared the life expectancy of individuals i

A recent study compared the life expectancy of individuals in the UK with individuals in the US. They found that the average life expectancy in the UK was 76 and the US had an average life expectancy of 72.5. But now they want to know if the social medicine helps the variance. For a sample of 15 US individuals the sample standard deviation of life expectancy is 5. The Sample standard deviation of a sample of 13 UK citizens was 3.5. At the significance level of 0.05 can we conclude that there is a difference in the variance of the life expectancy of the two countries?

What is the null hypothesis?

What is the alternative hypothesis?

What is the appropriate test statistic?

Z-score

t stat

F distribution

p-value

What is the calculated value of the test statistic?

What is the critical value of the test statistic?

What decision can you make based on the results?

Reject the null hypothesis

Fail to reject the null hypothesis

Reject the alternative hypothesis

Fail to reject the alternative hypothesis

Interpret the results:

Solution

Ho: sigma₁² = sigma₂²

Ha: sigma₁² not= sigma₂²

The appropriate test statistic is F distribution

The calculated value of the test statistic : S₁²/S₂² =25/12.25 =2.0408 df=14,12

The critical value of the test statistic is .328 >F or F> 3.206 ( from excel)

Fail to reject Ho.

There is not enough evidence to conclude that there is a difference in the variance of the life expectancy of the two countries.