AP Statistic Confidence Intervals

Please help!!! I cant figure out how to solve this

The following data is the age at death (in days) for infants who died from sudden infant death syndrome (SIDS). Construct a 95% confidence interval for the true difference in mean age at death for male and female SIDS victims.

Female 55 120 135 154 54 115

Male 56 60 60 106 140 147

Thanks for any help anyone can give

Re: AP Statistic Confidence Intervals

Hey serenamae1216.

What is the definition of your test statistic? (Hint: It will involve a t-distribution that involves the mean and standard deviation of the sample).

Re: AP Statistic Confidence Intervals

I did it originally by finding a mean difference of 10.333 and standard deviation of 59.24. So then I did 10.333 ± invT(.975,1)(59.24/√6). but then I got a huge confidence interval. I am having trouble figuring out where I am going wrong

Re: AP Statistic Confidence Intervals

You will have two standard errors: One for male and one for female.

Show us your working out for the standard errors (i.e. the standard deviations of the sample). Also remember that the sample variance needs to be divided by the size of the sample for a comparison of means.

Re: AP Statistic Confidence Intervals

Would you want to do (105.5-94.8333) ± InvT(.975,5)√((41.77^2/6)+(42.002^2/6)) to get the correct answer? Although I still get a huge confidence interval of (-51.498, 138.169)

Re: AP Statistic Confidence Intervals

Or could you do a 2-SampTInt on the calculator with the data and get (-43.22,64.55)

Re: AP Statistic Confidence Intervals

Using R, I got the following standard error for the difference:

> x

[1] 55 120 135 154 54 115

> y

[1] 56 60 60 106 140 147

> (((var(x)*6/5)/6) + ((var(y)*6/5)/6))^0.5

[1] 26.49251

Now the size of the interval for a 95% interval is 2*1.96*se which is

> 1.96*2*(((var(x)*6/5)/6) + ((var(y)*6/5)/6))^0.5

[1] 103.8507

You also have to remember that for estimating the variance you need to divide by (n-1) and not n which means that if you divided by n, your estimate will be smaller than it should be.