# Solved – Formula confidence interval for difference in means – one sample t-test

I am looking for the formula of the confidence interval for the difference between means in a one sample t-test. I have only been able to locate the formula for a two sample t-test.

Let me give an example: I have the following ten scores

``10,12,13,11.5,9,11,11.1,11.9,12.1,9.3 ``

I want to know if the mean of these scores is significantly different from my population mean of 11.5. When I conduct a one sample t-test in SPSS I get the following results:

``t obtained = -10.776    SIG (i.e., P) = 0.000   95% CI of the difference = -5.3363 to -3.4837.    ``

I know how SPSS calculated t and P but not how it calculated the 95% CI of the difference. This 95% CI of the difference is not the same as the CI for the mean. The CI for the mean I can obtain by

\$\$CI =bar{x} pm t S/sqrt{N}.\$\$

The confidence interval in this case is: 10.16,12.01. I get the same result if I calculate this in SPSS.

So my question is: what is the formula for the CI of the difference which SPSS produces? How do I get that range? I do not want the CI for the mean. Thanks.

Contents

``A = c(10, 12, 13, 11.5, 9, 11, 11.1, 11.9, 12.1, 9.3)  B = A - 11.5  t.test(A, mu=11.5)     ### One Sample t-test    ### data:  A    ### t = -1.0013, df = 9, p-value = 0.3428    ### alternative hypothesis: true mean is not equal to 11.5    ### 95 percent confidence interval:    ### 10.16374 12.01626    ### sample estimates:    ### mean of x     ###     11.09  t.test(B, mu=0)     ### One Sample t-test    ### data:  B    ### t = -1.0013, df = 9, p-value = 0.3428    ### alternative hypothesis: true mean is not equal to 0    ### 95 percent confidence interval:    ### -1.3362575  0.5162575    ### sample estimates:    ### mean of x     ###     -0.41  ``