A special study is conducted to test the hypothesis that people with glaucoma have higher blood pressure than average. In the study, 200 people with glaucoma are recruited with a mean systolic blood pressure of 140 mmHg and a standard deviation of 25 mmHg. If the average systolic blood pressure for people of comparable age is 130 mmHg, is there an association between glaucoma and blood pressure at the 0.05 significance level?

The answer key says from a t table that the p value is .002 but I am confused as to how that is found. The values on the table jump from 100 to 1000 and I cannot find a table with 200.

  • 1
    $\begingroup$ The true p-value in this instance is less than $10^{-7}$. That 0.002 will presumably be an upper bound on the p-value obtained from the table obtained by looking at the next smaller df and the smallest significance level in the tables you have. The best way to get the p-value is to use a program. Failing that, there's some discussion about how to get more accurate table values by using various forms of interpolation at this post: How do I find values not given in statistical tables? ... ctd $\endgroup$ – Glen_b Apr 22 '19 at 2:06
  • $\begingroup$ ctd ... but that would be of limited value in this specific instance, where you have run off the end of your table. There are tables with 200 df (which you could use interpolation with to get to 199, but there's no point in this instance) here - I hit that one in the top few hits on my first try at googling ("extensive t-tables"). If you look up 200 df it indicates that a two tailed p-value would be less than 0.0001. $\endgroup$ – Glen_b Apr 22 '19 at 2:06

(1) You do not give the $T$-statistic, which is required to find the P-value, independently of the computer output.

Furthermore, I cannot replicate your P-value from the numbers you give. Using Minitab's one-sample t test of $H_0: \mu = 130$ vs $H_a: \mu \ne 130.$

One-Sample T 

Test of μ = 130 vs ≠ 130

  N    Mean  StDev  SE Mean       95% CI          T      P
200  140.00  25.00     1.77  (136.51, 143.49)  5.66  0.000

Minitab shows a P-value of 0.000, which means less than $0.0005.$ A test against the one sided alternative $H_a: \mu > 130$ would be half as large---still less than $0.0005.$

(2) Usually, you cannot find a P-value from printed tables. The difficulty is not mainly that degrees of freedom (rows) jump from 100 to 1000. There is not much change in tabled values between those two rows.

The main difficulty is not most printed t tables show too few columns, so that you can only 'bracket' the P-value between probabilities given in two adjacent column headers.

(3) A right-sided P-value corresponding to test statistic $T$ is the Probability (area) beneath the density curve of $\mathsf{T}(\nu = 200-1 = 199)$ to the right of $T.$ One can use statistical software to find an exact P-value (very small). For $T = 5.66$ (as in the computer output above) is as shown in the computation from R below:

1 - pt(5.66, 199)
[1] 2.613768e-08

If you had $T = 2.40,$ then the one-sided P-value would be 0.0087$ (to four places).

1-pt(2.4, 199)
[1] 0.0086583

Note: Above I said there is not much change between degrees of freedom $\nu = 100$ and $\nu = 1000.$ For $T = 2.40,$ the corresponding P-values for sample sizes $n = 101$ and $n = 1001$ be about $0.009$ and $0.008,$ respectively.

1-pt(2.4, 100)
[1] 0.009122769
1-pt(2.4, 1000)
[1] 0.008288541

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.