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Hello Everyone,
I'm relatively new to statistics and this community so please bear with me and forgive me if I commit any mistake while asking my question. For starters, here are two hypotheses tests that I want to perform:

Test 1: If on an average, Girls score more than 600 in an exam?
Test 2: Mean score of girls = 600.

Given: sample mean(x) = 633.4, sample size(n) = 30, population std dev(σ) = 100

So first I performed Test 1 where (null hypothesis(H0): mean<= 600) and (alternate hypothesis(H1): mean > 600).

Z = (x – μ) / (σ / √n) = (633.4-600)/(100/√30) = 1.83.
Taking alpha = 0.05 and since this is a upper-tailed Z test, the decision rule is: Reject H0 if Z > 1.645.
Hence, we reject H0 and accept H1. Therefore, Girls score more than 600 in an exam.

Now, coming to Test 2
H0: mean score = 600 and H1: mean score != 600

Z = 1.83 and since this a Two-Tailed Z Test, the decision rule is: Reject H0 if Z < -1.960 or if Z > 1.960

However, we fail to reject H0. Therefore, mean score = 600.

Please explain why the 2 tests are giving contradictory conclusions?

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1 Answer 1

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The test aren't giving contradictory results. You are in a sense using contradictory significance levels.

If you want to require equivalent "evidence" (I'm putting that in quotation marks, because p-values do not exactly assess evidence) in a "particular direction" (i.e. in the "is higher" or "is lower" direction), then one uses one-sided tests with half of the significance level of two-sided tests. I.e. if you use a two-sided test for $\alpha = 0.05$, then the one-sided test would use $\alpha = 0.025$. If one does so, then the significance threshold for the test statistic is at $\pm 1.96$ in both cases.

Or, to put it in other words, in your example you suddenly say that in the one-sided case you are willing to tolerate a twice as large type I error rate in the "greater than"-direction under the null hypothesis as you were willing to tolerate in the two-sided case.

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  • $\begingroup$ Umm do you mind showing it mathematically so that the results match? That would be a great help!! $\endgroup$
    – Tanay
    Sep 8, 2022 at 10:55
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    $\begingroup$ How do you mean other than 0.05 / 2 = 0.025? The 2.5th / 97.5th percentile of a standard normal distribution is then $\approx \pm 1.96$, while the 5th/95th ones are $\approx \pm 1.65$. Another way to look at it: You get the two-sided t-test for mean = 600 at the 0.05 level by doing two one-sided t-tests (one for the < 600 alternative and one for the > 600 alternative) at the 0.025 level and rejecting the null hypothesis of mean = 0 when either of the two one-sided tests has rejected its null hypothesis. $\endgroup$
    – Björn
    Sep 8, 2022 at 13:43

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