Someone claims that adult koalas eat an average of 392 grams of eucalyptus leave

Question

Someone claims that adult koalas eat an average of 392 grams of eucalyptus leaves per day. In order to test that claim, we collect a sample of 50 koalas and find that the sample mean is 395 grams with a sample standard deviation of 53 grams. Using an alpha of 0.01, can we reject the null hypothesis that the population mean is 392 grams? Why or why not?

I think that we agreed that we cannot reject the null hypothesis. Formally, the critical values are +/- 2.58 (found via =NORMSINV(0.005) and =NORMSINV(0.995) since we have a two-sided test and an alpha of 0.01). The sample value is (395 – 392) / (53 / square root of 50) = -3 / 7.5 = -0.4, which is not outside the critical values, hence we do not reject the null hypothesis.

If we changed the original problem so that the claim is now that adult koalas eat an average of 398 grams of eucalyptus leaves per day. Using the same sample (sample size = 50, sample mean = 395 grams, sample standard deviation = 53 grams), can we reject that null hypothesis that the population mean is 398 grams?

Explanation / Answer

In that case,

z = (x-u)/(s/sqrt(n))

Thus, z = (395-398)/(53/sqrt(50))

z = -0.40

p value => P(Z< -0.40) = 0.3446 > alpha(0.01)

Thus, we failed to reject the null hypothesis that the population mean is 398 grams

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