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Students are given a smart pill and the grades they've received are below. The average mark for students not taking the smart pill was 56.

56 57 35 57 44 35 54 36 54 35 54

The question asked me to find the percentage of variance.

Wondering if I'm calculating percentage of variance correct

(56 - sum(grades)/count(students) ) / 56
(56 - 47) / 56
= 16%

Next part it ask me to state my hypothesis and use critical test statistic given alpha of 0.01 and state my conclusion

Does that mean just solve for the z score?

What I tried was looking up z table of .01 and finding the z = 2.33

Sorry if I rephrased the question poorly.

Here's the actual Question

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  • $\begingroup$ Um....the average of those score is $47$. Not entirely sure that the term "smart pill" is applicable. $\endgroup$
    – lulu
    Dec 1, 2016 at 17:27
  • $\begingroup$ Something seems off to me here as the sample mean is so far below the asserted population mean. If the null hypothesis is that the drug does nothing, then presumably you want to assess the probability that a sample mean would be that far below the population mean. To do that, you need to know the standard deviation of the population statistics. Presumably that data is available...though it doesn't appear in your question. $\endgroup$
    – lulu
    Dec 1, 2016 at 17:35
  • $\begingroup$ A word by word posting of your exercise would clarify a lot. $\endgroup$ Dec 1, 2016 at 17:41
  • $\begingroup$ @lulu how is the average not 47? $\endgroup$ Dec 1, 2016 at 17:45
  • $\begingroup$ @callculus Hi, i've added the question. $\endgroup$ Dec 1, 2016 at 17:45

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