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The question I was asked is as follows:

A website received 6802 page views last week. Does this provide evidence at a 1% significance level against the claim that the website receives on average 1000 page views per day?

I have calculated the mean as: 6802 / 7 = 971.71 (2 d.p)

H0 = 1000

H1 = 971.71

I figure that it is impossible to calculate a standard deviation without sample data, and therefore the inability to calculate a z score and a resulting p value. Is this is a trick question that is unanswerable or am I missing something? I figured I could try generating a random sample of n = 7 with a normal distribution based on that mean, and go from there? Building upon that, should I also perform bootstrapping on that sample? Or should I create another random sample with a mean of 1000 and compare the two using a randomisation distribution?

Help would be much appreciated, I'm racking my brains trying to figure this out.

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  • $\begingroup$ It is common to assume that counts and proportions come from distributions where the spread and locations are interrelated, so it is not necessary to specify a standard deviation. Search for the sampling distribution of a proportion. $\endgroup$ May 5, 2016 at 5:36
  • $\begingroup$ @MichaelLew okay so I assume the proportion in this case is 971.71/1000 = 0.97171? $\endgroup$ May 5, 2016 at 5:55
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    $\begingroup$ @Michael I completely agree with your first sentence, but this doesn't look like a proportions question to me. It's a comparison of a count per exposure time with a prespecified population rate. $\endgroup$
    – Glen_b
    May 5, 2016 at 7:09
  • $\begingroup$ I may have just fallen off the turnip truck, but to me (absent other assumptions or priors), this looks like a comparison between a single data point of views for week of 6802 vs. claim of average 7000 views per week. $\endgroup$ May 5, 2016 at 17:38
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    $\begingroup$ Given that a Poisson distribution with parameter $7000$ indicates the chance of an outcome of $6802$ or less is $0.9\%$, and that's very close to $1\%$, a shrewd guess is (a) this is a textbook question that (b) is asked in the context of material on the Poisson distribution. You will likely find useful clues in the related textbook material. The text would also indicate whether you are expected to approximate the p-value using a Normal distribution. $\endgroup$
    – whuber
    May 5, 2016 at 18:29

1 Answer 1

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Given the number of assumptions that would have to be made, I would answer this question with "no".

You could treat the one week average as the sample mean of 7 day-observations, but only if you assume that the web site daily page views follow a normal distribution with equal variance. To do this I would use a one sample t-test.

In the real-world, given the variance many websites have given day of week, time of year, or special events(like the NFL website during playoffs), I would want more than one week of data to form any conclusion.

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