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I struggle with following question:

It is now considered that house prices are rising again following the recent slump. Given that in 2008 the average price of a standard 3 bedroom house was €215,000. A sample of 25 similar houses in 2013 gave an average price of €275,000 with a standard deviation of €50,000. Test to determine if the average price for the houses in 2013 exceeds the average price in 2008 at the 5% level. Yes\No answer only, will receive zero marks.

NOTE: NO dataset is provided.

This is an assignment question which of course I don't want you to do it all for me!!

I'm bit confused as can't figure out what equation/formula to use. As reference I'm using my slides notes and Statistics boook (James McClave, 2014) 12th edition.

I'm using this formula: http://www.nursingplanet.com/biostatistics/inferential_statistics_t-test.html Student t-testing calculating t-statistic for comparing means.

Any help hints/tips would be appreciated! Thanks!

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  • $\begingroup$ What specific step are you struggling with? What specifically is your problem? $\endgroup$
    – jona
    Commented Jun 2, 2014 at 11:42
  • $\begingroup$ what would be my null hypothesis - if any. Giving is average and sd for 2013, how to calc this for 2008? $\endgroup$
    – Jakubee
    Commented Jun 2, 2014 at 11:47
  • $\begingroup$ It's not a two-sample t-test, since you're comparing a 2008 population with a 2013 sample. You're presumably meant to be doing a one-sample t-test. $\endgroup$
    – Glen_b
    Commented Jun 2, 2014 at 11:52

1 Answer 1

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First, consider what kind of t-test you want to use. There is the unpaired two-t-test, comparing two unmatched, independent samples; the two-sample paired t-test, comparing two matching samples (such as two value being measured in 10 subjects each); and the one-sample t-test, where one sample is compared to a target value. Here are free online implementations for each of these. Which of these is the correct one? Consider what data you are investigating:

Given that in 2008 the average price of a standard 3 bedroom house was €215,000. A sample of 25 similar houses in 2013 gave an average price of €275,000

First answer this for yourself, and only then read the next section!

Your null hypothesis is spelled out in this sentence:

Test to determine if the average price for the houses in 2013 exceeds the average price in 2008 at the 5% level.

What then is your null hypothesis?

Since the question requires you to test one sample against a reference value, you want to use the one-sample t-test. Find the correct formula for the one-sample test!

Once you've found the formula, you'll see that the t-statistic requires the standard deviation of the sample, the mean of the sample, and the value representing the null. The null hypothesis is that (sample mean) - (mean associated with H0) is zero.

Standard deviation, sample mean and the value to be tested against are all given in the text. What are they?

Once you know your t value, you compare it against your critical value you look up in a table, such as this one. You will see that your critical value depends on your "alpha" level or "significance" threshold on one hand, and on your degrees of freedom (which in this case is simply sample size minus one). Make sure you understand what critical values are and how they relate to p values and significance thresholds; see for example here.

If your critical value is exceeded by your test statistic (in this case, your t), your test is "significant" at the chosen level.

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  • $\begingroup$ so null hypothesis: H 0: μ = 215,000 alternative hypothesis: H a : μ > 215,000. Am I correct? $\endgroup$
    – Jakubee
    Commented Jun 2, 2014 at 12:04
  • $\begingroup$ Traditionally, we call the alternative hypothesis H1. (I hadn't notice the question implies a one-tailed test, silly me.) But yes, that is how I would interpret the question. Or to be precise: H0 is that the 2013 value is smaller or equal to the 2008 one. H1 is that it is not smaller or equal (but larger). $\endgroup$
    – jona
    Commented Jun 2, 2014 at 12:10
  • $\begingroup$ I also need to find the critical value (or Tz value). Significance level for me is 5% but I have no idea to which number in table this corresponds to $\endgroup$
    – Jakubee
    Commented Jun 2, 2014 at 12:57
  • $\begingroup$ @Jakubee, I have edited my answer, see if that helps you. $\endgroup$
    – jona
    Commented Jun 2, 2014 at 13:16
  • $\begingroup$ Of course. I have worked out my critical value which is -1.711 (but i might be wrong). $\endgroup$
    – Jakubee
    Commented Jun 2, 2014 at 13:32

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