Proportion testing


It is claimed that 40% of adults use biros. In a random survey of 400 adults, 260 said they did not use biros. Investigate whether the actual usage is less than that claimed.

edit - updated

I have the following:

$n = 400$

$p = 0.4$

$\hat{p} = 0.35$ , which is found from $\frac{400 - 260}{400} = 0.35$

$\sigma = \sqrt{p(1 - p)} = \sqrt{0.4(0.6)} \approx 0.49$

Then standard error SE is

$SE = \frac{\sigma}{\sqrt{n}} = \frac{\sigma}{20} = \frac{0.49}{20} = 0.0245$

And I want to find whether $0.65$ is significantly different from $0.4$, I do this using $z$ tables, finding $z$ as

$z = \frac{\hat{p} - p}{SE} = \frac{0.35 - 0.4}{0.0245} \approx -2.04$

This is a one-tailed test as it's asking whether the usage is less than claimed.

Value from table $0.0207$ which is less than $0.05$ and means that that $H_0$ is rejected at a $5\%$ significance level.

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  • $\begingroup$ Shouldn't $\hat{p} = 1-.65 = 0.35$ $\endgroup$
    – user145807
    Commented Apr 26, 2017 at 16:32
  • $\begingroup$ Hint: Do you want a one-sided or two-sided test? $\endgroup$
    – user145807
    Commented Apr 26, 2017 at 16:35
  • 2
    $\begingroup$ We frequently observe $z$-scores greater than 3. 3 is the limit that they include on tables because 1) there would need to be several more pages of probability tables and 2) the evidence can be ruled as highly inconsistent with the null hypothesis if the z is 3 or greater. $\endgroup$
    – AdamO
    Commented Apr 26, 2017 at 16:38

2 Answers 2


The rate of biro use is 140/400 or 35%, not 65%.

You then test whether the rate, .35 is less than the claimed rate of .40, which is the rate under the null hypothesis.

  • $\begingroup$ thanks - I have updated the post , hopefully the reasoning is correct. Would be appreciated if you could read it to see if there's any poor logic. $\endgroup$
    – baxx
    Commented Apr 26, 2017 at 17:02

Note that 260 responded negatively, and the null hypothesis is about a positive response. That alone should solve your problem in this case.

But to give a general answer, $z$ values are not bounded: a very high $z$ corresponds to a very significant p-value. The tables are limited just because textbook examples, which use such tables, do not require you to differentiate between $p=0.0001$ and $p=0.000001$.


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