Proportion testing

Question

Proportion testing

Question:

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.

Answer 1

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.

Answer 2

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 $p=0.000001$.