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.