Question

A consumer organisation wants to obtain information about , the mean number of drawing pins in the boxes of a certain brand which, according to the label, should contain 100 pins. In nine randomly chosen boxes this organisation finds the following numbers of drawing pins:

90 94 88 92 90 86 94 90 86



(a) (i) Test whether < 100. Take as level of significance a= 0:05.

(ii) Which assumption do you need?

(b) If the true value of u is 101, what type of error did you make in part (a)?

(c) (i) Determine a 90% confidence interval for u

(ii) Is it likely that u= 101?

Answer 1

Part (a)

(i)

$t=\frac{r\sqrt{n-2}}{\sqrt{1-r^2}}\\ t=\frac{(-0.9594)\sqrt{9-2}}{\sqrt{1-0.9594^2}}\\ t=-8.9996\\ df=n-2=9-2=7\\ (1-\alpha)=0.95\\ \alpha=0.05\\ \frac{\alpha}{2}=0.025\\$

The corresponding probability of 0.975 is -2.306

Test statistic is $-8.9996< t_s-2.306$

H_o is rejected

(ii) The population form which sample is drawn is normally distributed

(b) A Type II error (i.e. do not reject the null hypothesis H0 when it was in fact false)

(c) (i)

$90+-Z_{1-0.05}*\frac{6}{\sqrt{9}}\\ 90+-1.645*\frac{6}{3}\\ 90+-1.96*2\\ (86.71,93.29)$

(ii)Yes