Question

A researcher wanted to estimate the difference between the percentages of users of two 

tooth pastes who will never switch to another tooth paste. In a sample of 500 users of 

Toothpaste A taken by this researcher, 100 said that they will never switch to another 

toothpaste. In a sample of 400 users of Toothpaste B taken by the same researcher 68 said 

that they will never switch to another toothpaste.

(i) Let p1 and p2 be the populations of all users of Toothpastes A and B, respectively who 

will never switch to toothpaste. What is the point of p1-p2? (4 marks)

(ii) Construct a 98% confidence interval for the difference between the proportions of all 

users of the two toothpastes who will never switch (6 marks)

Answer 1

$i:\\\hat{p}_1=\frac{100}{500}=0.2\\\hat{p}_2=\frac{68}{400}=0.17\\\hat{p}_1-\hat{p}_2=0.03\\ii:\\\hat{p}=\frac{100+68}{500+400}=0.186667\\Confidence\,\,interval:\\\left( \hat{p}_1-\hat{p}_2-\sqrt{\hat{p}\left( 1-\hat{p} \right) \left( \frac{1}{n_1}+\frac{1}{n_2} \right)}z_{\frac{1+\gamma}{2}},\hat{p}_1-\hat{p}_2+\sqrt{\hat{p}\left( 1-\hat{p} \right) \left( \frac{1}{n_1}+\frac{1}{n_2} \right)}z_{\frac{1+\gamma}{2}} \right) =\\=\left( 0.03-\sqrt{0.186667\left( \frac{1}{400}+\frac{1}{500} \right)}\cdot 2.3263,0.03+\sqrt{0.186667\left( \frac{1}{400}+\frac{1}{500} \right)}\cdot 2.3263 \right) =\\=\left( -0.0374226,0.0974226 \right)$