Answer to Question #173597 in Statistics and Probability for ANJU JAYACHANDRAN

Question

8. a) A shop manufacturer is trying to determine whether or not to market a new type of

shop. It chooses a random samples of equivalent size in England and France. It asks

each of the people in each sample to try the new soap, and see whether he or she

likes it better than other soaps. The results are as follows:

United States England France

Prefer new soap 81 43 26

Do not prefer it 219 257 274

Test the hypothesis that there are international difference in the proportion of people

who prefer the new soap, using 5% level of significance.

Answer 1

The Data table is given below-

Let $H_o:\mu_1\neq \mu_2,$ there are international difference in the proportion of people

who prefer the new soap at 5% level of significance.

$H_a: \mu_1=\mu_2$ , there are not any international difference in the proportion of people

who prefer the new soap at 5% level of significance.

$n_1=3,n_2=3$

$\bar{x_1}=\dfrac{\sum x_1}{n_1}=\dfrac{150}{3}=50 \\[9pt] \bar{x_2}=\dfrac{\sum x_2}{n_2}=\dfrac{750}{3}=250$

Then The standard deviation $S=\sqrt{\dfrac{\sum(x_1-\bar{x_1})^2+\sum (x_2-\bar{x_2})^2}{n_1+n_2-2}}$

$=\sqrt{\dfrac{1586+1586}{3+3-2}}=\sqrt{\dfrac{3172}{4}}=28.16$

Using test statics for two independent samples-

$t=\dfrac{\bar{x_1}-\bar{x_2}}{S\sqrt{\dfrac{1}{n_1}+\dfrac{1}{n_2}}}$

$=\dfrac{50-250}{28.16\sqrt{\dfrac{1}{3}+\dfrac{1}{3}}}=\dfrac{-200\times \sqrt{3}}{28.61\times \sqrt{2}}=-8.566$

The value of t at 5% level of significance and degree of freedom is $n_1+n_2-2\text{ i.e. }4$ is 2.132

Conclusion: As calculated value of t is less than the standard value so Null hypothesis is accepted i.e. there are international difference in the proportion of people who prefer the new soap at 5% level of significance.