Answer to Question #314275 in Statistics and Probability for EDWIN

Question

b). Seven students went on a diet in an attempt to lose weight, with two of them losing weight while all the others added weight. Is the diet is an effective way to losing weight α=1% (5mks)

c) Two judges in a beauty contest rank the ten competitors in the following order.

x 6 4 3 9 2 7 1 5 10 8

 y 4 1 6 7 5 8 2 3 9 10

Do the two judges appear to agree in their standards?                                                        (6mks)

Answer 1

b) Let p denote the proportion of people who lose weight with the diet. The null hypothesis is that the diet is not effective, i.e. the proportion is 0.5, the alternative is that the proportion is greater than 0.5.

 

We have the

sample proportion

$\hat{p}=\frac{2}{7}$

The P- value

$P\left( \hat{p}\geqslant \frac{2}{7} \right) =P\left( \sqrt{n}\frac{\hat{p}-p}{\sqrt{p\left( 1-p \right)}}\geqslant \sqrt{n}\frac{\frac{2}{7}-p}{\sqrt{p\left( 1-p \right)}} \right) =\\=P\left( Z\geqslant \sqrt{7}\frac{\frac{2}{7}-0.5}{\sqrt{0.5\left( 1-0.5 \right)}} \right) =\\=P\left( Z\geqslant -1.13389 \right) =1-\Phi \left( -1.13389 \right) =0.87158$

c) The differences of ranks $d_i,i=1,...,10$ are:

2 3 -3 2 -3 -1 -1 2 1 -2

Then the Spearman correlation coefficient is

$r=1-\frac{6\sum{{d_i}^2}}{n\left( n^2-1 \right)}=0.7212$

Test this coefficient for significance:

$t=r\sqrt{\frac{n-2}{1-r^2}}=0.7212\sqrt{\frac{10-2}{1-0.7212^2}}=2.94468$

The P-value is

$P\left( \left| T_{n-2} \right|\geqslant 2.94468 \right) =2F_{T,8}\left( -2.94468 \right) =2\cdot 0.00929=0.01858$

Hence the coefficient is significant at $\alpha =0.05$ level. We conclude that there is a strong positive correlation, hence the judges agree in their standards.