Question

d) A random sample of size 36 was taken from a population distributed as Nμ,3.92.The value of the sample x was 15.6. i. Find a 90% confidence interval for μ.                                 (5mks)

It is believed that value of μ is 17. Use your confidence interval to comment on this belief.(2mks)

1. Using the provided information below,

x=33.2 , x2=131.67,  y=30.78 , y2=116.52, xy=119.8 , and n=10

1. Find the sample correlation coefficient r                                                                 (4mks)
2. Find the co-efficient of determination                                                                    (2mks)

Answer 1

d) The confidence interval is

$\left( \bar{x}-\frac{\sigma}{\sqrt{n}}z_{\frac{1+\gamma}{2}},\bar{x}+\frac{\sigma}{\sqrt{n}}z_{\frac{1+\gamma}{2}} \right) =\left( 15.6-\frac{3.92}{\sqrt{36}}\cdot 1.65,15.6+\frac{3.92}{\sqrt{36}}\cdot 1.65 \right) =\\=\left( 14.552,16.678 \right)$

Since 17 doesn’t lie within the confidence interval, the belief is false with confidence level 0.9.

1:

$x=33.2,x^2=131.67,y=30.78,y^2=116.52,xy=119.8,n=10\\r=\frac{nxy-x\cdot y}{\sqrt{\left( nx^2-\left( x \right) ^2 \right) \left( ny^2-\left( y \right) ^2 \right)}}=\frac{10\cdot 119.8-33.2\cdot 30.78}{\sqrt{\left( 10\cdot 131.67-33.2^2 \right) \left( 10\cdot 116.52-30.78^2 \right)}}=0.814846\\R^2=r^2=0.663974$