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) $CI=x±z*\frac{\sigma}{\sqrt{n}}$

The critical value for 90% confidence interval $z=±1.645$

Then $CI=15.6±1.645*\frac{3.92}{\sqrt{36}}$

The lower limit is $15.6-1.07=14.53$

The upper limit is $15.6+1.07=16.67$

The confidence interval is therefore $(14.53,16.67)$

Since the interval does not contain the value of $\mu=17$ , we can conclude that the value of the population mean $\mu≠17$ .

1)

$r=\frac{n∑xy-∑x∑y}{\sqrt{[n∑x^2-(∑x)^2][[n∑y^2-(∑y)^2]}}$

$=\frac{10(119.8)-33.2(30.78)}{\sqrt{[10(131.667)-33.2^2][10(116.52)-30.78^2}}$

$=\frac{1198-1024.884}{\sqrt{(214.46)(217.7916)}}$

$=\frac{173.116}{\sqrt{46707.58654}}$

$=\frac{173.116}{216.1194}$

$=0.801$

2) $r^2=0.6416$