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