Answer :
Answer:
(a) We conclude after testing that mean diameter is 8.2500 cm.
(b) P-value of test = 2 x 0.0005% = 1 x [tex]10^{-5}[/tex] .
(c) 95% confidence interval on the mean diameter = [8.2525 , 8.2545]
Step-by-step explanation:
We are given with the population standard deviation, [tex]\sigma[/tex] = 0.0020 cm
Sample Mean, Xbar = 8.2535 cm and Sample size, n = 15
(a) Let Null Hypothesis, [tex]H_0[/tex] : Mean Diameter, [tex]\mu[/tex] = 8.2500 cm
Alternate Hypothesis, [tex]H_1[/tex] : Mean Diameter,[tex]\mu[/tex] [tex]\neq[/tex] 8.2500 cm{Given two sided}
So, Test Statistics for testing this hypothesis is given by;
[tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] follows Standard Normal distribution
After putting each value, Test Statistics = [tex]\frac{8.2535-8.2500}{\frac{0.0020}{\sqrt{15} } }[/tex] = 6.778
Now we are given with the level of significance of 5% and at this level of significance our z score has a value of 1.96 as it is two sided alternative.
Since our test statistics does not lie in the rejection region{value smaller than 1.96} as 6.778>1.96 so we have sufficient evidence to accept null hypothesis and conclude that the mean diameter is 8.2500 cm.
(b) P-value is the exact % where test statistics lie.
For calculating P-value , our test statistics has a value of 6.778
So, P(Z > 6.778) = Since in the Z table the highest value for test statistics is given as 4.4172 and our test statistics has value higher than this so we conclude that P - value is smaller than 2 x 0.0005% { Here 2 is multiplied with the % value of 4.4172 because of two sided alternative hypothesis}
Hence P-value of test = 2 x 0.0005% = 1 x [tex]10^{-5}[/tex] .
(c) For constructing Two-sided confidence interval we know that:
Probability(-1.96 < N(0,1) < 1.96) = 0.95 { This indicates that at 5% level of significance our Z score will lie between area of -1.96 to 1.96.
P(-1.96 < [tex]\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < 1.96) = 0.95
P([tex]-1.96\frac{\sigma}{\sqrt{n} }[/tex] < [tex]Xbar - \mu[/tex] < [tex]1.96\frac{\sigma}{\sqrt{n} }[/tex] ) = 0.95
P([tex]-Xbar-1.96\frac{\sigma}{\sqrt{n} }[/tex] < [tex]-\mu[/tex] < [tex]1.96\frac{\sigma}{\sqrt{n} }-Xbar[/tex] ) = 0.95
P([tex]Xbar-1.96\frac{\sigma}{\sqrt{n} }[/tex] < [tex]\mu[/tex] < [tex]Xbar+1.96\frac{\sigma}{\sqrt{n} }[/tex]) = 0.95
So, 95% confidence interval for [tex]\mu[/tex] = [[tex]Xbar-1.96\frac{\sigma}{\sqrt{n} }[/tex] , [tex]Xbar+1.96\frac{\sigma}{\sqrt{n} }[/tex]]
= [[tex]8.2535-1.96\frac{0.0020}{\sqrt{15} }[/tex] , [tex]8.2535+1.96\frac{0.0020}{\sqrt{15} }[/tex]]
= [8.2525 , 8.2545]
Here [tex]\mu[/tex] = mean diameter.
Therefore, 95% two-sided confidence interval on the mean diameter
= [8.2525 , 8.2545] .