Answer :
Answer:
95% confidence interval for the variance of the bowling ball weight is between a lower limit of 13.328 pounds and an upper limit of 14.672 pounds.
Step-by-step explanation:
Confidence interval is given as weight +/- margin of error (E)
weight = 14 pounds
sample sd = 0.94 pound
n = 10
degree of freedom = n - 1 = 10 - 1 = 9
confidence level (C) = 95% = 0.95
significance level = 1 - C = 1 - 0.95 = 0.05 = 5%
critical value (t) corresponding to 9 degrees of freedom and 5% significance level is 2.262
E = t × sample sd/√n = 2.262×0.94/√10 = 0.672 pounds
Lower limit of weight = weight - E = 14 - 0.672 = 13.328 pounds
Upper limit of weight = weight + E = 14 + 0.672 = 14.672 pounds.
95% confidence interval is (13.328, 14.672)
Answer:
95% confidence interval for the variance of the bowling ball weight is (0.418 , 2.945).
Step-by-step explanation:
We are given that the company has a problem with the variability of the weight. For this, a sample of 10 of the bowling balls, the sample standard deviation was found to be 0.94 pounds.
So, firstly the pivotal quantity for 95 % confidence interval for the population standard deviation is given by;
P.Q. = [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] ~ [tex]\chi^{2} __n_-_1[/tex]
where, [tex]s^{2}[/tex] = sample variance = [tex]0.94^{2}[/tex]
[tex]\sigma^{2}[/tex] = population variance
n = sample of bawling balls = 10
So, 95% confidence interval for population standard deviation, is;
P(2.7 < [tex]\chi^{2} __1_0_-_1[/tex] < 19.02) = 0.95 {As the table of [tex]\chi^{2}[/tex] at 9 degree of freedom
gives critical values of 2.7 & 19.02}
P(2.7 < [tex]\frac{(n-1)s^{2} }{\sigma^{2} }[/tex] < 19.02) = 0.95
P( [tex]\frac{2.7}{(n-1)s^{2} }[/tex] < [tex]\frac{1}{\sigma^{2} }[/tex] < [tex]\frac{19.02}{(n-1)s^{2} }[/tex] ) = 0.95
P([tex]\frac{ (n-1)s^{2}}{19.02}[/tex] < [tex]\sigma^{2}[/tex] < [tex]\frac{ (n-1)s^{2}}{2.7}[/tex] ) = 0.95
95% confidence interval for = ( [tex]\frac{ (n-1)s^{2}}{19.02}[/tex] , [tex]\frac{ (n-1)s^{2}}{2.7}[/tex] )
= ( [tex]\frac{ (10-1) \times 0.94^{2}}{19.02}[/tex] , [tex]\frac{ (10-1) \times 0.94^{2}}{2.7}[/tex] )
= (0.418 , 2.945)
Therefore, 95% confidence interval for the variance of the bowling ball weight is (0.418 , 2.945).