ater botling company\'s specificstione sate th sampla maan amount oe watar Far 1
ID: 3337462 • Letter: A
Question
ater botling company's specificstione sate th sampla maan amount oe watar Far 1.galon bom 0 ses galon Compiata par (a) thrash [C) a c rstnict 99% connsenca intervaltímata to, the population maan amount of watar included n 1-galon batla b. On the bais of these rezu, do you hink that the distribunor has a right to complain to the water betting campany?Why? because, 1 galon oo de contining cxacti, 1 gallon cfw tries | te 99% confidence interval c. Must you ame tharthepoplaticn amount of waer per bettie ncmaly diztnibuted here? Loplain O C. No, because the Central umit Theurer" uhmost ah,sy5 ensures lht t X normalt, distributed ehen n is small In this case, the value of " rnal. O D. No, because the Central umt Thcarem host Jhoys cnsures that X normalt, dlatributed ihon n is large this case te value of n is large d. Construct %% conidence interval csdmate. I low does this change your rower to prt (b)? Hound 1o ive dacima placas as neadad) Huw does Ihis charge your wer to pert b? A 1-gelon bottle containing exactly 1-galon o water les the 96% coridence nterval. The dertutor right to complain 10 the botting company.Explanation / Answer
Solution:
a) A confidence interval for population mean is x ± Z/2 /n
For 99% confidence interval, Z/2 = 2.58
Therefore, a 99% confidence interval estimate for the population mean amount of water included in a 1-gallon is given by-
= 0.969 ± (2.58) 0.02/50 = 0.969 ± 0.0073
=(0.9617, 0.9763)
b) Thus, with 99% confidence, it can be concluded that the population mean is between 0.9617 gallon and 0.9763 gallon. As this interval includes 1, it indicates that the water bottling compnay is woeking properly and hence the distributor has no right to complain to the water bottling compnay.
c) Yes, it can be assumed that the population amount of water per bottle is normally distributed. This is because under the assumption the sample was selected.
B) Yes, since nothing is known about the distribution of the population it must be assumed that the population is normally distributed.
The amount of water per bottle is normally distributed.
d) The 95% confidence interval is computed as follows:
x ± Z/2 /n
For 95% confidence interval Z/2 = 1.96
Therefore, a 95% confidence interval estimate for the population mean amount of water included in a 1-gallon is given by
x ± Z/2 /n = 0.969 ± (1.96) 0.02/50 = 0.969 ± 0.0055
= (0.9635, 0.9745)
The 95% confidence interval is (0.9635, 0.9745). The width of the confidence interval decreases with the decrease in confidence level.
The width of the confidence interval decreases with the decrease in confidence level and vice-versa. At 95% confidence level the estimate for the population mean of water is between 0.9635 and 0.9745 gallons.
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