A bottled water distributor wants to determine whether the mean amount of water contained in 1-gallon bottles purchased from a nationally known water bottling company is actually 1 gallon. You select a random sample of 100 bottles, the mean amount of water per 1-gallon bottle is 0.994 gallon, and standard deviation of the amount of water per bottle is 0.03 gallon. You select a random sample of 50 bottles, and the mean amount of water per 1-gallon bottle is 0.995 gallon.
a. Is there evidence that the mean amount is different from 1.0 gallon? (Use α-0.01.)
b. Compute the p-value and interpret its meaning
c. Construct a 99% confidence interval estimate of the population mean amount of water per bottle.
d. Compare the results of (a) and (c). What conclusions do you reach?

We fail to reject H0; Hence, we conclude that there is no significant evidence that the mean amount of water per gallon is different from 1.0 gallon

Pvalue = - 2

(0.98626 ; 1.00174)

Since, 1.0 exist within the confidence interval, then we can conclude that mean amount of water per gallon is 1.0 gallon.

Step-by-step explanation:

H0 : μ= 1

H1 : μ < 1

The test statistic :

(xbar - μ) / (s / sqrt(n))

(0.994 - 1) / (0.03/sqrt(100))

-0.006 / 0.003

= - 2

The Pvalue :

Pvalue form Test statistic :

P(Z < - 2) = 0.02275

At α = 0.01

Pvalue > 0.01 ; Hence, we fail to reject H0.

The confidence interval :

Xbar ± Margin of error

Margin of Error = Zcritical * s/sqrt(n)

Zcritical at 99% confidence level = 2.58

Margin of Error = 2.58 * 0.03/sqrt(100) = 0.00774

Confidence interval :

0.994 ± 0.00774

Lower boundary = (0.994 - 0.00774) = 0.98626

Upper boundary = (0.994 + 0.00774) = 1.00174

(0.98626 ; 1.00174)

step-by-step explanation:

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