ore: 0.8 of 4 pts 3017(6 complete) HW Score: 49.45%, 12.86 of: 9.1.16-T E Questi
ID: 3310297 • Letter: O
Question
ore: 0.8 of 4 pts 3017(6 complete) HW Score: 49.45%, 12.86 of: 9.1.16-T E Question Help 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 know from the water bottling company specifications that the standard deviation of the amount of water is 0.03 gallon. You select random sample of 45 bottles, and the mean amount of water per 1-gallon bottle is 0.994 gallon. Complete parts (a) through (d) below a. Is there evidence that the mean amount is different from 1.0 gallon? (Use =0.05.) Let be the population mean. Determine the null hypothesis, Ho, and the alternative hypothesis, H1. What is the test statistic? ZSTAT-l(Round to two decimal places as needed)Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 1
Alternative hypothesis: 1
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n)
S.E = 0.00447
DF = n - 1 = 45 - 1
D.F = 44
t = (x - ) / SE
t = - 1.34
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
Since we have a two-tailed test, the P-value is the probability that the t statistic having 44 degrees of freedom is less than -1.34 or greater than 1.34.
Thus, the P-value = 0.1872.
Interpret results. Since the P-value (0.1872) is greater than the significance level (0.05), we cannot reject the null hypothesis.
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