B. Given x ~ N(u, 4), state the 95% two-sided commence interval on u Assume u =
ID: 3233138 • Letter: B
Question
B. Given x ~ N(u, 4), state the 95% two-sided commence interval on u Assume u = 10 and use computer software to plot the distribution & Explain the graph. (Objective: One Sided Hypothesis Testing- Right Sided) The shelf life of a carbonated beverage is studied. Ten bottles are randomly selected and tested, and the following results are obtained: a) We would like to demonstrate that the mean shelf life exceeds 110 days. Set up appropriate hypotheses for investigating this claim. H_0 : mu = H_1 : mu > b) Test the hypotheses at alpha = 0.01. State your conclusion? c) Output: Use Minitab to test the same above hypothesis d) Calculate P = P (t > to)=? Probability of finding a random t value that has greater value than the t test statistics. (e) is PExplanation / Answer
Solution:-
(a) State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: = 110
Alternative hypothesis: > 110
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too large.
(b) Formulate an analysis plan. For this analysis, the significance level is 0.01. The test method is a one-sample t-test. Critical t0.01,9 = 2.8214
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) = 19.5448 / sqrt(10) = 6.181
DF = n - 1 = 10 - 1 = 9
t = (x - ) / SE = (131 - 110)/6.181 = 3.398
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.
The observed sample mean produced a t statistic test statistic of 3.398. We use the t Distribution Calculator to find P(t < 3.398) = 0.003949.
Conclusion. Reject the null hypothesis, that is the mean shelf life exceeds 110 days.
99% confidence interval - 110.91404 x 151.08596 (Use formula CI = x ± Z/2 × (/n) or calculator)
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