An alcoholic beverage manufacturer claims that on average the alcohol content of

Question

An alcoholic beverage manufacturer claims that on average the alcohol content of their new brand is less than 10.5%. It would be unwise to reject their claim without strong contradictory evidence. A random sample of 5 bottles were tested for alcohol content. The sample mean and standard deviation were 10.67 and 0.21 respectively. It is known that the alcohol content is normally distributed but for sample size 5: delta sigma. (a) Is it clear that on average the alcohol content of the new brand is greater than 10.5%? An appropriate problem formulation is to test H_0: mu = 10.5 versus H_alpha: mu > 10.5 What is the formula for the appropriate test statistic and what is its value for this sample. (b) If we test with significance level alpha = .05, decide whether or not H_0 should be rejected.

Explanation / Answer

Solution:-

(a) No, we can no assure that the alcohol content on average is greater than 10.5 as we are considering a small sample of 5 bottles and the initial claim by the alcohol beverage manufacturer is that on an average the alcohol content is LESS THAN 10.5%.

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: = 10.5
Alternative hypothesis: > 10.5

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean 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) = 0.21 / sqrt(5) = 0.0939
DF = n - 1 = 5 - 1 = 4
t = (x - ) / SE = (10.67 - 10.5)/0.0939 = 1.81

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.

Here is the logic of the analysis: Given the alternative hypothesis ( > 10.5), we want to know whether the observed sample mean is large enough to cause us to reject the null hypothesis.

The observed sample mean produced a t statistic test statistic of 1.81.

We use the t Distribution Calculator to find P(t < 1.81)

The P-Value is 0.072274.
The result is not significant at p < 0.05.

Interpret results. Since the P-value (0.07) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Conclusion. Fail to reject the null hypothesis. initial claim by the alcohol beverage manufacturer is true, that is on an average the alcohol content is LESS THAN or equal to 10.5%.

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