Question

3. Z-test for the population mean u (o known) Hosmer Winery (on Cayuga Lake, New York) manufactures and bottles wine for sale. They have a machine that automatically fills the bottles. The amount of wine that the machine dispenses will naturally vary slightly from bottle to bottle. To determine whether the machine is working correctly, bottles are occasionally sampled and the volume of wine is measured. One particular sample of eight bottles showed the following amounts of wine (in milliliters): 745 744 750 745 756 753 748 751 If the machinery is working correctly, bottles should contain 753 ml of wine, on average. Either too little or too much could be bad. To decide if the machinery is working correctly, we want to test whether the mean volume, u, differs from this stated value. Suppose the volume of wine dispensed by the machine is known to have a normal distribution with standard deviation o =4.3 ml. (a) State the appropriate Null Hypothesis and Alternative Hypothesis. (b) Before proceeding with the test, it is important to check whether the conditions that allow us to safely use this inference procedure are met. Even though it does not say so in the problem description, we can assume that the sample of n = 8 bottles is random. The second condition ensures the normality of the sampling distribution of X. Explain why this condition is satisfied despite the very small sample size. (c) Now we can safely proceed with the test... • Find the Z-test statistic. (round to two decimal places) (Note: you'll first need to calculate m).

Answer 1

(a) The Hypothesis

H0: $\mu$ = 753

Ha: $\mu$ $\neq$ 753

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(b) By the property of Normality, since the population from which the sample comes from is given to be normal, therefore the distribution of the sample means can also be considered to be normal, regardless of the size of N. This is called the First Known Property.

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(c) $\bar{x}$ = (745 + 744 + 750 + 745 + 756 + 753 + 748 + 751) / 8 = 5992/8 = 749

The Test Statistic:

$Z = \frac{\bar{x}-\mu }{\frac{\sigma}{\sqrt{n}}} = \frac{749-753}{\frac{4.3}{\sqrt{8}}} = -2.63$

The p value (2 tail) : Since z is -ve, we find the p value for -2.63 and then multiply the value by 2 = 0.0043 * 2 = 0.0086

Since p value is < 0.05, reject H0.

The Conclusion: There is sufficient evidence at the 95% significance level to conclude that the amount of wine the machine dispenses is different from 253 ml.

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(d) The Zcritical (2 tail) for $\alpha$ = 0.05, is 1.96

The Confidence Interval is given by $\bar{x}$ $\pm$ ME, where

$ME = Zcritical * \frac{\sigma}{\sqrt{n}} = 1.96 * \frac{4.3}{\sqrt{8}} = 2.98$

The Lower Limit = 749 - 2.98 = 746.02

The Upper Limit = 749 + 2.98 = 751.98

The Confidence Interval is (746.02 , 751.98)

The confidence interval does not contain the value of 753, and hence we can reject the null hypothesis that mean is equal to 753 ml, and conclude that the mean is different from 753 ml.

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