Question

A coin-operated drink machine was designed to discharge a mean of 7 ounces of coffee per cup. In a test of the machine, the discharge amounts in 21 randomly chosen cups of coffee from the machine were recorded. The sample mean and sample standard deviation were 6.84 ounces and 0.3 ounces, respectively. If we assume that the discharge amounts are normally_distributed, is there enough evidence, at the 0.1 level of significance, to conclude that the true mean discharge, μ, differs from 7 ounces? Perform a two-tailed test. Then fill in the table below Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.) The null hypothesis The alternative hypothesis: FH The type of test statistic: | | (Choose one) ク The value of the test statistic: (Round to at least three decimal places.) The two critical values at the 0.1 level of significance: (Round to at least three decimal places.) and At the 0.1 level of significance, can we conclude that the true mean discharge differs from 7 ounces? Yes No

Answer 1

We want to test if the true mean discharge, $\mu$ 'differs' from 7 ounces, i.e., we want to test if $\mu \ne 7$ .

Thus, our null and alternative hypothesis are given by:

7 11メ /o

Now, we are given the following details about the population :

1. The discharge amounts (population) is normally distributed.

2. Size of sample, n = 21

3. Sample mean, $\bar{x} = 6.84$

4. Sample standard deviation, s = 0.3

5. Level of significance, $\alpha$ = 0.1

Now, since the population is normally distributed and we don't have the value of the true population standard deviation, the test statistic which we use here is:

Under H₀ : $\frac{\bar{x} - \mu}{s/\sqrt{n}} \sim t_{n-1}$ , i.e., the t-statistic.

Thus, the value of the t-statistic is given by:
$\begin{align*} \frac{\bar{x} - \mu}{s/\sqrt{n}} &= \frac{6.84 - 7}{0.3/\sqrt{21}} \ \ \ \ \ \text{(We use the value of } \mu \text{ under }H_0) \\ &= -2.44404 \\ &= -2.444 \ \ \ \ \text{(Rounded to 3 decimal places)} \end{align*}$

Now, the critical values at 0.1 level of significance are the values of the t₂₀ distribution the areas to the left of which is $\alpha/2 = 0.05$ and $1-\alpha/2 = 0.95$ .

Thus, from the table of t₂₀ distribution, we get:

$\\ t_{20,0.05} = -1.725 \ \ \ \text{(Rounded to 3 decimal places)} \\ t_{20,0.95} = 1.725 \ \ \ \text{(Rounded to 3 decimal places)}$

which are our critical values.

Now, since our test statistic = -2.444 lies outside the interval (-1.725,1.725) we can reject $H_0 : \mu =7$ at 0.1 level of significance and conclude that $\mu \ne 7$ .

Thus, Yes, we can conclude that the true mean discharge differs from 7 ounces.

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