A coffee-dispensing machine is supposed to deliver 12 ounces of liquid into a large paper cup, but the costumer believes that the actual amount is less. As a test he plans to obtain a sample of 30 cups of the dispensed liquid. The mean of his sample is 11.5 ounces. The machine operates with a known standard deviation of σ = 1.1 ounces. Is there enough evidence at the α = .05 level for him to claim the machine is broken?
Z = (mean-mean)/SEm
SEm = SD/sqrt n
Use table in back of text labeled area under normal distribution.
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To determine whether there is enough evidence for the customer to claim that the machine is broken, we will conduct a hypothesis test using the given information.
Step 1: State the hypotheses.
The null hypothesis (H0): The machine is not broken, and the mean amount of liquid delivered equals 12 ounces.
The alternative hypothesis (Ha): The machine is broken, and the mean amount of liquid delivered is less than 12 ounces.
Step 2: Set the significance level.
The significance level (α) is given as 0.05.
Step 3: Calculate the test statistic.
We will use a one-sample t-test to compare the sample mean to the population mean.
t = (x̄ - μ) / (σ / sqrt(n))
Where:
x̄ = sample mean (11.5 ounces)
μ = population mean (12 ounces)
σ = standard deviation (1.1 ounces)
n = sample size (30 cups)
Substituting the values, we get:
t = (11.5 - 12) / (1.1 / sqrt(30))
t ≈ -0.932
Step 4: Determine the critical value.
Since the alternative hypothesis is one-tailed (less than), we will find the critical value corresponding to a left-tailed t-distribution with (n-1) degrees of freedom and α = 0.05.
df = (n - 1) = 30 - 1 = 29
Using a t-table or calculator, the critical value for a left-tailed test with 29 degrees of freedom and α = 0.05 is approximately -1.699.
Step 5: Compare the test statistic with the critical value.
Since the test statistic (-0.932) is greater than the critical value (-1.699), we fail to reject the null hypothesis.
Step 6: State the conclusion.
There is not enough evidence at the α = 0.05 level to support the claim that the machine is broken. The customer's sample mean of 11.5 ounces does not provide sufficient evidence to conclude that the machine's mean liquid delivery is less than 12 ounces.
To answer this question, we can perform a hypothesis test using the sample mean and the known population standard deviation.
Step 1: State the hypotheses:
- Null hypothesis (H0): The machine is not broken, and the true mean amount of liquid dispensed is 12 ounces.
- Alternative hypothesis (H1): The machine is broken, and the true mean amount of liquid dispensed is less than 12 ounces.
Step 2: Select a significance level (alpha):
In this case, the significance level (alpha) is given as 0.05 or 5%.
Step 3: Compute the test statistic:
The test statistic for this problem is a one-sample z-test since we know the population standard deviation. We can calculate it using the formula:
z = (x̄ - μ) / (σ / sqrt(n))
where x̄ is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
In this case:
x̄ = 11.5 ounces
μ = 12 ounces
σ = 1.1 ounces
n = 30
Plugging the values into the formula, we get:
z = (11.5 - 12) / (1.1 / sqrt(30))
z = -0.5 / (1.1 / sqrt(30))
z ≈ -1.21
Step 4: Determine the critical value:
Since the alternative hypothesis is one-tailed (less than 12 ounces), we need to find the critical value corresponding to a 5% level of significance (alpha) in the left tail of the standard normal distribution.
This critical value can be looked up in a standard normal distribution table or calculated using statistical software. For a significance level of 0.05 (alpha = 0.05), the critical value is approximately -1.645.
Step 5: Make a decision:
Compare the calculated test statistic (z = -1.21) with the critical value (-1.645). If the calculated test statistic is less than the critical value, we can reject the null hypothesis in favor of the alternative hypothesis. Otherwise, we fail to reject the null hypothesis.
In this case, -1.21 > -1.645, which means the calculated test statistic does not fall in the rejection region. Therefore, we fail to reject the null hypothesis. There is not enough evidence, at the α = .05 level, to claim that the machine is broken.