1. Conduct a hypothesis test for the following scenario. SHOW ALL YOUR WORK THE WAY WE DID IN CLASS!
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?
I don't know how you did it in class.
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability for the Z score. How does it compare to your alpha level?
To conduct a hypothesis test for this scenario, we need to set up the null and alternative hypotheses, determine the test statistic, calculate the p-value, and make a decision based on the p-value.
Null Hypothesis (H0): The mean amount of liquid dispensed by the machine is 12 ounces.
Alternative Hypothesis (Ha): The mean amount of liquid dispensed by the machine is less than 12 ounces.
Next, we need to determine the test statistic. In this case, we are comparing a sample mean to a population mean with a known standard deviation. Therefore, we will use the one-sample z-test.
The formula for the test statistic (z) is:
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.
Given values:
x̄ = 11.5 ounces (sample mean)
μ = 12 ounces (population mean)
σ = 1.1 ounces (population standard deviation)
n = 30 (sample size)
Plugging in the values:
z = (11.5 - 12) / (1.1 / sqrt(30))
Calculating the test statistic:
z ≈ -2.081
The next step is to calculate the p-value, which represents the probability of obtaining a test statistic as extreme as the observed, assuming the null hypothesis is true.
To find the p-value, we will use a standard normal distribution table or calculator. We are conducting a one-tailed test in the left tail, since the alternative hypothesis is that the mean amount is less than 12 ounces.
Looking up the z-score (-2.081) in the standard normal distribution table, we find that the corresponding p-value is approximately 0.0188.
The p-value (0.0188) is less than the significance level (α = 0.05). This means that there is enough evidence to reject the null hypothesis.
Therefore, based on the data, there is enough evidence to support the customer's claim that the coffee-dispensing machine is broken and delivering less than 12 ounces of liquid.