The Phillies were accused of cheating their beer customers by not putting enough beer in their 12 oz cups. When 15 cups were selected at random, they were found to have a mean of 11.7 ounces with a standard deviation of 0.5 ounces. At the .01 significance level, test the claim that the customers are being cheated.
Z = (score-mean)/SEm = (11.7-12)/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 (.01) related to the Z score.
To test the claim that the customers are being cheated, we can set up a hypothesis test. The null hypothesis (H₀) would state that the mean amount of beer in a 12 oz cup is equal to 12 ounces. The alternative hypothesis (H₁) would state that the mean amount of beer in a 12 oz cup is less than 12 ounces.
H₀: μ = 12 (Mean amount of beer in a 12 oz cup is equal to 12 ounces)
H₁: μ < 12 (Mean amount of beer in a 12 oz cup is less than 12 ounces)
To conduct this hypothesis test, we will use a one-sample t-test since the population standard deviation is unknown.
First, calculate the test statistic (t-score) using the given information:
Sample mean (x̄) = 11.7 ounces
Sample standard deviation (s) = 0.5 ounces
Sample size (n) = 15
t-score = (x̄ - μ) / (s / √n)
t-score = (11.7 - 12) / (0.5 / √15)
Next, we need to find the critical value to determine the rejection region. Since the significance level is 0.01 and we are testing if the mean is less than 12 ounces, we want to find the critical value for a one-tailed test at a 0.01 level of significance.
To find the critical t-value, we need to look it up in the t-distribution table or use statistical software. Assuming a one-tailed test with a significance level of 0.01 and degrees of freedom (df) = n-1 = 14, the critical t-value is approximately -2.624.
Finally, we can compare the test statistic (t-score) with the critical t-value to make a decision. If the t-score is less than the critical t-value, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.
If the t-score is less than -2.624, we reject the null hypothesis and conclude that there is evidence to support the claim that the customers are being cheated. If the t-score is greater than or equal to -2.624, we fail to reject the null hypothesis and do not have sufficient evidence to claim that the customers are being cheated.
Now, you can calculate the test statistic (t-score) and compare it with the critical t-value to make a decision.