Reginald Brown an inspector from the department of weights and measures, weighs 15 eighteen-ounce cereal boxes of corn flakes. he finds their mean weight to be 17.78 ounces with a standard deviation of .4 ounces. are the cereal boxes lighter than they should be? let Alpha = .01
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 related to the Z score.
I still am not sure how to proceed with this with the information you gave me. Confused on this one
thank you
To determine whether the cereal boxes are lighter than they should be, we can perform a hypothesis test. In this case, we want to test if the mean weight of the cereal boxes is less than the expected weight.
Let's set up the null and alternative hypotheses:
Null hypothesis (H0): The mean weight of the cereal boxes is equal to the expected weight.
Alternative hypothesis (H1): The mean weight of the cereal boxes is less than the expected weight.
Now, let's calculate the test statistic using the given information:
Sample mean (x̄) = 17.78 ounces
Population mean (μ) = 18 ounces (expected weight)
Standard deviation (σ) = 0.4 ounces
Sample size (n) = 15
The test statistic for this case is the z-score, which measures how many standard deviations the sample mean (x̄) is away from the population mean (μ). The formula to calculate the z-score is as follows:
z = (x̄ - μ) / (σ / √n)
Substituting the values into the formula:
z = (17.78 - 18) / (0.4 / √15)
z = (-0.22) / (0.1039)
z = -2.12 (rounded to two decimal places)
Next, we need to determine the critical value at a significance level (α) of 0.01. This critical value corresponds to the left-tail test because we are testing if the cereal boxes are lighter than the expected weight.
Using a standard normal distribution table or a statistical calculator, we find that the critical value for α = 0.01 (one-tailed test) is approximately -2.33 (rounded to two decimal places).
Comparing the calculated z-score (-2.12) with the critical value (-2.33), we see that -2.12 falls to the left of -2.33. Therefore, the p-value (probability of obtaining a test statistic as extreme as the observed result) is smaller than the significance level (p-value < α).
Since the p-value is less than the significance level, we reject the null hypothesis. Thus, we can conclude that the cereal boxes are lighter than they should be.
In summary, based on the hypothesis test, we can say that the cereal boxes are lighter than the expected weight at a significance level of 0.01.