30 randomly selected cans of Coke are measured for the amount of cola in ounces. The sample values have a mean of 12.18 ounces and a standard deviation of 0.118 ounces. Test the claim that cans of coke have a mean amount of cola greater than 12 ounces.
What is the p-value?
A) 8.532
B) 1.000
C) 0.9999
D) ≤ 0.0001
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 your Z score.
To find the p-value, we need to conduct a hypothesis test. In this case, we are testing the claim that cans of Coke have a mean amount of cola greater than 12 ounces.
Here are the steps to perform the hypothesis test:
1. State the hypotheses:
Null hypothesis (H₀): The mean amount of cola in cans of Coke is equal to 12 ounces.
Alternative hypothesis (H₁): The mean amount of cola in cans of Coke is greater than 12 ounces.
2. Choose the significance level (α). Let's assume α = 0.05 (5%).
3. Compute the test statistic. In this case, we will use the t-test since we have the sample mean and the population standard deviation is unknown.
The test statistic formula is:
t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))
t = (12.18 - 12) / (0.118 / sqrt(30))
4. Determine the critical value(s). The critical value(s) depend on the test statistic and the desired significance level. Since we are testing whether the mean is greater than 12, we will use a right-tailed test. We can find the critical value from the t-distribution table or using statistical software. Let's assume the critical value is t_critical = 1.697 (for a one-tailed test with 29 degrees of freedom at α = 0.05).
5. Calculate the p-value. The p-value is the probability of observing a test statistic as extreme as, or more extreme than, the observed test statistic, assuming that the null hypothesis is true. Since we are performing a right-tailed test, we calculate the area to the right of the observed test statistic under the t-distribution curve. We can find this value using statistical software or a t-distribution table.
6. Compare the p-value with the significance level. If the p-value is less than the significance level (α), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.
Now, let's calculate the p-value using the provided information.
t = (12.18 - 12) / (0.118 / sqrt(30))
t = 5.084
The p-value can be found using statistical software or a t-distribution table for a one-tailed test at 29 degrees of freedom. Assuming that the p-value is approximately 0.00001, which is less than 0.0001, the correct answer for the p-value is:
D) ≤ 0.0001