Questions: Velma conducted a study with 49 coffee drinkers who add chicory and found the sample mean of the amount of caffeinated coffee they drink to be 14.36 ounces, with a sample standard deviation of 2.46. She wants to test if this is less than the accepted average of 15 ounces. At a significance level of α = 0.04, what is the p-value for this test? Round to 4 decimal places.

Solution Steps

To find the p-value for this test, we need to perform a one-sample t-test. The steps are as follows:

1. Calculate the t-statistic using the sample mean, population mean, sample standard deviation, and sample size.
2. Use the t-distribution to find the p-value corresponding to the calculated t-statistic.

Given:

• Sample mean (\(\bar{x}\)) = 14.36
• Population mean (\(\mu\)) = 15
• Sample standard deviation (\(s\)) = 2.46
• Sample size (\(n\)) = 49

The t-statistic is calculated using the formula: \[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \]

Substituting the given values: \[ t = \frac{14.36 - 15}{2.46 / \sqrt{49}} = -1.8211 \]

The p-value is found using the cumulative distribution function (CDF) of the t-distribution with \(df = n - 1\).

Given:

• \(t\)-statistic = -1.8211
• Degrees of freedom (\(df\)) = 49 - 1 = 48

Using the t-distribution CDF: \[ p\text{-value} = \text{CDF}(t, df) = 0.0374 \]

Final Answer