Questions: A manufacturer fills soda bottles. Periodically the company tests to see if there is a difference between the mean amounts of soda put in bottles of regular cola and diet cola. A random sample of 14 bottles of regular cola has a mean of 500.6 mL of soda with a standard deviation of 3.8 mL. A random sample of 16 bottles of diet cola has a mean of 502.9 mL of soda with a standard deviation of 4.7 mL. Test the claim that there is a difference between the mean fill levels for the two types of soda using a 0.10 level of significance. Assume that both populations are approximately normal and that the population variances are not equal since different machines are used to fill bottles of regular cola and diet cola. Let bottles of regular cola be Population 1 and let bottles of diet cola be Population 2. Compute the value of the test statistic. Round your answer to three decimal places.

Solution Steps

The standard error is calculated using the formula:

\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]

Substituting the values:

\[ SE = \sqrt{\frac{3.8^2}{14} + \frac{4.7^2}{16}} = \sqrt{\frac{14.44}{14} + \frac{22.09}{16}} = \sqrt{1.03 + 1.38} = \sqrt{2.41} \approx 1.55 \]

The test statistic is calculated using the formula:

\[ t = \frac{\bar{x}_1 - \bar{x}_2}{SE} \]

Substituting the values:

\[ t = \frac{500.6 - 502.9}{1.55} = \frac{-2.3}{1.55} \approx -1.48 \]

The degrees of freedom for Welch's t-test is calculated using the formula:

\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \]

Substituting the values:

\[ df = \frac{\left(\frac{3.8^2}{14} + \frac{4.7^2}{16}\right)^2}{\frac{\left(\frac{3.8^2}{14}\right)^2}{14 - 1} + \frac{\left(\frac{4.7^2}{16}\right)^2}{16 - 1}} = \frac{(1.03 + 1.38)^2}{\frac{(1.03)^2}{13} + \frac{(1.38)^2}{15}} \approx 25.0 \]

The P-value is calculated using the formula:

\[ P = 2(1 - T(|t|)) \]

Given that \( t \approx -1.48 \):

\[ P = 2(1 - T(1.48)) \approx 0.15 \]

For a significance level of \( \alpha = 0.10 \) and \( df \approx 25 \), the critical value from the t-distribution table is approximately:

\[ \text{Critical Value} \approx 1.711 \]

Final Answer