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 colà. A random sample of 18 bottles of regular cola has a mean of $500.5 \mathrm{~mL}$ of soda with a standard deviation of $3.9 \mathrm{~mL}$. A random sample of 12 bottles of diet cola has a mean of $497.2 \mathrm{~mL}$ of soda with a standard deviation of $4.3 \mathrm{~mL}$. Test the claim that there is a difference between the mean fill levels for the two types of soda using a 0.02 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. Step 2 of 3: Compute the value of the test statistic. Round your answer to three decimal places. Answer Tables Keypad Keyboard Shortcuts
Solution
Step 1 ：The problem is asking for the value of the test statistic in a two-sample t-test. The formula for the test statistic in a two-sample t-test is: \[ t = \frac{{\bar{X}_1 - \bar{X}_2}}{{\sqrt{\frac{{s_1^2}}{{n_1}} + \frac{{s_2^2}}{{n_2}}}}} \] where \(\bar{X}_1\) and \(\bar{X}_2\) are the sample means, \(s_1^2\) and \(s_2^2\) are the sample variances, and \(n_1\) and \(n_2\) are the sample sizes.
Step 2 ：In this case, we have: \(\bar{X}_1 = 500.5\), \(\bar{X}_2 = 497.2\), \(s_1 = 3.9\), \(s_2 = 4.3\), \(n_1 = 18\), and \(n_2 = 12\).
Step 3 ：We can substitute these values into the formula to calculate the test statistic: \[ t = \frac{{500.5 - 497.2}}{{\sqrt{\frac{{3.9^2}}{{18}} + \frac{{4.3^2}}{{12}}}}} \]
Step 4 ：The value of the test statistic, rounded to three decimal places, is \(\boxed{2.136}\).
