Question

question 6 of 8, step 2 of 3
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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 501.3 ml of soda with a standard deviation of 3.4 ml. a random sample of 18 bottles of diet cola has a mean of 497.7 ml of soda with a standard deviation of 3.6 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.
step 2 of 3: compute the value of the test statistic. round your answer to three decimal places.
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Explanation:

Step1: Identify the formula

The formula for the two - sample t - test statistic when variances are unequal is $t=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}}$. Here, $\mu_1-\mu_2 = 0$ (null hypothesis), $\bar{x}_1 = 501.3$, $s_1=3.4$, $n_1 = 14$, $\bar{x}_2=497.7$, $s_2 = 3.6$, $n_2=18$.

Step2: Substitute the values

\[

$$\begin{align*} t&=\frac{(501.3 - 497.7)-0}{\sqrt{\frac{3.4^{2}}{14}+\frac{3.6^{2}}{18}}}\\ &=\frac{3.6}{\sqrt{\frac{11.56}{14}+\frac{12.96}{18}}}\\ &=\frac{3.6}{\sqrt{0.8257+0.72}}\\ &=\frac{3.6}{\sqrt{1.5457}}\\ &=\frac{3.6}{1.2432}\\ &\approx 2.909 \end{align*}$$

\]

Answer:

$2.909$