# Is there a difference in the weights of contents in a can of Diet Coke versus a regular can of Coke?  In a random sample of n=36 cans of Diet Coke, the mean weight of the contents of a can was 0.78479 lbs with a standard deviation of 0.00439 lbs. In an independent random sample of n=36 cans of regular Coke, the mean weight of the contents of a can was 0.81682 lbs with a standard deviation of 0.00751 lbs.Assuming that both populations are normally distributed and that sigma1 to the power of 2 = sigma2 to the power of 2, determine whether there is a difference in the mean weight of the contents of a can of Diet Coke vs. regular Coke with a 5% significance level.

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Is there a difference in the weights of contents in a can of Diet Coke versus a regular can of Coke?  In a random sample of n=36 cans of Diet Coke, the mean weight of the contents of a can was 0.78479 lbs with a standard deviation of 0.00439 lbs. In an independent random sample of n=36 cans of regular Coke, the mean weight of the contents of a can was 0.81682 lbs with a standard deviation of 0.00751 lbs.

• Assuming that both populations are normally distributed and that sigma1 to the power of 2 = sigma2 to the power of 2, determine whether there is a difference in the mean weight of the contents of a can of Diet Coke vs. regular Coke with a 5% significance level.
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Step 1

Solution:

Null and alternative hypotheses:

Null hypothesis: µ1−µ2 = 0

Alternative hypothesis: µ1−µ2 ≠ 0

Test statistic:

The population standard deviation is unknown. Hence, the test statistic for this problem is shown below.

Step 2

Calculation of test statistic value:

Here, for diet coke, x1-bar=0.78479, s1=0.00439, n1=36

For regular coke, x2-bar=0.81682, s2=0.00751, n2=36

Step 3

Degrees of freedom:

df = n1 + n2 – 2 = 36 + 36 – 2 = 70.

Rejection region:

The rejection region for 5% significance level can be obtained using the EXCEL formula. “=T.INV.2T(0.05,70)”

Thus, the critical value of Student’s t is ttab= 1.9944.

Decision rule for two-tailed t...

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