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Maine Mountain Dairy claims that its 8-ounce low-fat yogurt cups contain, on average, fewer calories than the 8 -ounce low-fat yogurt cups produced by a competitor. A consumer agency wanted to check this claim. A sample of 27 such yogurt cups produced by this company showed that they contained an average of 141 calories per cup. A sample of 25 such yogurt cups produced by its competitor showed that they contained an average of 144 calories per cup. Assume that the two populations are approximately normally distributed with population standard deviations of \(5.5\) and \(6.4\) calories, repectively. a. Make a \(98 \%\) confidence interval for the difference between the mean number of calories in the 8 -ounce low-fat yogurt cups produced by the two companies. b. Test at a \(1 \%\) significance level whether Maine Mountain Dairy's claim is true. c. Calculate the \(p\) -value for the test of part b. Based on this \(p\) -value, would you reject the null hypothesis if \(\alpha=.05 ?\) What if \(\alpha=.025 ?\)

Step by step solution

01

Compute the mean difference and standard error

First, calculate the difference in sample means. This is \(141 - 144 = -3\) calories. Next, calculate the standard error (SE) using the formula \(SE = \sqrt{(s1^2/n1) + (s2^2/n2)}\), where s1 and s2 are the standard deviations and n1 and n2 are the sample sizes. Substituting the given values, we get \(SE = \sqrt{(5.5^2/27) + (6.4^2/25)} = 1.6178\).

02

Calculate the 98% confidence interval

The 98% confidence interval for the difference between the two means is given by \((d - Zα/2*SE, d + Zα/2*SE)\), where d is the mean difference, Zα/2 is the critical value from the standard normal distribution for a 98% confidence level, and SE is the standard error. The critical value Zα/2 for a 98% confidence level is 2.33. Substituting the values, we get the 98% confidence interval as \((-3 - 2.33*1.6178, -3 + 2.33*1.6178) = (-6.77, 0.77)\).

03

Run a hypothesis test at 1% significance level

To test the hypothesis at a 1% significance level, compute the Z statistic using the formula Z = (d - μ)/SE, where μ is the difference stated in the null hypothesis (in our case 0). The calculated Z is -3/1.6178 = -1.855. The critical value for a two-tailed test at a 1% significance level is approximately ±2.58. Since our computed Z falls within the critical region, we would not reject the null hypothesis that the mean number of calories in the yogurt cups produced by the two companies is the same.

04

Calculate the p-value

The p-value is the smallest level of significance at which we would be able to reject the null hypothesis. In this instance, it is calculated from the Z statistic obtained in step 3 using a standard normal distribution table. The p-value is two times the area to the left of -1.855, which is 2(0.0314) = 0.0628.

05

Interpret the p-value

For α = 0.05, because the p-value of 0.0628 is greater than 0.05, we would not reject the null hypothesis. Therefore, the data do not provide enough evidence to support the claim that Maine Mountain Dairy's yogurt cups contain fewer calories. And for α = 0.025, because the p-value of 0.0628 is still greater than 0.025, we would not reject the null hypothesis at this level either.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Interval

A confidence interval provides a range of values which we believe contain the true difference in population means. Here, we use a 98% confidence level to find the range for the caloric difference between yogurt cups from two companies.

First, calculate the mean difference, which is \(-3\) calories. Then, utilize the standard error \(SE = 1.6178\) to determine the confidence interval using the critical value \(Z_{\alpha/2} = 2.33\). The resulting interval is \((-6.77, 0.77)\).

This interval suggests that while the average difference may vary, there's a possibility the true mean difference is negative, indicating fewer calories in Maine Mountain Dairy's yogurt compared to the competitor's.

Significance Level

The significance level, denoted as \(\alpha\), is the probability of rejecting the null hypothesis when it is true. It sets the threshold for determining whether a test result is statistically significant.

In this problem, we test the claim at a 1% significance level (\(\alpha = 0.01\)). This means that we must obtain very strong evidence against the null hypothesis to consider its rejection.

The critical value for a two-tailed test at this level is approximately \(\pm2.58\). If our computed test statistic is within this region, we do not reject the null hypothesis. Therefore, the result indicates insufficient evidence to claim a caloric difference in favor of Maine Mountain Dairy's yogurts.

p-value

The p-value helps determine the significance of the results. It represents the probability of observing the calculated data, or something more extreme, assuming the null hypothesis is true.

In this case, the p-value is calculated to be \(0.0628\), which suggests that there is a \(6.28\%\) chance of finding the observed or more extreme difference in calorie counts if the null hypothesis is correct.

For \(\alpha = 0.05\) or \(\alpha = 0.025\), since the p-value is greater, we do not reject the null hypothesis. This implies that the evidence is not strong enough to support Maine Mountain Dairy's claim when relying on these significance levels.

Null Hypothesis

The null hypothesis (\(H_0\)) is a default assumption that there is no difference or effect. It serves as a statement to test against the alternative hypothesis (\(H_a\)).

In the scenario provided, the null hypothesis is that there is no difference in the average number of calories between the yogurt cups from the two companies. This implies the mean difference is zero.

Conducting a hypothesis test involves comparing this hypothesis to the calculated test statistic. If it falls within the critical range, we retain the null hypothesis. Here, it means the evidence was not sufficient to confirm that Maine Mountain Dairy's yogurts contain fewer calories than the competitor's.