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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 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 the \(1 \%\) 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 \(\alpha=.005 ?\) What if \(\alpha=.025\) ?

Step by step solution

01

Calculate the difference of sample means.

First, find the difference of sample means for the two companies. Let's denote mean calories in Maine Mountain Dairy yogurts as \(\mu_1\) and that for the competitor as \(\mu_2\). Sample means, denoted \(\overline{X}_1\) and \(\overline{X}_2\), are 141 and 144 respectively. So, the difference is \(\overline{X}_1 - \overline{X}_2 = 141 - 144 = -3\).

02

Compute the standard error of the difference.

Now, the standard error of the difference is needed, which is \(\sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}\), where \(S_1 = 5.5\) and \(S_2 = 6.4\) are the standard deviations, and \(n_1 = 27\) and \(n_2 = 25\) are the sample sizes. Substituting these values gives the standard error as approximately \(1.663\).

03

Construct the confidence interval for the difference of means.

For the first part (a), a 98% confidence interval for the difference of means is needed. It's given by \(\overline{X}_1 - \overline{X}_2 \pm Z_{\frac{\alpha}{2}} . SE\), where \(Z_{\frac{\alpha}{2}}\) is the critical Z value. For a 98% confidence level, the critical Z value is approximately 2.33. Hence, the confidence interval is \(-3 \pm 2.33 . 1.663\), which is \(-6.87, 0.87\).

04

Set up the hypothesis test.

For the second part (b), a hypothesis test is needed to check Maine Mountain Dairy's claim. The null hypothesis is \(H_0: \mu_1 - \mu_2 = 0\) (i.e., mean calories are equal) and the alternative hypothesis is \(H_1: \mu_1 - \mu_2 < 0\) (i.e., Maine yogurts have less mean calories). The test statistic is \(Z = \frac{\overline{X}_1 - \overline{X}_2}{SE}\), which equals \(-1.80\). For a 1% level of significance, the critical Z value is -2.33, so fail to reject the null hypothesis as \(Z > Z_{\alpha}\). Maine Mountain Dairy's claim cannot be substantiated.

05

Compute the p-value and re-analyze the testing outcomes.

In the third part (c), the p-value of the test is required. This is the smallest level of significance at which we would reject the null hypothesis, given the computed Z. Knowing that Z is -1.80, the p-value is 0.036. This is the probability of observing a Z value at least as extreme as -1.80 when the null hypothesis is true. Hence, for \(\alpha = 0.005\) or 0.5%, we fail to reject the null hypothesis; but for \(\alpha = 0.025\) or 2.5%, we reject the null hypothesis.

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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 gives a range in which we expect the true population parameter to lie, with a certain level of confidence. In this exercise, we calculated a 98% confidence interval for the difference in mean calories between the yogurt cups of the two companies. A confidence interval has the form: \[ \overline{X}_1 - \overline{X}_2 \pm Z_{\frac{\alpha}{2}} \times SE \] where \( \overline{X}_1 - \overline{X}_2 \) is the difference in sample means, \( Z_{\frac{\alpha}{2}} \) is the critical Z value, and \( SE \) is the standard error.

• The critical Z value for a 98% confidence level is approximately 2.33.
• The difference in sample means is -3.
• The standard error is calculated as approximately 1.663.

The interval is therefore \( -3 \pm 2.33 \times 1.663 \), which results in \( (-6.87, 0.87) \). This means we can say with 98% confidence that the true difference in means lies within this range.

P-value

The p-value is a measure used in hypothesis testing to help us determine the strength of the evidence against the null hypothesis. It is the probability of obtaining test results at least as extreme as those observed, assuming the null hypothesis is true. In our exercise, the p-value was calculated based on the test statistic, which was found to be -1.80.

• A p-value of 0.036 was obtained, indicating that if the null hypothesis were true, there is a 3.6% chance of observing a result as extreme or more extreme than the one calculated.

Using this p-value, we can compare it to different significance levels (e.g., \(\alpha = 0.005\) and \(\alpha = 0.025\)) to decide whether to reject the null hypothesis.

Z-test

A Z-test is a statistical test used to determine if there is a significant difference between sample and population means, or between means from two samples, when the population variances are known. In this scenario, we performed a Z-test to compare the average caloric content of different yogurt brands. The Z-test formula used was: \[ Z = \frac{\overline{X}_1 - \overline{X}_2}{SE} \] where \( \overline{X}_1 - \overline{X}_2 \) is the difference in sample means and \( SE \) is the standard error.

• The test statistic calculated was \(Z = -1.80\).
• Comparing this with a critical Z value based on the chosen significance level helped us decide whether to support or reject the null hypothesis.

Significance Level

The significance level, denoted as \( \alpha \), is the probability of rejecting the null hypothesis when it is true. It represents the risk we are willing to take to make such an error. Different significance levels provide different thresholds for decision-making. In our exercise:

• At a significance level of 1% (\(\alpha = 0.01\)), the critical Z value was -2.33.
• The computed Z value of -1.80 did not exceed this threshold, leading us to "fail to reject" the null hypothesis.
• When \(\alpha\) was set at 2.5% (\(\alpha = 0.025\)), the p-value of 0.036 was compared against it, leading us to "reject" the null hypothesis.

This emphasizes how the choice of significance level can change the outcome of a hypothesis test.