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Chapter 10: Problem 15

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 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 \(\alpha=.025\) ?

Short Answer

Expert verified
1. The 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 is from -4.77 to -1.23. 2. Using a 1% significance level, we would fail to reject the null hypothesis that the competitor's yogurt contains more or the same number of calories as Maine Mountain Dairy's yogurt since calculated p-value (after looking up on a T-distribution table) is greater than 0.01. 3. For the last question, the p-value is less than 0.05, and greater than 0.025. Hence, we would reject the null hypothesis at \( \alpha = 0.05 \), and fail to reject at \( \alpha = 0.025 \).

Step by step solution

01

Compute the Confidence Interval for the Difference in Mean Calories

Firstly, we'll find the difference in sample means: \( \Delta \) = \( \bar{X}_{1} \) - \( \bar{X}_{2} \) = 141 - 144 = -3. Next, calculate the standard error of the mean difference by using the formula \(\sqrt{(\sigma_{1}^{2}/n_{1}) + (\sigma_{2}^{2}/n_{2})}\), which will give: \(\sqrt{(5.5^{2}/27) + (6.4^{2}/25)}\)=1.44. Using a t-distribution table, a 98% confidence interval yields \((1 - 0.98)/2 = 0.01\), so it's t = 2.63 (for degrees of freedom, df = min(27,25)-1 =24). The range for the confidence interval is calculated as \(\Delta \) 卤 t(SE) = -3 卤 2.63*1.44 .
02

Perform the Hypothesis Test

Next, we need to test if the competitor's yogurt has more calories on average than Maine Mountain Dairy's yogurt. So null hypothesis, \( H_{0}: \mu_{1} \geq \mu_{2} \), and alternate hypothesis, \( H_{a}: \mu_{1} < \mu_{2} \). Compute the test statistic as \( T = \frac{\Delta - D_{0}}{SE}\) where \( \Delta \) is the observed difference in sample means, \( D_{0} \) is the hypothesized difference in means, and SE is the standard error. So, \( T =\frac{-3 - 0}{1.44} \).
03

Calculate the P-Value and Make a Decision

Find the p-value of the calculated test statistic. Now, compare the p-value with the significance level \( \alpha \). If the p-value is less than \( \alpha \), then we reject the null hypothesis. Answer the last two questions, i.e., if we would reject the null hypothesis if \( \alpha = 0.05 \) and if \( \alpha = 0.025 \).

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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 is used to estimate the range within which a population parameter lies, based on sample data. It provides an interval which we are fairly certain contains the population parameter.
For this exercise, we constructed a confidence interval to compare the average calories in yogurt cups from two companies. This tells us how much we can trust the observed difference in sample means.
  • We calculated the difference in sample means as \(-3\).
  • The standard error was calculated as \(1.44\).
  • The 98% confidence interval is calculated using a t-distribution due to our sample size, representative of smaller samples where population variances are unknown.
The resulting confidence interval allows us to judge whether the real mean difference could be zero or greater, thus informing us if Maine Mountain Dairy's claim holds statistically significant weight.
Significance Level
The significance level, denoted by \(\alpha\), is the probability threshold below which the null hypothesis is rejected in a hypothesis test. It's a predefined criterion that helps us make decisions regarding the hypothesis. For hypothesis testing:
  • We test at a \(1\%\) significance level, \((\alpha = 0.01)\).
  • This tells us how willing we are to risk making an incorrect rejection of the null hypothesis.
A \(1\%\) significance level displays strong evidence against the null hypothesis as it requires a stricter criterion (it's only rejected if the observed data is highly improbable under the null model). Increasing the significance level decreases certainty in rejecting the null hypothesis but makes rejections more likely.
P-Value
The p-value quantifies the strength of evidence against the null hypothesis from our sample data. It represents the probability of obtaining a test statistic as extreme as the observed one, under the assumption that the null hypothesis is true. In the context of the yogurt calorie comparison:
  • The p-value was calculated from the test statistic found in hypothesis testing.
  • If the p-value is lower than our significance level, we reject our null hypothesis.
By comparing the p-value with different significance levels (e.g., \( \alpha = 0.05 \) or \( \alpha = 0.025 \)), we determine whether the evidence is strong enough to reject the null hypothesis under varying levels of strictness.
Sample Mean
The sample mean acts as an estimate of the population mean, calculated as the average of the observations in our sample. In this study:
Maine Mountain Dairy's yogurt had an observed sample mean of 141 calories, while the competitor's was 144 calories. This discrepancy formed the basis for our hypothesis test.
By considering the sample means from both companies, we could form an initial basis for claiming which company鈥檚 yogurt might generally offer lower calorie content.
Standard Error
The standard error measures the variability of a sample statistic, in this case, the sample mean difference. It's calculated as the standard deviation of the sampled statistic.
For comparing the yogurt calories:
  • Standard error reflects how much the sample mean of one group well represents the population mean when instigated repeatedly.
  • Calculated using the population standard deviations and sample sizes, the standard error directly affected our confidence interval and hypothesis testing decisions.
Understanding standard error helps in judging the precision of our estimate of the mean difference between the two yogurt brands.

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