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The article "Evaluating Variability in Filling Operations" (Food Tech., 1984: 51-55) describes two different filling operations used in a ground-beef packing plant. Both filling operations were set to fill packages with \(1400 \mathrm{~g}\) of ground beef. In a random sample of size 30 taken from each filling operation, the resulting means and standard deviations were \(1402.24 \mathrm{~g}\) and \(10.97 \mathrm{~g}\) for operation 1 and \(1419.63 \mathrm{~g}\) and \(9.96 \mathrm{~g}\) for operation \(2 .\) a. Using a 05 significance level, is there sufficient evidence to indicate that the true mean weight of the packages differs for the two operations? b. Does the data from operation 1 suggest that the true mean weight of packages produced by operation 1 is higher than \(1400 \mathrm{~g}\) ? Use a \(.05\) significance level.

Short Answer

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(a) Not enough evidence to conclude means differ. (b) Evidence suggests mean is higher than 1400g.

Step by step solution

01

Identify the Hypotheses for Part (a)

To determine if there's a difference in the true mean weights for the two operations, we commence with the null hypothesis. The null hypothesis states that there is no difference in the means of the two groups \(H_0: \mu_1 = \mu_2\). The alternative hypothesis suggests there is a difference: \(H_a: \mu_1 eq \mu_2\). This will be a two-tailed test.
02

Gather Test Statistic for Part (a)

We need to calculate the test statistic for the difference in means. The formula for the test statistic when two independent samples are involved is given by: \\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]where \(\bar{x}_1 = 1402.24\), \(\bar{x}_2 = 1419.63\), \(s_1 = 10.97\), \(s_2 = 9.96\), and \(n_1 = n_2 = 30\).
03

Calculate Degrees of Freedom and Critical Value for Part (a)

The degrees of freedom for a two-sample t-test using unequal variances can be estimated with \( u \approx \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \). With these values, calculate the degrees of freedom \\[ u \approx \] Use a t-distribution table to find the critical value for a two-tailed test at \( \alpha = 0.05\).
04

Compare Test Statistic to Critical Value for Part (a)

Calculate the test statistic for the means difference using the formula from Step 2. If the calculated \( t \) is greater than the critical value from Step 3, reject the null hypothesis. Otherwise, do not reject it.
05

Identify the Hypotheses for Part (b)

For operation 1, the null hypothesis states the true mean is equal to \(1400\): \(H_0: \mu_1 = 1400\). The alternative hypothesis claims that the mean is greater than \(1400\): \(H_a: \mu_1 > 1400\). This is a one-tailed test.
06

Calculate Test Statistic for Part (b)

For a one-sample t-test, use the formula: \\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \]where \(\bar{x} = 1402.24\), \(\mu = 1400\), \(s = 10.97\), \(n = 30\).
07

Find Critical Value for Part (b)

Using the t-distribution table, find the critical value for a one-tailed test with \( \alpha = 0.05\) and \( n - 1 = 29 \) degrees of freedom.
08

Compare Test Statistic to Critical Value for Part (b)

Compute the test statistic from Step 6. Compare it to the critical value from Step 7. If the test statistic is greater, reject the null hypothesis. Otherwise, do not reject it.

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Key Concepts

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

Significance Level
The significance level, commonly represented by the Greek letter alpha (\( \alpha \)), is a threshold we set for determining whether a result is statistically significant. In hypothesis testing, the significance level indicates the probability of rejecting the null hypothesis when it is actually true. This is known as a Type I error.

For example, a significance level of \(0.05\) means there's a \(5\%\) probability of making a Type I error. The choice of \( \alpha \) often depends on the context and consequences of the potential outcomes.
  • If the consequences of a false positive are severe, a smaller significance level (like \(0.01\)) might be chosen.
  • If errors carry minor consequences, a larger \( \alpha \) (such as \(0.1\)) may suffice.
In general, \(0.05\) is a conventional choice in many scientific disciplines, offering a balanced approach between sensitivity and specificity. When conducting tests, if the p-value obtained is less than the chosen significance level, we reject the null hypothesis, suggesting evidence for the alternative hypothesis.
Two-Sample T-Test
A two-sample t-test is a statistical method used to compare the means of two independent groups to determine if they are significantly different from one another. This test is ideal when you have two separate groups and you want to see if they come from the same population.

For instance, in the exercise given, the two-sample t-test can help identify any significant difference in the mean weights of packages between the two operations. Key steps in performing a two-sample t-test include:
  • Formulating the null hypothesis (\(H_0\)), which states that the group means are equal, and the alternative hypothesis (\(H_a\)), which asserts that they are not.
  • Calculating the test statistic using the formula: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]
  • Finding the degrees of freedom using the formula provided, which can often be approximated due to its complexity.
  • Comparing the calculated test statistic to a critical value determined using a t-distribution table at the chosen significance level for a two-tailed test.
If the absolute value of the test statistic exceeds the critical value, the null hypothesis is rejected, suggesting a difference in means. The test is versatile and essential for evaluating differences, provided the data sets are independent and approximately normally distributed.
One-Sample T-Test
A one-sample t-test is used when you want to compare the sample mean to a known value or population mean to determine if they are significantly different. It's helpful in situations where you're evaluating hypotheses about a single data set in comparison to a standard or expected value.

In the problem provided, operation 1's mean is compared to the expected mean of 1400g using a one-sample t-test. Here's how it is conducted:
  • State the null hypothesis (\(H_0\)): that the sample mean equals the population mean (e.g., \(\mu_1 = 1400\)).
  • Define the alternative hypothesis (\(H_a\)): for a one-tailed test, that the sample mean is greater (or less) than the population mean (e.g., \(\mu_1 > 1400\)).
  • Calculate the test statistic with: \[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \]
  • Compare the test statistic to a critical value from the t-distribution table for the specified \( \alpha \) and degrees of freedom (\( n-1 \)).
A significant result, where the test statistic exceeds the critical value, would lead to rejecting the null hypothesis, indicating that the sample mean is significantly different from the comparison mean. It's especially useful when the sample size is small and variability is unknown.
Degrees of Freedom
Degrees of freedom (df) reflect the number of values in a statistical calculation that are free to vary. It forms an essential part of various statistical analyses, including t-tests, by defining the exact form of the sampling distribution that underpins the test.

In t-tests, degrees of freedom are critical in determining the shape of the t-distribution, which consequently affects the critical values and p-values. The formula for degrees of freedom differs between statistical tests:
  • For a two-sample t-test with unequal variances, a complicated formula is often used. An approximation, known as the Welch-Satterthwaite equation, can be applied when calculating df, simplifying interpretation.
  • In a one-sample t-test, the degrees of freedom are straightforward, calculated as \( n-1 \), where \( n \) is the sample size.
Understanding degrees of freedom helps in correctly interpreting results from critical value tables and offers insight into how sample size impacts statistical power and significance. More degrees of freedom usually mean a more reliable estimate of the population parameter, impacting how rigorous or lenient a hypothesis test can be.

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Most popular questions from this chapter

A mechanical engineer wishes to compare strength properties of steel beams with similar beams made with a particular alloy. The same number of beams, \(n\), of each type will be tested. Each beam will be set in a horizontal position with a support on each end, a force of 2500 lb will be applied at the center, and the deflection will be measured. From past experience with such beams, the engineer is willing to assume that the true standard deviation of deflection for both types of beam is \(.05\) in. Because the alloy is more expensive, the engineer wishes to test at level \(.01\) whether it has smaller average deflection than the steel beam. What value of \(n\) is appropriate if the desired type II error probability is .05 when the difference in true average deflection favors the alloy by 04 in.?

Olestra is a fat substitute approved by the FDA for use in snack foods. Because there have been anecdotal reports of gastrointestinal problems associated with olestra consumption, a randomized, double-blind, placebo- controlled experiment was carried out to compare olestra potato chips to regular potato chips with respect to Gl symptoms ("Gastrointestinal Symptoms Following Consumption of Olestra or Regular Triglyceride Potato Chips, J. of the Amer. Med. Assoc., 1998: 150-152). Among 529 individuals in the TG control group, 17.6\% experienced an adverse GI event, whereas among the 563 individuals in the olestra treatment group, \(15.8 \%\) experienced such an event. a. Carry out a test of hypotheses at the \(5 \%\) significance level to decide whether the incidence rate of GI problems for those who consume olestra chips according to the experimental regimen differs from the incidence rate for the TG control treatment. b. If the true percentages for the two treatments were \(15 \%\) and \(20 \%\), respectively, what sample sizes \((m=n)\) would be necessary to detect such a difference with probability \(90 ?\)

Headability is the ability of a cylindrical piece of material to be shaped into the head of a bolt, screw, or other cold-formed part without cracking. The article "New Methods for Assessing Cold Heading Quality" (Wire J. Intl., Oct. 1996: 66-72) described the result of a headability impact test applied to 30 specimens of aluminum killed steel and 30 specimens of silicon killed steel. The sample mean headability rating number for the steel specimens was \(6.43\), and the sample mean for aluminum specimens was 7.09. Suppose that the sample standard deviations were \(1.08\) and \(1.19\), respectively. Do you agree with the article's authors that the difference in headability ratings is significant at the \(5 \%\) level (assuming that the two headability distributions are normal)?

Arsenic is a known carcinogen and poison. The standard laboratory procedures for measuring arsenic concentration \((\mu \mathrm{g} / \mathrm{L})\) in water are expensive. Consider the accompanying summary data and Minitab output for comparing a laboratory method to a new relatively quick and inexpensive field method (from the article "Evaluation of a New Field Measurement Method for Arsenic in Drinking Water Samples," J. of Envir. Engr, 2008: 382-388). Two-Sample T-Test and CI \(\begin{array}{lcccr}\text { Sample } & \text { N } & \text { Mean } & \text { StDev } & \text { SE Mean } \\ 1 & 3 & 19.70 & 1.10 & 0.64 \\ 2 & 3 & 10.90 & 0.60 & 0.35 \\ \text { Estimate for difference: } 8.800 & \end{array}\) 95? CI for difference: \((6.498,11.102)\) T-Test of difference \(=0\) (vs not \(=\) ): \(\mathrm{T}\)-Value \(=12.16 \mathrm{P}\)-Value \(=0.001 \mathrm{DF}=3\) What conclusion do you draw about the two methods, and why? Interpret the given confidence interval. [Note: One of the article's authors indicated in private communication that they were unsure why the two methods disagreed.]

Suppose a level \(.05\) test of \(H_{0}: \mu_{1}-\mu_{2}=0\) versus \(H_{a}: \mu_{1}-\mu_{2}>0\) is to be performed, assuming \(\sigma_{1}=\sigma_{2}=\) 10 and normality of both distributions, using equal sample sizes \((m=n)\). Evaluate the probability of a type II error when \(\mu_{1}-\mu_{2}=1\) and \(n=25,100,2500\), and 10,000 . Can you think of real problems in which the difference \(\mu_{1}-\mu_{2}=1\) has little practical significance? Would sample sizes of \(n=10,000\) be desirable in such problems?

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