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For Exercises 7 through 20 , assume that all variables are normally distributed, that the samples are independent, that the population variances are equal, and that the samples are simple random samples, one from each of the populations. Also, for each exercise, perform the following steps. The amount of sodium (in milligrams) in one serving for a random sample of three different kinds of foods is listed. At the 0.05 level of significance, is there sufficient evidence to conclude that a difference in mean sodium amounts exists among condiments, cereals, and desserts?$$ \begin{array}{ccc} \text { Condiments } & \text { Cereals } & \text { Desserts } \\ \hline 270 & 260 & 100 \\ 130 & 220 & 180 \\ 230 & 290 & 250 \\ 180 & 290 & 250 \\ 80 & 200 & 300 \\ 70 & 320 & 360 \\ 200 & 140 & 300 \\ & & 160 \end{array}$$

Short Answer

Expert verified
Perform an ANOVA test; reject or fail to reject the hypothesis based on the F-statistic compared to the critical F-value.

Step by step solution

01

State the Hypotheses

In hypothesis testing, we begin by stating the null and alternative hypotheses. Here, the null hypothesis \(H_0\) is that there is no difference in the mean amount of sodium across the three food categories (condiments, cereals, and desserts). The alternative hypothesis \(H_1\) is that at least one category mean is different. Mathematically, \(H_0: \mu_1 = \mu_2 = \mu_3\) versus \(H_1: \) not all means are equal.
02

Calculate Means and Variances

Calculate the sample means and variances for each food category. Sum the values for each category and divide by the number of samples to find the mean. For variance, use the formula for sample variance: \(s^2 = \frac{1}{n-1}\sum_{i=1}^{n}(x_i-\bar{x})^2\). This helps us understand spread or dispersion in each group.
03

ANOVA Test Setup

We use the ANOVA test to compare means from more than two groups. Compute the total sum of squares (SST), sum of squares between groups (SSB), and sum of squares within groups (SSW). SST is calculated by summing the squared differences between each observation and the grand mean. SSB is calculated by summing \(n_k(\bar{x}_k - \bar{x}_{ ext{grand}})^2\), and SSW is the total variability within each group.
04

Calculate F-statistic

Calculate the F-statistic using the ANOVA formulas. The F-statistic is given by \(F = \frac{SSB/(k-1)}{SSW/(N-k)}\), where \(N\) is the total number of observations and \(k\) is the number of groups. This F-value will be compared against the F-distribution critical value.
05

Determine Critical Value and Conclusion

Using the F-distribution table and the degrees of freedom for the groups and the errors (\(k-1\) and \(N-k\) respectively), find the critical F-value at the 0.05 significance level. If the computed F-statistic is greater than the critical value, we reject the null hypothesis. Otherwise, we fail to reject it, indicating insufficient evidence to conclude a difference exists.

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Key Concepts

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

Hypothesis Testing
Hypothesis testing is a fundamental statistical technique used to make decisions about data. In this exercise, we start by setting up two competing hypotheses:
  • Null Hypothesis (\(H_0\)): Assumes that there is no difference in the mean sodium content among the three food categories: condiments, cereals, and desserts. Mathematically, this is expressed as \(H_0: \mu_1 = \mu_2 = \mu_3\).
  • Alternative Hypothesis (\(H_1\)): Assumes that at least one group mean is different from the others. This is denoted as \(H_1\): not all means are equal.
The goal of hypothesis testing is to use sample data to determine which of these hypotheses is more likely to be true. We use statistical tests, like ANOVA, to make these determinations based on sample data characteristics.
F-statistic
The F-statistic is a vital component in ANOVA testing, serving to determine if the variability between group means is more significant than the variability within the groups. To calculate the F-statistic, we first find the sums of squares:
  • Sum of Squares Between (SSB): Represents the variability due to the interaction between different groups. It reflects the difference between each group mean and the overall mean.
  • Sum of Squares Within (SSW): Captures the variability within each group.
The F-statistic calculation involves these components and is described by the formula:\[F = \frac{SSB/(k-1)}{SSW/(N-k)}\]where \(k\) is the number of groups and \(N\) is the total number of observations. A higher F-value indicates that the difference among group means may be more substantial than would be expected by chance alone.
Degrees of Freedom
Degrees of freedom are essential for understanding statistical tests as they indicate the number of values in a calculation that are free to vary. In our ANOVA test, we work with two types of degrees of freedom:
  • Between Groups (\(df_{between}\)): Calculated as \(k - 1\), where \(k\) is the number of groups. It reflects the number of comparisons being made between the group means.
  • Within Groups (\(df_{within}\)): Given by \(N - k\), where \(N\) is the total sample size. This applies to data variability within each group.
Understanding degrees of freedom helps in interpreting the F-statistic and finding the critical value from F-distribution tables.
Sample Variance
Sample variance measures how data values differ from the sample mean. It provides insight into the data's spread or dispersion. The formula for computing sample variance is:\[s^2 = \frac{1}{n-1}\sum_{i=1}^{n}(x_i-\bar{x})^2\]where \(n\) is the sample size, \(x_i\) represents each data point, and \(\bar{x}\) is the sample mean. By calculating variances for each of the food categories, we understand how sodium amounts are distributed within each category.
Sample variance is a critical step in ANOVA since it is used in the calculation of the F-statistic and helps decide whether to accept or reject the null hypothesis.

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Most popular questions from this chapter

For Exercises 7 through 20 , assume that all variables are normally distributed, that the samples are independent, that the population variances are equal, and that the samples are simple random samples, one from each of the populations. Also, for each exercise, perform the following steps. Annual child care costs for infants are considerably higher than for older children. At \(\alpha=0.05,\) can you conclude a difference in mean infant day care costs for different regions of the United States? (Annual costs per infant are given in dollars. $$ \begin{array}{ccc} \text { New England } & \text { Midwest } & \text { Southwest } \\ \hline 10,390 & 9,449 & 7,644 \\ 7,592 & 6,985 & 9,691 \\\ 8,755 & 6,677 & 5,996 \\ 9,464 & 5,400 & 5,386 \\ 7.328 & 8.372 & \end{array} $$

For Exercises 7 through 20 , assume that all variables are normally distributed, that the samples are independent, that the population variances are equal, and that the samples are simple random samples, one from each of the populations. Also, for each exercise, perform the following steps. Americans appear to be eating healthier. Between 1970 and 2013 the per capita consumption of broccoli increased \(1200.5 \%\) from 0.5 to 6.4 pounds. A nutritionist followed a group of people randomly assigned to one of three groups and noted their monthly broccoli intake (in pounds). At \(\alpha=0.05,\) is there a difference in means? $$ \begin{array}{ccc} \text { Group A } & \text { Group B } & \text { Group C } \\ \hline 2.0 & 2.0 & 3.7 \\ 1.5 & 1.5 & 2.5 \\ 0.75 & 4.0 & 4.0 \\ 1.0 & 3.0 & 5.1 \\ 1.3 & 2.5 & 3.8 \\ 3.0 & 2.0 & 2.9 \end{array} $$.

Explain the difference between the two tests used to compare two means when the null hypothesis is rejected using the one-way ANOVA \(F\) test.

For Exercises 7 through 20 , assume that all variables are normally distributed, that the samples are independent, that the population variances are equal, and that the samples are simple random samples, one from each of the populations. Also, for each exercise, perform the following steps. The average local cell phone monthly bill is \(\$ 50.07\). A random sample of monthly bills from three different providers is listed below. At \(\alpha=0.05\), is there a difference in mean bill amounts among providers? $$ \begin{array}{ccc} \text { Provider X } & \text { Provider Y } & \text { Provider Z } \\ \hline 48.20 & 105.02 & 59.27 \\ 60.59 & 85.73 & 65.25 \\ 72.50 & 61.95 & 70.27 \\\ 55.62 & 75.69 & 42.19 \\ 89.47 & 82.11 & 52.34 \end{array} $$

For Exercises 7 through 20 , assume that all variables are normally distributed, that the samples are independent, that the population variances are equal, and that the samples are simple random samples, one from each of the populations. Also, for each exercise, perform the following steps. The following data show the yearly budgets for leading business sectors in the United States. At \(\alpha=0.05\), is there a significant difference in the mean budgets of the business sectors? The data are in thousands of dollars. $$ \begin{array}{cccc} & \text { Food } & \text { Supportive } \\ \text { Beverages } & \text { Electronics } & \text { producers } & \text { services } \\\ \hline 170 & 46 & 59 & 56 \\\128 & 24 & 58 & 37 \\ 19 & 18 & 33 & 19 \\\ 16 & 14 & 31 & 19 \\ 12 & 13 & 28 & 17 \\ 11 & 12 & 22 & 15 \\ 10 & 10 & 16 & 15 \end{array} $$.

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