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Suppose that a random sample of size \(n=5\) was selected from the vineyard properties for sale in Sonoma County, California, in each of three years. The following data are consistent with summary information on price per acre (in dollars, rounded to the nearest thousand) for disease-resistant grape vineyards in Sonoma County (Wines and Vines, November 1999). $$\begin{array}{llllll} 1996 & 30,000 & 34,000 & 36,000 & 38,000 & 40,000 \\ 1997 & 30,000 & 35,000 & 37,000 & 38,000 & 40,000 \\ 1998 & 40,000 & 41,000 & 43,000 & 44,000 & 50,000 \end{array}$$ a. Construct boxplots for each of the three years on a common axis, and label each by year. Comment on the similarities and differences. b. Carry out an ANOVA to determine whether there is evidence to support the claim that the mean price per acre for vineyard land in Sonoma County was not the same for the three years considered. Use a significance level of \(.05\) for your test.

Short Answer

Expert verified
The boxplots will give a visual representation of the distribution, center and spread of the prices for each year. The outcome of the ANOVA test will give evidence whether to accept or reject the claim that the mean prices are the same for all three years.

Step by step solution

01

Construct Boxplots

Boxplots can be constructed using any statistical software or even manually. The minimum, first quartile, median, third quartile, and maximum are calculated for each year and a boxplot is sketched for the computed values. Draw your scale and create a box with three vertical lines that represent the first quartile, median and third quartile for the price values. Finally, create lines (whiskers) to the minimum and maximum value, originating from the box.
02

Comment on Boxplots

Study the boxplots constructed. Comment on the range, interquartile range, median, any possible outliers and similarities and differences between the plots for each year. The interpretation will be based on the distribution, center and spread of the boxplots constructed.
03

Calculate ANOVA

Firstly, calculate the sum of squares within (SSW) and sum of squares between (SSB). The SSW is the variation due to differences within each year, while the SSB is the variation due to differences between years. Next, the degrees of freedom between (dfb) and degrees of freedom within (dfw) are calculated. Then, The Mean of Sum Squares Between (MSB) and Mean of Sum Squares Within (MSW) are calculated. Finally, the F statistic is calculated by dividing MSB by MSW.
04

Hypothesis Testing

Set up the null hypothesis that all means are equal and the alternative hypothesis that at least one mean is different. The computed F statistic value is then compared with the F critical value from the F distribution table for dfb and dfw at the significance level of 0.05. If the calculated F statistic is greater than the F critical value, reject the null hypothesis. Otherwise, do not reject the null hypothesis.

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Most popular questions from this chapter

The article "The Soundtrack of Recklessness: Musical Preferences and Reckless Behavior Among Adolescents" (Journal of Adolescent Research [1992]: \(313-331\) ) described a study whose purpose was to determine whether adolescents who preferred certain types of music reported higher rates of reckless behaviors, such as speeding, drug use, shoplifting, and unprotected sex. Independently chosen random samples were selected from each of four groups of students with different musical preferences at a large high school: (1) acoustic/pop, (2) mainstream rock, $$\begin{array}{ccccccccc} \text { Type of Box } & & & {\text { Compression Strength (Ib) }} & & & & {\text { Sample Mean }} & {\text { Sample SD }} \\ \hline 1 & 655.5 & 788.3 & 734.3 & 721.4 & 679.1 & 699.4 & 713.00 & 46.55 \\ 2 & 789.2 & 772.5 & 786.9 & 686.1 & 732.1 & 774.8 & 756.93 & 40.34 \\ 3 & 737.1 & 639.0 & 696.3 & 671.7 & 717.2 & 727.1 & 698.07 & 37.20 \\ 4 & 535.1 & 628.7 & 542.4 & 559.0 & 586.9 & 520.0 & 562.02 & 39.87 \\ & & & & & & & \overline{\bar{x}} =682.50 & \end{array}$$ (3) hard rock, and (4) heavy metal. Each student in these samples was asked how many times he or she had engaged in various reckless activities during the last year. The following table lists data and summary quantities on driving over \(80 \mathrm{mph}\) that is consistent with summary quantities given in the article (the sample sizes in the article were much larger, but for the purposes of this exercise, we use \(\left.n_{1}=n_{2}=n_{3}=n_{4}=20\right)\) $$\begin{array}{rrrr} \text { Acoustic/Pop } & \text { Mainstream Rock } & \text { Hard Rock } & \text { Heavy Metal } \\ \hline 2 & 3 & 3 & 4 \\ 3 & 2 & 4 & 3 \\ 4 & 1 & 3 & 4 \\ 1 & 2 & 1 & 3 \\ 3 & 3 & 2 & 3 \\ 3 & 4 & 1 & 3 \\ 3 & 3 & 4 & 3 \\ 3 & 2 & 2 & 3 \\ 2 & 4 & 2 & 2 \\ 2 & 4 & 2 & 4 \\ 1 & 4 & 3 & 4 \\ 3 & 4 & 3 & 5 \\ 2 & 2 & 4 & 4 \\ 2 & 3 & 3 & 5 \\ 2 & 2 & 3 & 3 \\ 3 & 2 & 2 & 4 \\ 2 & 2 & 3 & 5 \\ 2 & 3 & 4 & 4 \\ 3 & 1 & 2 & 2 \\ 4 & 3 & 4 & 3 \\ 20 & 20 & 20 & 20 \\ 2.50 & 2.70 & 2.75 & 3.55 \\ .827 & .979 & .967 & .887 \\ 6830 & 0584 & 0351 & 7868 \end{array}$$ Also, \(N=80\), grand total \(=230.0\), and \(\overline{\bar{x}}=230.0 / 80=\) 2.875. Carry out an \(F\) test to determine if these data provide convincing evidence that the true mean number of times driving over 80 mph varies with musical preference.

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Consider the accompanying data on plant growth after the application of different types of growth hormone. $$\begin{array}{llrlrl} & \mathbf{1} & 13 & 17 & 7 & 14 \\ & \mathbf{2} & 21 & 13 & 20 & 17 \\ \text { Hormone } & \mathbf{3} & 18 & 14 & 17 & 21 \\ & \mathbf{4} & 7 & 11 & 18 & 10 \\ & \mathbf{5} & 6 & 11 & 15 & 8 \end{array}$$ a. Carry out the \(F\) test at level \(\alpha=.05\). b. What happens when the T-K procedure is applied? (Note: This "contradiction" can occur when is "barely" rejected. It happens because the test and the multiple comparison method are based on different distributions. Consult your friendly neighborhood statistician for more information.)

Samples of six different brands of diet or imitation margarine were analyzed to determine the level of physiologically active polyunsaturated fatty acids (PAPUFA, in percent), resulting in the data shown in the accompanying table. (The data are fictitious, but the sample means agree with data reported in Consumer Reports.) $$\begin{array}{llllll} \text { Imperial } & 14.1 & 13.6 & 14.4 & 14.3 & \\ \text { Parkay } & 12.8 & 12.5 & 13.4 & 13.0 & 12.3 \\ \text { Blue Bonnet } & 13.5 & 13.4 & 14.1 & 14.3 & \\ \text { Chiffon } & 13.2 & 12.7 & 12.6 & 13.9 & \\ \text { Mazola } & 16.8 & 17.2 & 16.4 & 17.3 & 18.0 \\ \text { Fleischmann's } & 18.1 & 17.2 & 18.7 & 18.4 & \end{array}$$ a. Test for differences among the true average PAPUFA percentages for the different brands. Use \(\alpha=.05\). b. Use the T-K procedure to compute \(95 \%\) simultaneous confidence intervals for all differences between means and interpret the resulting intervals.

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