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Per-pupil costs (in thousands of dollars) for cyber charter school tuition for school districts in three areas are shown. Test the claim that there is a difference in means for the three areas, using an appropriate parametric test. $$\begin{array}{r|cccccc} \hbox{Area 1} &6.2 &9.3 &6.8 &6.1 &6.7 &7.5 \cr \hbox{Area 2} &7.5 &8.2 &8.5 &8.2 &7.0 &9.3 \cr \hbox{Area 3} &5.8 &6.4 &5.6 &7.1 &3.0 &3.5 \cr \end{array}$$
Interval or ratio data, independent samples, each distribution is approximately normal, and all variances are equal (not significantly different).
Null hypothesis: $\mu_1=\mu_2=\mu_3$
$$\begin{array}{ccccc}
&\hbox{SS}&\hbox{df}&\hbox{MS}&\hbox{F}\cr
\hbox{Between}&25.7&2&12.8&8.17\cr
\hbox{Within}&23.5&15&1.57\cr
\hbox{Total}&49.2&17
\end{array}$$
Critical value: 3.68
Reject the null hypothesis. The test statistic is in the rejection region.
There is enough evidence to support the claim that there is a difference in mean tuition for the three areas.
Critical value: 7.36
$H_0: \mu_1=\mu_2,\ F_S=1.99$, Fail to reject $H_0$
$H_0: \mu_1=\mu_3,\ F_S=6.68$, Fail to reject $H_0$
$H_0: \mu_2=\mu_3,\ F_S=15.96$, Reject $H_0$
There is a significant difference in mean tuition between Area 2 and Area 3.
Random samples of employees were selected from three different types of stores at the mall, and their ages recorded. Assume the assumptions for the parametric test are met. At the 0.05 significance level, test the claim that there is a difference in mean ages for the three groups. If you find a difference, perform pairwise Scheffe tests to determine where the difference lies. $$\begin{array}{ccccc } \hbox{Department Store}&\hbox{Music}&\hbox{Sporting Goods}\cr \bar x = 47.2 &\bar x = 26.8 &\bar x = 29.8 & \qquad & S_{}S_B=1446\cr n=6 & n=6 & n=6 & & S_{}S_W = 1377 \end{array}$$
Null hypothesis: $\mu_D=\mu_M=\mu_S$
$$\begin{array}{ccccc}
&\hbox{SS}&\hbox{df}&\hbox{MS}&\hbox{F}\cr
\hbox{Between}&1446&2&723&7.876\cr
\hbox{Within}&1377&15&91.8\cr
\hbox{Total}&2823&17
\end{array}$$
Critical value: 3.68
Reject the null hypothesis. The test statistic is in the rejection region.
There is enough evidence to support the claim that there is a difference in mean ages.
Scheffe tests:
Critical value: 7.36
$H_0: \mu_D=\mu_M,\ F_S=13.60$, Reject $H_0$
$H_0: \mu_M=\mu_S,\ F_S=0.29$, Fail to reject $H_0$
$H_0: \mu_D=\mu_S,\ F_S=9.89$, Reject $H_0$
The mean age of department store employees is significantly different from employees of music and sporting goods stores.
The number of grams of fat per serving for three different varieties of pizza from several manufacturers is measured and a partial ANOVA table is provided below. Complete the ANOVA table. At the 0.05 significance level, is there a significant difference in mean fat content for the three varieties? Assume the assumptions for the test are met. $$\begin{array}{ccccc} &\hbox{SS}&\hbox{df}&\hbox{MS}&\hbox{F}\cr \hbox{Between}&23.0&2\cr \hbox{Within}&316.9&18\cr \hbox{Total}&339.9&20 \end{array}$$
Null hypothesis: $\mu_1=\mu_2=\mu_3$
$$\begin{array}{ccccc}
&\hbox{SS}&\hbox{df}&\hbox{MS}&\hbox{F}\cr
\hbox{Between}&23.0&2&11.5&.653\cr
\hbox{Within}&316.9&18&17.6\cr
\hbox{Total}&339.9&20
\end{array}$$
Critical value: 3.55
Fail to reject the null hypothesis. The test statistic is not in the rejection region.
There is not enough evidence to support the claim that there is a difference in mean fat content.
Do not do Scheffe tests because no significant difference was found.
Prices (rounded to the closest dollar amount) of men's women's, and children's athletic shoes were compared by obtaining random samples of 18 shoes each. The variances of the samples are homogeneous and the distributions are assumed to be approximately normal.
Use the information provided to test at $\alpha=.05$ whether there is a difference is mean shoe prices for the three types of shoes.
The means for the three groups are as follows: Women's shoes = \$48.78; Men's shoes = \$60.33; Children's shoes = \$45.39. Perform the necessary tests to find where the difference lies. State your overall inference.
Null hypothesis: $\mu_M=\mu_W=\mu_C$
$$\begin{array}{ccccc}
&\hbox{SS}&\hbox{df}&\hbox{MS}&\hbox{F}\cr
\hbox{Between}&2210.1&2&1105.05&11.09\cr
\hbox{Within}&5083.4&51&99.67\cr
\hbox{Total}&7293.5&53
\end{array}$$
Critical value: 3.23
Reject the null hypothesis. The test statistic is in the rejection region.
There is a significant difference in the prices of men's women's, and children's athletic shoes.
Critical value: 6.46
$H_0: \mu_M=\mu_W,\ F_S=12.05$, Reject $H_0$
$H_0: \mu_W=\mu_C,\ F_S=1.04$, Fail to reject $H_0$
$H_0: \mu_M=\mu_C,\ F_S=20.16$, Reject $H_0$
Men's athletic shoe prices are significantly different from both women's and children's.
A researcher wishes to try three different techniques to lower the blood pressure of individuals diagnosed with high blood pressure. The subjects are randomly assigned to three groups; the first group takes medication, the second group exercises, and the third group follows a special diet. After four weeks, the reduction in each person's blood pressure is recorded. Assume that the data in each group is approximately normal. Is there a significant difference between the techniques used to lower blood pressure, at $\alpha=0.05$? If you find a difference, perform pairwise tests to determine where the difference lies. $$\begin{matrix} \hline \hbox{Medication group} &\quad \hbox{Exercise group} \quad &\quad \hbox{Diet group} \quad \cr \hline 9 &0 & 4\cr 10& 2 &5\cr 12 &3 &8\cr 13& 6& 9\cr 15& 8& 12\cr \hline\hline\bar x_1=11.8 & \bar x_2= 3.8 &\bar x_3= 7.6\cr s_1^2= 5.7 & s_2^2=10.2 & s_3^2=10.3\\\\ \qquad S_{}S_B=160.13 &\qquad S_{}S_W=104.80 \end{matrix} $$
Null hypothesis: $\mu_M=\mu_E=\mu_D$
$$\begin{array}{ccccc}
&\hbox{SS}&\hbox{df}&\hbox{MS}&\hbox{F}\cr
\hbox{Between}&160.13&2&80.07&9.17\cr
\hbox{Within}&104.80&12&8.73\cr
\hbox{Total}&264.93&14
\end{array}$$
Critical value: 3.89
Reject the null hypothesis. The test statistic is in the rejection region.
There is enough evidence to support the claim that there is a difference between the techniques used.
Scheffe tests:
Critical value: 7.78
$H_0: \mu_M=\mu_E,\ F_S=18.33$, Reject $H_0$
$H_0: \mu_M=\mu_D,\ F_S=5.05$, Fail to reject $H_0$
$H_0: \mu_E=\mu_D,\ F_S=4.14$, Fail to reject $H_0$
The medication group is significantly different from the exercise group.
A turkey farmer tested three kinds of poultry feeds with the weights (in pounds) of the grown turkeys in each sample given below. Test at $\alpha=.05$ whether there is a difference in the mean weights of turkeys consuming the different feeds. If you find a difference, perform pairwise tests to determine where the difference lies. Assume the distributions are normal and variances are equal. $$\begin{array}{ccc} \hbox{Feed A}&\hbox{Feed B}&\hbox{Feed C}\cr 12.3&12.1&11.5\cr 11.4&13.4&12.4\cr 13.4&14.0&10.8\cr 12.5&13.6&12.6\cr 12.0&12.8&11.8\cr 13.1&14.2&11.9\cr \end{array}$$
Null hypothesis: $\mu_A=\mu_B=\mu_C$
$$\begin{array}{ccccc}
&\hbox{SS}&\hbox{df}&\hbox{MS}&\hbox{F}\cr
\hbox{Between}&6.98&2&3.49&6.69\cr
\hbox{Within}&7.82&15&0.5216\cr
\hbox{Total}&14.8&17
\end{array}$$
Critical value: 3.68
Reject the null hypothesis. The test statistic is in the rejection region.
There is a significant difference in the mean weights of turkeys consuming the different feeds.
Scheffe tests:
Critical value: 7.36
$H_0: \mu_A=\mu_B,\ F_S=4.66$, Fail to reject $H_0$
$H_0: \mu_B=\mu_C,\ F_S=13.29$, Reject $H_0$
$H_0: \mu_A=\mu_C,\ F_S=2.21$, Fail to rejecteject $H_0$
B is significantly different from C. Recommend using B, or possibly A, but not C.
A researcher wishes to see whether there is any difference in the weight gains of athletes following one of three special diets. Athletes are randomly assigned to three groups and placed on the diet for 6 weeks. The weight gains (in pounds) are given. Assume weight gains are normally distributed and the variances are equal. At a $0.05$ significance level, can the researcher conclude that there is a difference in the diets?
Three classes of ten students each were taught using the following methodologies: traditional, inquiry-oriented and a mixture of both. At the end of the term, the students were tested, their scores were recorded and this yielded the following partial ANOVA table. Assume distributions are normal and variances are equal. $$\begin{array}{l|c|c|c|c|} & SS & df & MS & F\\\hline \textrm{Between} & 136 & & & \\\hline \textrm{Within} & 432 & & & \\\hline \textrm{Total} & & & & \\\hline \end{array}$$
Complete the above table and use it to test the claim that there is a difference in the mean score of the students taught with the three different methodologies. Use $\alpha = 0.05$.
Suppose that the students taught by the traditional method had a mean score of $80$ and the ones taught by the inquiry-oriented method had a mean score of $89$. Do a Scheffe test to determine whether the difference in means is significant when we compare the traditional method with the inquiry-based one.
The mean for the mixed approach is $85$. Test statistics are $F_S = 7.8125$ for between traditional and mixed methods and $F_S = 5.000$ for between inquiry oriented and mixed methods. Give an overall inference for the teaching methods.
Below are HIC measures of the severity of head injuries collected from crash test dummies placed in different types of vehicles. Use a 0.05 significance level to test the claim that the shown car categories have the same mean.
$$\begin{array}{lcccccccccc} \textrm{Small Cars} & 290 & 406 & 371 & 544 & 374 & 501 & 376 & 499 & 479 & 475\\ \textrm{Medium Cars} & 245 & 502 & 474 & 505 & 393 & 264 & 368 & 510 & 296 & 349\\ \textrm{Large Cars} & 342 & 216 & 335 & 698 & 216 & 169 & 608 & 432 & 510 & 332 \end{array}$$Test statistic: $F=0.3974$. Critical value of $F \doteq 3.3541$. $p$-value $ = 0.6759$. Fail to reject the null hypothesis that the means for each car type all agree. There is no evidence of a difference in the mean HIC measure for each car type.
Below are times (in minutes and seconds) five bicyclists took to complete each mile of a 3 mile race. Test the claim that it takes the same time to ride each mile at $\alpha = 0.05$.
$$\begin{array}{lccccc} \textrm{Mile 1} & \textrm{3:15} & \textrm{3:24} & \textrm{3:23} & \textrm{3:22} & \textrm{3:21}\\ \textrm{Mile 2} & \textrm{3:19} & \textrm{3:22} & \textrm{3:21} & \textrm{3:17} & \textrm{3:19}\\ \textrm{Mile 3} & \textrm{3:34} & \textrm{3:31} & \textrm{3:29} & \textrm{3:31} & \textrm{3:29}\\ \end{array}$$Test statistic $F=27.2488$. Critical value $F=3.8853$. $p$-value is incredibly small and certainly less than $\alpha = 0.05$. Reject the null hypothesis that the mean times for each mile agree. There is evidence the third mile takes longer.
The below data shows the nicotine content of king size, menthol, and filtered 100mm non-menthol cigarettes for 25 different brands. Test at $\alpha = 0.05$ the claim that the three categories of cigarettes yield the same mean amount of nicotine.
$$\begin{array}{ccc} \textrm{King Size} & \textrm{Menthol} & \textrm{Filtered}\\ 1.1 & 1.1 & 0.4\\ 1.7 & 0.8 & 1.0\\ 1.7 & 1.0 & 1.2\\ 1.1 & 0.9 & 0.8\\ 1.1 & 0.8 & 0.8\\ 1.4 & 0.8 & 1.0\\ 1.1 & 0.8 & 1.1\\ 1.4 & 0.8 & 1.1\\ 1.0 & 0.9 & 1.1\\ 1.2 & 0.8 & 0.8\\ 1.1 & 0.8 & 0.8\\ 1.1 & 1.2 & 0.8\\ 1.1 & 0.8 & 0.8\\ 1.1 & 0.8 & 1.0\\ 1.1 & 1.3 & 0.2\\ 1.8 & 0.7 & 1.1\\ 1.6 & 1.4 & 1.0\\ 1.1 & 0.2 & 0.8\\ 1.2 & 0.8 & 1.0\\ 1.5 & 1.0 & 0.9\\ 1.3 & 0.8 & 1.1\\ 1.1 & 0.8 & 1.1\\ 1.3 & 1.2 & 0.6\\ 1.1 & 0.6 & 1.3\\ 1.1 & 0.7 & 1.1 \end{array}$$Test statistic $F=18.9931$. Critical value: $F \doteq 3.1504$. $p$-value is incredibly small and easily $ \lt 0.05$. Reject the null hypothesis that all three means agree. There is evidence of a difference in nicotine content between these three categories of cigarettes.