Find the confidence interval for the difference in means of two groups, given that the sample means of the groups are $\bar x_1 = 35$ and $\bar x_2 = 39$ and the margin of error is 2.5. Should the null hypothesis $\mu_1=\mu_2$ be rejected? Why or why not?

$(\bar x_1-\bar x_2) \pm E=(35-39)\pm 2.5$

Confidence interval: $(-6.5,-1.5)$

Reject $H_0,\ \mu_1-\mu_2=0$, because 0 is not in the confidence interval.A random sample of 9-year-olds yielded the data below for their heights (in centimeters).

$$\begin{array}{ccc } \hbox{Boys}&\qquad&\hbox{Girls}\cr \bar x = 123.5 && \bar x = 126.2 \cr s=9.90 &&s=10.95\cr n=60 && n=50 \end{array}$$Construct a 95% confidence interval for the difference in mean heights of boys and girls at that age.

Use your confidence interval to respond to the claim that at age 9 the average height for boys and girls is the same.

Assumptions: $60,\ 50 \geq30$

$E=3.94$ Confidence interval: $-6.6 \lt \mu_B-\mu_G \lt 1.2$

We are 95% confident that the difference in mean heights of boys and girls at age 9 is between $-6.6$ cm and $1.2$ cm.

Since 0 is in the confidence interval, we do not reject $\mu_B=\mu_G$.

We do not have enough evidence to reject the claim that at age 9 the average height for boys and girls is the same.An advertisement claims that children who use Smile toothpaste have fewer cavities than children who use any other brand. To test this claim, a consumer's group selected 100 children and divided them into two groups of 50 each. The children of group I were told to brush daily with only Smile toothpaste. The children of group II were told to brush daily with Vanish toothpaste. The experiment lasted one year. The children who used Smile toothpaste had a mean of 2.31 cavities with a standard deviation of 0.6. For the children who used Vanish, there was a mean of 2.68 cavities with a standard deviation of 0.4. Is Smile significantly more effective than Vanish at preventing cavities? Use the $P$-value method with a 0.05 significance level.

Assumptions: $50\geq 30$

$H_0: \mu_S=\mu_V \qquad H_1: \mu_S \lt \mu_V $

Test Statistic: $z=-3.63$

$P$-value: 0.0001 (from the table)

Reject the null hypothesis because the $P$-value is less than the significance level.

There is enough evidence to support the claim that Smile is significantly more effective than Vanish at preventing cavities.It is commonly felt by people in northwest Ohio that on Interstate 75 drivers from Michigan drive faster than drivers from Ohio. To examine this claim, a random sample of 50 Michigan drivers is found to have a mean speed of 67 mph with a standard deviation of 8 mph. A random sample of 50 Ohio drivers is found to have a mean speed of 64 mph with a standard deviation of 7 mph. Test the claim that Michigan drivers are faster. Use the $P$-value method at significance level 0.05.

Assumptions: $50\geq30$

$H_0: \mu_M=\mu_O \qquad H_1: \mu_M \gt \mu_O $

Test Statistic: $z=1.996$ $P$-value$= .0228$ Reject the null hypothesis because the $P$-value is less than the significance level.

There is evidence that Michigan drivers are faster than Ohio drivers.Recorded actual high temperatures and predicted high temperatures from 5 days earlier are recorded below. Test the claim that there is no difference between actual and predicted temperature. Use the critical value method with significance level 0.05. $$\begin{array}{r|cccccccc} \hbox{Date}&9/1&9/5&9/12&9/15&9/22&9/23&9/27&9/30\cr \hbox{Actual High Temperature ($^\circ F$)}&81&73&78&73&82&81&74&62\cr \hbox{Predicted High Temperature ($^\circ F$)}&80&79&79&78&73&79&70&69 \end{array}$$

Assumptions: $(-18.25, 15.75)$ No outliers; $I=-2.05$ Not significantly skewed.

$H_0: \mu_D=0 \qquad H_1: \mu_D\not=0 $

Test Statistic: $t=-.193$

Critical values: $\pm 2.365$

Fail to reject the null hypothesis because the test statistic is not in the rejection region.

There is not enough evidence to reject the claim that there is no difference between actual and predicted temperature.A farmer is considering increasing the amount of time the lights in his hen house are on. Ten hens were selected, and the number of eggs each produced over a week was recorded under normal and increased lighting conditions. At $\alpha=.05$, can it be concluded that the increased light time changed egg production? Use the confidence interval method. $$\begin{array}{r|cccccccccc} \hbox{Hen}&A&B&C&D&E&F&G&H&I&J\cr \hbox{Normal Light:}&4&3&8&7&6&4&9&7&6&5\cr \hbox{Increased Light:}&6&5&9&7&4&5&10&6&9&6 \end{array}$$

Assumptions: $(-5,3)$ No outliers; $I=0.405$ Not significantly skewed.

$H_0: \mu_D=0 \qquad H_1: \mu_D\not=0 $

$E=1.06$ Confidence interval: $-1.86 \lt \mu_D \lt 0.26$

Fail to reject the null hypothesis because 0 is in the confidence interval.

There is not enough evidence to support the claim that the increased light time changed egg production.A certain soccer camp advertised that participants would become better ball handlers. As one piece of evidence students were asked to juggle a soccer ball at the beginning of camp and at the end of the camp. The number of times each student was able to juggle the soccer ball with their knees and feet were recorded. Data from a random sample of these campers are given below: $$\begin{array}{r|ccccccccccc} \hbox{Player}&A&B&C&D&E&F&G&H&I&J&K\cr \hbox{Pre:}&2&5&3&1&9&7&5&4&6&2&3\cr \hbox{Post:}&4&4&6&4&11&7&10&3&7&4&5 \end{array}$$

Construct a 95% confidence interval for the mean difference in juggling ability.

Is there an increase in juggling ability associated with this camp? Explain your reasoning, based on the confidence interval.

Assumptions: $(-7.5,4.5)$ No outliers; $I=0.605$ Not significantly skewed.

$E=1.212$ Confidence interval: $-2.848 \lt \mu_D \lt -.424$

We are 95% confident that the mean change in juggling ability is between $-2.848$ and $-0.424$.

Since 0 is not in the confidence interval, there is a significant change in juggling ability.

Since the confidence interval is negative, "Pre"-"Post"$ \lt 0$, the change represents an increase in juggling ability associated with the camp.A survey recorded the amount of weight change (in pounds) over the Thanksgiving holidays for a random sample of men and women. Assume the samples are approximately normal. $$\begin{array}{rcc} \hbox{Men: }& s= 1.3& n=21\cr \hbox{Women: }& s= 2.1& n=25 \end{array}$$ Are the variances of the samples significantly different at $\alpha=.05$?

Based on your results above, determine the appropriate test to use in testing whether there was a difference between men's and women's weight changes.

Samples are approx. normal.

$H_0: \sigma^2_M=\sigma^2_W; H_1: \sigma^2_M\not=\sigma^2_W $

Test statistic: $F=2.61$

Critical value: $2.41$ (dfN=24, dfD=20, .025 in each tail)

Reject the null hypothesis. The test statistic is in the rejection region.

The variances are significantly different.

Use the non-parametric Wilcoxon Rank Sum test.The times (in minutes) it took ten white mice and ten brown mice to learn to run a simple maze are given below. Assume samples are approximately normal. $$\begin{array}{r|ccccccccccccc} \hbox{White mice: }&13&15&17&18&19&21&22&24&25&28&\qquad&\bar x_W=20.2 &s_W=4.69\cr \hbox{Brown mice: }&10&14&15&16&17&17&19&20&23&26&&\bar x_B=17.7 &s_B=4.57 \end{array}$$

Are the variances of the samples significantly different at $\alpha=.05$?

Test at $\alpha=.05$ the claim that there is no difference in the learning rates of the two kinds of mice.

Samples are approx. normal.

$H_0: \sigma^2_W=\sigma^2_B \qquad H_1: \sigma^2_W\not=\sigma^2_B $

Test statistic: $F=1.053$

Critical value: $4.03$ (dfN=9, dfD=9, .025 in each tail)

Fail to reject the null hypothesis. The test statistic is not in the rejection region.

The variances are not significantly different. (Use the pooled-variance t-test.)Assumptions: Variances are not significantly different, samples approximately normal.

$H_0: \mu_W=\mu_B \qquad H_1: \mu_W\not=\mu_B $

$s_p^2=21.44$

Test statistic: $t=1.207$

Critical values: $\pm2.101$

Fail to reject the null hypothesis. The test statistic is not in the rejection region.

There is no significant difference in the learning rates of the two kinds of mice.

The number of grams of carbohydrates contained in 1-ounce servings of randomly selected chocolate and non-chocolate candy is listed below. Assume samples are approximately normal. $$\begin{array}{ r|ccccccccccccc} \hbox{Chocolate: }&17&24&25&25&27&29&29&29&32&34&36&38&41\cr \hbox{Non-chocolate: }&10&12&29&29&30&37&38&39&41&41&55 \end{array}$$

Are the variances of the samples significantly different at $\alpha=.05$?

Test at $\alpha=.05$ the claim that there is no difference in carbohydrate content in the two kinds of candy.

Samples are approx. normal.

$H_0: \sigma^2_C=\sigma^2_N \qquad H_1: \sigma^2_C\not=\sigma^2_N $

Test statistic: $F=4.02$

Critical value: $3.37$ (dfN=10, dfD=12, .025 in each tail)

Reject the null hypothesis. The test statistic is in the rejection region.

The variances are significantly different.Use the Wilcoxon Rank Sum test. (See exercises for non-parametric tests.)

Body temperatures of men and women were measured at 8 am. The information is given below. $$\begin{array}{ccc} \hbox{Men}&\qquad &\hbox{Women}\cr\hline n=11&&n=59\cr \bar x=97.69^\circ{\rm F}&&\bar x=97.45^\circ{\rm F}\cr s=0.89^\circ{\rm F}&&s=0.66^\circ{\rm F} \end{array}$$

Are the variances of the samples significantly different at $\alpha=.05$? Assume both samples are normally distributed.

Test at $\alpha=.05$ the claim that there is no difference between men's and women's body temperatures.

Samples are approx. normal.

$H_0: \sigma^2_M=\sigma^2_W \qquad H_1: \sigma^2_M\not=\sigma^2_W $

Test statistic: $F=1.818$

Critical value: $2.39$ (dfN=10, dfD=58, .025 in each tail)

Fail to reject the null hypothesis. The test statistic is not in the rejection region.

The variances are not significantly different. (Use the pooled-variance t-test.)Assumptions: Variances are not significantly different, samples approximately normal.

$H_0: \mu_M=\mu_W \qquad H_1: \mu_M\not=\mu_W $

$s_p^2=.4880$

Test statistic: $t=1.046$

Critical values: $\pm1.995$ (from calculator)

Fail to reject the null hypothesis. The test statistic is not in the rejection region.

There is no significant difference between men's and women's body temperatures.

To determine whether a significant difference exists in the lengths of fish from two hatcheries, 11 fish were randomly selected from hatchery A, and 10 fish were randomly selected from hatchery B. Their lengths, in centimeters, are given below. Assume that both data sets are approximately normal. Test the claim that there is no difference in the fish lengths for the two hatcheries using the critical value method with $\alpha=.05$.

$$\begin{array}{cccccccccccc} \textrm{Hatchery A:} & 12.4 & 12.7 & 12.9 & 13.3 & 14.2 & 14.3 & 14.3 & 14.8 & 14.8 & 15.3 & 15.3\\ \textrm{Hatchery B:} & 10.7 & 12.2 & 12.8 & 13.9 & 14.1 & 14.3 & 14.6 & 15.6 & 16.8 & 18.1 \end{array}$$Samples are approx. normal.

$H_0: \sigma^2_A=\sigma^2_B \qquad H_1: \sigma^2_A\not=\sigma^2_B $

Test statistic: $F=4.354$

Critical value: $3.78$ (dfN=9, dfD=10, .025 in each tail)

Reject the null hypothesis. The test statistic is in the rejection region.

The variances are significantly different.

Use the Wilcoxon Rank Sum test. (See exercises for non-parametric tests.)A tax collector wishes to compare the values of tax-exempt properties for two large cities. The values for two random samples are shown. Assume samples are approximately normal. Test the claim that the property values are the same in the two cities at $\alpha=.05$. $$\begin{array}{r|ccccccccccccccc} \hbox{City A: }&2&5&7&8&11&14&19&19&22&23&23&25&30&31&44\cr \hbox{City B: }&2&4&5&5&9&11&12&17&17&19&20&40&51&52&68 \end{array}$$

Samples are approx. normal.

$H_0: \sigma^2_A=\sigma^2_B \qquad H_1: \sigma^2_A\not=\sigma^2_B $

Test statistic: $F=3.29$

Critical value: $3.05$ (dfN=14, dfD=14, .025 in each tail)

Reject the null hypothesis. The test statistic is in the rejection region.

The variances are significantly different.

Use the Wilcoxon Rank Sum test. (See exercises for non-parametric tests.)Parents suspect that a certain school places students into first grade according to ability. The kindergarten ITBS test scores were obtained for the two first grade classes. One first grade class had a mean of 114 and a standard deviation of 83 while the second class had a mean of 158 and a standard deviation of 75. Each class has 20 students. Assume the distributions are approximately normal. Based on the kindergarten ITBS test, is there reason to believe that the two first- grade classes are different? $ ( \alpha = .05 )$

Samples are approx. normal.

$H_0: \sigma^2_1=\sigma^2_2 \qquad H_1: \sigma^2_1\not=\sigma^2_2 $

Test statistic: $F=1.22$

Critical value: $2.62$ (dfN=19, dfD=19, .025 in each tail)

Fail to reject the null hypothesis. The test statistic is not in the rejection region.

The variances are not significantly different.

Use the pooled-variance t-test.

Assumptions: Variances are not significantly different, samples approximately normal.

$H_0: \mu_1=\mu_2 \qquad H_1: \mu_1\not=\mu_2 $

$s_p^2=6257$

Test statistic: $t=-1.759$

Critical values: $\pm2.024$

Fail to reject the null hypothesis. The test statistic is not in the rejection region.

There is no significant difference between the ITBS scores of the two first-grade classes.