Chapter 9.2, Problem 24E

Damage to grapes from bird predation is a serious problem for grape growers. The article “Experimental Method to Investigate and Monitor Bird Behavior and Damage to Vineyards” (Amer. J. of Enology and Viticulture, 2004: 288–291) reported on an experiment involving a bird-feeder table, time-lapse video, and artificial foods. Information was collected for two different bird species at both the experimental location and at a natural vineyard setting. Consider the following data on time (sec) spent on a single visit to the location.

Species	Location	n	X	SE mean
Blackbirds	Exptl	65	13.4	2.05
Blackbirds	Natural	50	9.7	1.76
Silvereyes	Exptl	34	49.4	4.78
Silvereyes	Natural	46	38.4	5.06

a. Calculate an upper confidence bound for the true average time that blackbirds spend on a single visit at the experimental location.

b. Does it appear that true average time spent by blackbirds at the experimental location exceeds the true average time birds of this type spend at the natural location? Carry out a test of appropriate hypotheses.

c. Estimate the difference between the true average time blackbirds spend at the natural location and true average time that silvereyes spend at the natural location, and do so in a way that conveys information about reliability and precision. [Note: The sample medians reported in the article all seemed significantly smaller than the means, suggesting substantial population distribution skewness. The authors actually used the distribution-free test procedure presented in Section 2 of Chapter 15.]

To determine

Obtain 95% upper confidence interval for the average time that the black birds spend on a single Visit at the experimental location.

Answer to Problem 24E

The 95% confidence interval for the average time that the black birds spend on a single Visit at the experimental location is $\underset{\_}{16.83\text{ seconds}}$

Explanation of Solution

Given info:

The information is based on the time spent by the birds on a single visit on the experimental location.

Calculation:

Here, $\mu$ represents the average time spent by the black birds on a single visit.

The following assumptions are required for the confidence interval about mean using t- distribution to be valid.

Requirements for t-distribution to construct confidence interval about mean:

• The sample must be drawn using simple random sampling.
• The population standard deviation is unknown.
• Either the population must be approximately normal or the sample size must be greater than 30.

Thus, the requirements must be satisfied for the confidence interval about mean using t- distribution to be valid.

Critical value:

For 95% level of significance,

$\begin{array}{c}1-\alpha =1-0.95\\ \alpha =0.05\end{array}$

Degrees of freedom:

$\begin{array}{c}n-1=65-1\\ =64\end{array}$

From Table A.5 of the critical values for t distribution in Appendix, the critical value $\left(t_{0.05}\right)$ with 64 is $\underset{\_}{1.671}$.

Thus, the critical value is $\underset{\_}{\left(t_{0.05}\right)=1.671}$.

Confidence interval:

$\overline{x}+t_{\alpha ,\left(n-1\right)}.SE\left(\overline{x}\right)$

The confidence interval for the average time spent is,

$\begin{array}{l}13.4+t_{0.05,\left(65-1\right)}.\left(2.05\right)=13.4+\mathrm{1.671.}\left(2.05\right)\\ =13.4+3.4256\\ =16.83\end{array}$

Thus, the 95% confidence interval for the average time that the black birds spend on a single Visit at the experimental location is $\underset{\_}{16.83\text{ seconds}}$.

To determine

Check whether the average time that the black birds spend on a single Visit at the experimental location is greater than that of natural location.

Answer to Problem 24E

The conclusion is that, there is no enough evidence to infer that the average time that the black birds spend on a single Visit at the experimental location is greater than that of natural location.

Explanation of Solution

Given info:

Let $\overline{x}$ denotes the sample mean time that the black birds spend on a single Visit at the experimental location and $\overline{y}$ denotes the sample mean time that the black birds spend on a single Visit at the natural location.

$\overline{x}=13.4$,$\overline{y}=9.7$, $SE\left(\overline{x}\right)=1.76$, $SE\left(\overline{y}\right)=1.76$, $\alpha =0.05$, $m=65$, and $n=50$.

Calculation:

Here, ${\mu }_1$ represents the average time that the black birds spend on a single Visit at the experimental location and ${\mu }_2$ represents the average time that the black birds spend on a single Visit at the natural location.

The test hypotheses are,

Null hypothesis:

$H_0:{\mu }_1={\mu }_2$

That is, the average time that the black birds spend on a single Visit at the experimental location is same as the natural location.

Alternative hypothesis:

$H_a:{\mu }_1>{\mu }_2$

That is, the average time that the black birds spend on a single Visit at the experimental location is greater than that of natural location.

Assumption for the two sample t-test:

• The sample X and Y taken from the population is selected at random.
• The samples X and Y are independent of each other.
• Samples must be distributed to normal.

Here, the samples selected from the older and younger adults were selected at random and independent. Moreover, the sample size is large and distributed to normal. Hence, the assumptions are satisfied.

The degrees of freedom is,

$\begin{array}{c}\text{df}=\frac{{\left[{\left(SE\left(\overline{x}\right)\right)}^2+{\left(SE\left(\overline{y}\right)\right)}^2\right]}^2}{\frac{{\left(SE\left(\overline{x}\right)\right)}^2}{n_1-1}+\frac{{\left(SE\left(\overline{y}\right)\right)}^2}{n_2-1}}\\ =\frac{{\left[\left({2.05}^2+{1.76}^2\right)\right]}^2}{\frac{{\left({2.05}^2\right)}^2}{65-1}+\frac{{\left({1.76}^2\right)}^2}{50-1}}\\ =\frac{{\left[\left(4.2025+3.0976\right)\right]}^2}{\frac{{\left(4.2025\right)}^2}{64}+\frac{{\left(3.0976\right)}^2}{49}}\end{array}$

$\begin{array}{l}=\frac{53.29}{0.276+0.196}\\ =\frac{53.29}{0.472}\\ =112.9\end{array}$

Rounding the degrees of freedom as 120.

Test statistic for two-sample t-test:

Test statistic:

Hence, the test statistic is obtained as:

$t=\frac{\left(\overline{x}-\overline{y}\right)-{\Delta }_0}{\sqrt{SE\left(\overline{x}\right)+SE\left(\overline{y}\right)}}$

The difference between the average time spend ${\Delta }_0=0$

The test statistic value is obtained below:

$\begin{array}{c}t=\frac{\left(13.4-9.7\right)-\left(0\right)}{\sqrt{{2.05}^2+{1.76}^2}}\\ =\frac{3.7}{\sqrt{4.2025+3.0976}}\\ =\frac{3.7}{2.70}\\ =1.37\end{array}$

Thus, the test statistic value is 1.37.

From the Appendix “Table A.5 Critical values for t Distributions”, the critical value $\left(z_{0.05}\right)$ at 120 degrees of freedom is $\underset{\_}{1.658}$.

Decision rule:

Rejection region for a Right-tailed test:

If $t>t_0\left(=1.714\right)$ , then reject the null hypothesis $\left(H_0\right)$.

Conclusion:

The level of significance is, $\alpha =0.05$ and the critical value is $\underset{\_}{1.658}$.

Here, the test statistic is less than the critical value. That is,$t\left(=1.37\right)<t_0\left(=1.658\right)$.

Thus, the decision is “fail to reject the null hypothesis”.

Thus, it can be concluded that there is no enough evidence to infer that the average time that the black birds spend on a single Visit at the experimental location is greater than that of natural location.

To determine

Estimate the difference between the true average time that the black birds spend on a single Visit at the natural location and time spend by the silvereyes at natural location.

Answer to Problem 24E

The difference between the true average time that the black birds spend on a single Visit at the natural location and time spends by the silvereyes at natural location is (17.99 Sec, 39.41 Sec).

Explanation of Solution

Calculation:

Critical value:

For 95% level of significance,

$\begin{array}{c}1-\alpha =1-0.95\\ \alpha =0.05\end{array}$

From the Appendix “Table A.5 Critical values for t Distributions”, the critical value $\left(z_{0.025}\right)$ is $\underset{\_}{2.00}$.

Thus, the critical value is $\underset{\_}{\left(t_{0.025}\right)=2.00}$.

Confidence interval:

$\left(\overline{x}-\overline{y}\right)\pm t_{\alpha ,\left(n-1\right)}.\sqrt{SE\left(\overline{x}\right)+SE\left(\overline{y}\right) }$

The confidence interval for the average time spent is,

$\begin{array}{c}\left(\overline{x}-\overline{y}\right)\pm t_{\alpha ,\left(n-1\right)}.\sqrt{SE\left(\overline{x}\right)+SE\left(\overline{y}\right) }=\left(38.4-9.7\right)\pm 2\sqrt{{1.76}^2+{5.06}^2}\\ =28.7\pm 2\sqrt{3.0976+25.60}\\ =28.7\pm 10.71\\ =\left(17.99,39.41\right)\end{array}$

The 95% confidence interval for the difference between the average time that the black birds spend on a single Visit at the natural location and time spends by the silvereyes at natural location is $\underset{\_}{\left(17.99<\left({\mu }_1-{\mu }_2\right)<39.41\right)}$.

Interpretation:

Hence, the 95% confidence interval for the difference between the average time that the black birds spend on a single Visit at the natural location and time spend by the silvereyes at natural location is (17.99 Sec, 39.41 Sec).