Question

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.

\(\begin{array}{l}Species Location n \overline x SEmean\\Blackbirds Exptl 65 13.4 2.05 \\Blackbirds Natural 50 9.7 1.76\\Silvereyes Exptl 34 49.4 4.78\\Silvereyes Natural 46 38.4 5.06\end{array}\)

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

Short answer

(a) \(95\% \)upper confidence bound: \(16.82555\)

(b) There is not sufficient evidence to support the claim that the true average time spent by blackbirds at the experimental location exceeds the true average time birds of this type spend at the natural location.

(c) \(( - 10.8421,10.6837)\)

Step-by-step solution

a)Step 1: Find the upper confident bound

Given:

\(\begin{array}{l}n = 65\\\bar x = 13.4\\{\sigma _{\bar x}} = 2.05\end{array}\)

Let us assume that we want to calculate the confidence bound with\(95\% \)confidence (other confidence levels work similarly).

\(c = 95\% = 0.95\)

Determine the t-value by looking in the row starting with degrees of freedom\(df = n - 1 = 65 - 1 = 64 > 60\)and in the column with\(1 - c = \)\(0.05\)in the table of the Student's T distribution:

\({t_\alpha } = 1.671\)

The margin of error is then:

\(E = {t_{\alpha /2}} \times \frac{s}{{\sqrt n }} = 1.671 \times 2.05 \approx 3.42555\)

The boundaries of the confidence interval then become:

\(\bar x + E = 13.4 + 3.42555 = 16.82555\)

Thus the \(95\% \) upper confidence bound is \(16.82555.\)

b)Step 2: Determine the test statistic

\(\begin{array}{l}{{\bar x}_1} = 13.4\\{{\bar x}_2} = 9.7\\{n_1} = 65\\{n_2} = 50\\{\sigma _{{{\bar x}_1}}} = 2.05 \Rightarrow {s_1} = {\sigma _{{{\bar x}_1}}}\sqrt n = 2.05\sqrt {65} \approx 16.5276\\{\sigma _{{{\bar x}_2}}} = 1.76 \Rightarrow {s_2} = {\sigma _{{{\bar x}_2}}}\sqrt n = 1.76\sqrt {50} \approx 12.4451\end{array}\)

Let us assume: \(\alpha = 0.05\)

Given claim: exceeds

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain the value mentioned in the claim.

\(\begin{array}{l}{H_0}:{\mu _1} = {\mu _2}\\{H_a}:{\mu _1} > {\mu _2}\end{array}\)

Determine the test statistic:

\(t = \frac{{{{\bar x}_1} - {{\bar x}_2}}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }} = \frac{{13.4 - 9.7}}{{\sqrt {\frac{{{{16.5276}^2}}}{{65}} + \frac{{{{12.4451}^2}}}{{50}}} }} \approx 1.369\)

Determine the degrees of freedom (rounded down to the nearest integer):

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{{{16.5276}^2}}}{{65}} + \frac{{{{12.4451}^2}}}{{50}}} \right)}^2}}}{{\frac{{{{\left( {{{16.5276}^2}/65} \right)}^2}}}{{65 - 1}} + \frac{{{{\left( {{{12.4451}^2}/50} \right)}^2}}}{{50 - 1}}}} \approx 112 > 60\)

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The \({\rm{P}}\)-value is the number (or interval) in the column title of Student's \(T\) distribution in the appendix containing the \(t\)-value in the row \(df = 60\) :

\(0.05 < P < 0.10\)

If the P-value is less than or equal to the significance level, then the null hypothesis is rejected:

\(P > 0.05 \Rightarrow {\rm{ Fail to reject }}{H_0}\)

There is not sufficient evidence to support the claim that the true average time spent by blackbirds at the experimental location exceeds the true average time birds of this type spend at the natural location.

C)Step 3: The end point of confidence

\({\bar x_1} = 9.7\)

\(\begin{array}{l}{{\bar x}_2} = 38.4\\{n_1} = 50\\{n_2} = 46\\{\sigma _{{{\bar x}_1}}} = 1.76 \Rightarrow {s_1} = {\sigma _{{{\bar x}_1}}}\sqrt n = 1.76\sqrt {50} \approx 12.4451\\{\sigma _{{{\bar x}_2}}} = 5.06 \Rightarrow {s_2} = {\sigma _{{{\bar x}_2}}}\sqrt n = 5.06\sqrt {46} \approx 34.3186\end{array}\)

Let us assume that we want to calculate the confidence interval with $95 \%$ confidence (other confidence levels work similarly).

\(c = 95\% = 0.95\)

Determine the degrees of freedom (rounded down to the nearest integer):

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{{{12.4451}^2}}}{{50}} + \frac{{{{34.3186}^2}}}{{46}}} \right)}^2}}}{{\frac{{{{\left( {{{12.4451}^2}/50} \right)}^2}}}{{50 - 1}} + \frac{{{{\left( {{{34.3186}^2}/46} \right)}^2}}}{{46 - 1}}}} \approx 55 > 50\)

Determine the t-value by looking in the row starting with degrees of freedom\(df = 50\)and in the column with\(1 - c/2 = 0.025\)in the Student's distribution table in the appendix:

\({t_{\alpha /2}} = 2.009\)

The margin of error is then:

\(E = {t_{\alpha /2}} \cdot \sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} = 2.009 \cdot \sqrt {\frac{{{{12.4451}^2}}}{{50}} + \frac{{{{34.3186}^2}}}{{46}}} \approx 10.7629\)

The endpoints of the confidence interval for \({\mu _1} - {\mu _2}\) are: \(\begin{array}{l}\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E = (1.5083 - 1.5875) - 10.7629 = - 0.0792 - 10.7629 = - 10.8421\\\left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E = (1.5083 - 1.5875) + 10.7629 = - 0.0792 + 10.7629 = 10.6837\end{array}\)