Q.7 Birds in the trees Researchers s... [FREE SOLUTION]

91影视

The Practice of Statistics for AP

David Moore,Daren Starnes,Dan Yates

$Math Studyset 91影视 Explanations$ Math

4th Edition 路 809 pages

ISBN: 9781319113339

Chapter 11: Q.7 (page 692)

Birds in the trees Researchers studied the behavior of birds that were searching for seeds and insects in an Oregon forest. In this forest, $54$ % of the trees were Douglas 铿乺s, $40$ % were ponderosa pines, and $6$ % were other types of trees. At a randomly selected time during the day, the researchers observed $156$ red-breasted nuthatches: $70$ were seen in Douglas 铿乺s, $79$ in ponderosa pines, and $7$ in other types of trees.^$2$ Do these data suggest that nuthatches prefer particular types of trees when they鈥檙e searching for seeds and insects? Carry out a chi-square goodness-of-铿乼 test to help answer this question.

Short Answer

Expert verified

From the given information, We cannot reject the null hypothesis because the chi-square value is relatively insignificant, thus we conclude that there is no significant difference between the observed and expected pea count. In other words, after performing the chi-square goodness of fit test, we can conclude that Mendel's prediction is correct.

Step by step solution

Given Information

It is given in the question that, In this forest, $54$ % of the trees were Douglas 铿乺s, $40$ % were ponderosa pines, and $6$ % were other types of trees. At a randomly selected time during the day, the researchers observed $156$ red-breasted nuthatches: $70$ were seen in Douglas 铿乺s, $79$ in ponderosa pines, and $7$ in other types of trees.

Explanation

Given: $n = 156$

Let us assume: $伪 = 0.05$

The null hypothesis states that the population proportions are equal to the mentioned proportions:

localid="1650617560505" $H_{0} : p_{1} = 54 % = 0.54,$

$H_{0} : p_{2} = 40 % = 0.4,$

$H_{0} : p_{3} = 6 % = 0.06$

The alternative hypothesis states the opposite of the null hypothesis:

$H_{a} : At least one of the p_{i}'s is different.$

The expected frequency E is the product of the sample size n and the probabilities pi

localid="1650617756516" $E_{1} = n p_{1} = 156 脳 0.54 = 84.24$

localid="1650617773816" $E_{2} = n p_{2} = 156 脳 0.4 = 62.4$

localid="1650617793473" $E_{3} = n p_{3} = 156 脳 0.06 = 9.36$

The squared differences between the actual and predicted frequencies, divided by the expected frequency, make up the chi-square subtotals.

Explanation

Thevalue of the test statisticis then the sum of the chi-square subtotals: $蠂^{2} = 鈭� \frac{(O 鈭� E)^{2}}{E} = \frac{(70 鈭� 84.24)^{2}}{84.24} + \frac{(79 鈭� 62.4)^{2}}{62.4} + \frac{(7 鈭� 9.36)^{2}}{9.36} = 7.4182$

The degree of freedom is the number of categories decreased by 1.

$d f = c 鈭� 1 = 3 鈭� 1 = 2$

Explanation

The $P$ -value is the probability of obtaining the value of the test statistic, or a value more extreme. The $P$ -value is the number (or interval) in the column title of the chi-square distribution table in the appendix containing the X^$2$-value in row d f= $2$

$0.01 < P < 0.05$

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

$P < 0.05 鈬� Reject H_{0}$

There is sufficient evidence to support the claim that nuthatches prefer particular types of trees.