Question

Improving the productivity of chickens. Refer to the Applied Animal Behaviour Science (October 2000) study of the color of string preferred by pecking domestic chickens, Exercise 6.124 (p. 376). Recall that n = 72 chickens were exposed to blue string and the number of pecks each chicken took at the string over a specified time interval had a mean of\(\overline x = 1.13\,\)pecks and a standard deviation of s = 2.21 pecks. Also recall that previous research had shown that m = 7.5 pecks if chickens are exposed to white string.

a. Conduct a test (at\(\alpha = 0.01\)) to determine if the true mean number of pecks at blue string is less than\(\mu = 7.5\)pecks.
b. In Exercise 6.122, you used a 99% confidence interval as evidence that chickens are more apt to peck at white string than blue string. Do the test results, part a, support this conclusion? Explain

Short answer

The null hypothesis is rejected, and it may conclude that there is sufficient evidence to support the claim that the true mean number of pecks at a blue string is less than\(\mu = 7.5\)pecks.

Step-by-step solution

Given information.

Given the summary statistics as follows:

Let n denotes the number of chickens, that is, n=72.

The sample mean is\(\overline x = 1.13\,pecks\)

The sample standard deviation is\(s = 2.21\,pecks\)

Here \(\mu = 7.5\,pecks\) if chickens are expressed to white string.

 Concept of t-test statistic

If the sample size is less than 30 or the population standard deviation is unknown, the t-test statistic is used to test the hypothesis.

 Testing the hypothesis

a.

The claim is to find if the true mean number of pecks at the blue string is less than\(\mu = 7.5\,pecks\)

From this information, the null and the alternative hypothesis are as follows:

\({H_0}:\mu = 7.5\,pecks\)

i.e., the true mean number of pecks at a blue string is 7.5.

\({H_a}:\mu < 7.5\,pecks\)

i.e., the true mean number of pecks at a blue string is 7.5.

Here, the level of significance\(\alpha = 0.01\)

Here sample size is less than 72 and has the sample standard deviation, and then use 1- sample t test to analyze the data.

Since, the t-test statistic is given by,

\(t = \frac{{\overline x - \mu }}{{\frac{s}{{\sqrt n }}}}\)

Therefore,

\(\begin{aligned}t &= \frac{{1.130 - 7.5}}{{\frac{{2.210}}{{\sqrt {72} }}}}\\ &= \frac{{ - 6.37}}{{0.2605}}\\ &= - 24.45\end{aligned}\)

From the table of t distribution, the tabulated value at a 1% level of significance is 0.000.

Since the calculated value is less than the tabulated value of the test statistic. Therefore, the null hypothesis is rejected, and it may conclude that there is sufficient evidence to support the claim that the true mean number of pecks at blue string is less than\(\mu = 7.5\)pecks.