Chapter 11.1, Problem 17E

A newspaper story headline reads “Gender Plays Part in Monkeys’ Toy Choices, Research Finds—Like Humans, Male Monkeys Choose Balls and Cars, While Females Prefer Dolls and Pots” (Knight Ridder Newspapers, December 8, 2005). The article goes on to summarize findings published in the paper “Sex Differences in Response to Children’s Toys in Nonhuman Primates” (Evolution and Human Behavior [2002]: 467–479). Forty-four male monkeys and 44 female monkeys were each given a variety of toys, and the time spent playing with each toy was recorded. The table below gives means and standard deviations (approximate values read from graphs in the paper) for the percentage of time that a monkey spent playing with a particular toy. Assume that it is reasonable to regard these two samples of 44 monkeys as representative of the populations of male monkeys and female monkeys. Use a 0.05 significance level for any hypothesis tests that you carry out when answering the various parts of this exercise.

1. a. The police car was considered a “masculine toy.” Do these data provide convincing evidence that the mean percentage of the time spent playing with the police car is greater for male monkeys than for female monkeys?
2. b. The doll was considered a “feminine toy.” Do these data provide convincing evidence that the mean percentage of time spent playing with the doll is greater for female monkeys than for male monkeys?
3. c. The furry dog was considered a “neutral toy.” Do these data provide convincing evidence that the mean percentage of time spent playing with the furry dog is not the same for male and female monkeys?
4. d. Based on the conclusions from the hypothesis tests of Parts (a)–(c), is the quoted newspaper story headline a reasonable summary of the findings? Explain.
5. e. Explain why it would be inappropriate to use the two-sample t test to decide if there was evidence that the mean percentage of time spent playing with the police car and the mean percentage of the time spent playing with the doll is not the same for female monkeys.

To determine

Check whether mean percentage of time spent playing with police car is greater for male monkeys than for female monkeys or not.

Answer to Problem 17E

There is convincing evidence that mean percentage of time spent playing with police car is greater for male monkeys than for female monkeys.

Explanation of Solution

Calculation:

Given table provides the percentage of time spent playing with toys.

The assumptions for the two-sample t-test:

• The random samples should be collected independently.
• The sample sizes should be large. That is, each sample size is at least 30.

The assumptions in this particular problem:

• Two samples of 44 monkeys are selected randomly and independently from the population.
• The sample sizes are large enough.

Here, both sample sizes are equal to 44 and which are greater than 30.

Therefore, the assumptions are satisfied.

Let ${\mu }_1$ be the mean percentage of time playing with police car for male monkeys.

Let ${\mu }_2$ be the mean percentage of time playing with police car for female monkeys.

Hypotheses:

Null hypothesis:

$H_0:{\mu }_1-{\mu }_2=0$

That is, the mean percentage of time playing with police car is same for both male and female monkeys.

Alternative hypothesis:

$H_a:{\mu }_1-{\mu }_2>0$

That is, the mean percentage of time spent playing with police car is greater for male monkeys than for female monkeys.

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain the P-value and test statistic by using MINITAB software is as follows:

• Choose Stat > Basic Statistics > 2 sample t.
• Choose Summarized data.
• In sample 1, enter Sample size as 44, Mean as 18, Standard deviation as 5.
• In sample 2, enter Sample size as 44, Mean as 8, Standard deviation as 4.
• Choose Options.
• In Confidence level, enter 95.
• In Alternative, select greater than.
• Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

Therefore, the P-value is 0 and the test statistic is 10.359.

Decision rule:

• If P-value is less than or equal to the level of significance, reject the null hypothesis.
• Otherwise fail to reject the null hypothesis.

Conclusion:

Here, the level of significance is 0.05.

Here, P-value is less than the level of significance.

That is, $P\text{-value}\left(=0\right)<0.05$ .

Therefore, reject the null hypothesis.

Hence, there is proper evidence to support the claim that the mean percentage of time spent playing with police car is greater for male monkeys than for female monkeys.

To determine

Check whether mean percentage of time spent playing with doll is greater for female monkeys than for male monkeys or not.

Answer to Problem 17E

There is convincing evidence that mean percentage of time spent playing with doll is greater for female monkeys than for male monkeys.

Explanation of Solution

Calculation:

The assumption for the two-sample t-test:

• The random samples should be collected independently.
• The sample sizes should be large. That is, each sample size is at least 30.

The assumptions in this particular problem:

• Two samples of 44 monkeys are selected randomly and independently from the population.
• The sample sizes are large enough.

Here, both sample sizes are equal to 44 and which are greater than 30.

Therefore, the assumptions are satisfied.

Let ${\mu }_1$ be the mean percentage of time playing with doll for male monkeys.

Let ${\mu }_2$ be the mean percentage of time playing with doll for female monkeys.

Hypotheses:

Null hypothesis:

$H_0:{\mu }_1-{\mu }_2=0$

That is, the mean percentage of time playing with doll is same for both male and female monkeys.

Alternative hypothesis:

$H_a:{\mu }_1-{\mu }_2<0$

That is, the mean percentage of time spent playing with doll is greater for female monkeys than for male monkeys.

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain the P-value and test statistic by using MINITAB software is as follows:

• Choose Stat > Basic Statistics > 2 sample t.
• Choose Summarized data.
• In sample 1, enter Sample size as 44, Mean as 9, Standard deviation as 2.
• In sample 2, enter Sample size as 44, Mean as 20, Standard deviation as 4.
• Choose Options.
• In Confidence level, enter 95.
• In Alternative, select less than.
• Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

Therefore, the P-value is 0 and the test statistic is –16.316.

Decision rule:

• If P-value is less than or equal to the level of significance, reject the null hypothesis.
• Otherwise fail to reject the null hypothesis.

Conclusion:

Here, the level of significance is 0.05.

Here, P-value is less than the level of significance.

That is, $P\text{-value}\left(=0\right)<0.05$ .

Therefore, reject the null hypothesis.

Hence, there is proper evidence to support the claim that the mean percentage of time spent playing with doll is greater for female monkeys than for male monkeys.

To determine

Check whether mean percentage of time spent playing with furry dog is not the same for female and male monkeys or not.

Answer to Problem 17E

There is convincing evidence that mean percentage of time spent playing with furry dog is not the same for female and male monkeys.

Explanation of Solution

Calculation:

The assumption for the two-sample t-test:

• The random samples should be collected independently.
• The sample sizes should be large. That is, each sample size is at least 30.

The assumptions in this particular problem:

• Two samples of 44 monkeys are selected randomly and independently from the population.
• The sample sizes are large enough.

Here, both sample sizes are equal to 44 and which are greater than 30.

Therefore, the assumptions are satisfied.

Let ${\mu }_1$ be the mean percentage of time playing with furry dog for male monkeys.

Let ${\mu }_2$ be the mean percentage of time playing with furry dog for female monkeys.

Hypotheses:

Null hypothesis:

$H_0:{\mu }_1-{\mu }_2=0$

That is, the mean percentage of time playing with furry dog is same for both male and female monkeys.

Alternative hypothesis:

$H_a:{\mu }_1-{\mu }_2\ne 0$

That is, the mean percentage of time spent playing with furry dog is not the same for female and male monkeys.

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain the P-value and test statistic by using MINITAB software is as follows:

• Choose Stat > Basic Statistics > 2 sample t.
• Choose Summarized data.
• In sample 1, enter Sample size as 44, Mean as 25, Standard deviation as 5.
• In sample 2, enter Sample size as 44, Mean as 20, Standard deviation as 5.
• Choose Options.
• In Confidence level, enter 95.
• In Alternative, select not equal.
• Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

Therefore, the P-value is 0 and the test statistic is 4.690.

Decision rule:

• If P-value is less than or equal to the level of significance, reject the null hypothesis.
• Otherwise fail to reject the null hypothesis.

Conclusion:

Here, the level of significance is 0.05.

Here, P-value is less than the level of significance.

That is, $P\text{-value}\left(=0\right)<0.05$ .

Therefore, reject the null hypothesis.

Hence, there is proper evidence to support the claim that the mean percentage of time spent playing with furry dog is not the same for female and male monkeys.

To determine

Explain whether the newspaper head line a reasonable summary of the findings.

Explanation of Solution

From parts (a), (b), and (c), it is clear that, the male monkeys spent more time with masculine toy than female monkeys, the female monkeys spent more time with feminine toy than male monkeys, mean percentage of time spent playing with neutral toy is not same for female and male monkeys. Therefore the result shows convincing evidence of a gender basis in the monkeys’ choice of how much time to spend playing with three of the toys.

To determine

Explain why two sample t test is inappropriate in this situation.

Explanation of Solution

Here, the population under consideration is female monkeys. The mean percentage of time spent playing with the police car and mean time spent playing with the doll is not same for the population. That is samples are taken from the same population. That is, samples are dependent. Two sample t test is appropriate only when samples are independent. Therefore, two sample t test is inappropriate in this situation.