Question

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."† The article goes on to summarize findings published in the paper "Sex Differences in Response to Children's Toys in Nonhuman Primates."† 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 gives means and standard deviations (approximate values read from graphs in the paper) for the percentage of the 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 of female monkeys. Use a 0.05 significance level for any hypothesis tests that you carry out when answering the various parts of this exercise. (Use μ₁ for male monkeys and μ₂ for female monkeys.)

		Percent of Time
		Female Monkeys			Male Monkeys
		n	Sample Mean	Sample Standard Deviation	n	Sample Mean	Sample Standard Deviation
Toy	Police Car	44	7	4	44	19	6
	Doll	44	19	4	44	10	2
	Furry Dog	44	20	5	44	26	5

(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?

State the appropriate null and alternative hypotheses.

Find the test statistic and P-value. (Use technology. Round your test statistic to one decimal place and your P-value to three decimal places.)

State the conclusion in the problem context.

(b)

The doll was considered a "feminine" toy. Do these data provide convincing evidence that the mean percentage of the time spent playing with the doll is greater for female monkeys than for male monkeys?

State the appropriate null and alternative hypotheses.

Find the test statistic and P-value. (Use technology. Round your test statistic to one decimal place and your P-value to three decimal places.)

State the conclusion in the problem context.

(c)

The furry dog was considered a "neutral" toy. Do these data provide convincing evidence that the mean percentage of the time spent playing with the furry dog is not the same for male and female monkeys?

State the appropriate null and alternative hypotheses.

Find the test statistic and P-value. (Use technology. Round your test statistic to one decimal place and your P-value to three decimal places.)

State the conclusion in the problem context.

Answer 1

Assume two data sets come from normal populations.

Test and CI for Two Variances

Method

Null hypothesis Sigma(1) / Sigma(2) = 1
Alternative hypothesis Sigma(1) / Sigma(2) not = 1
Significance level Alpha = 0.05

Statistics

Sample N StDev Variance
Female Monkey 44 4.000 16.000
Male Monkey 44 6.000 36.000

Ratio of standard deviations = 0.667
Ratio of variances = 0.444

Test
Method DF1 DF2 Statistic P-Value
F Test (normal) 43 43 0.44 0.009

Since p-value<0.05 so we can't assume two population variances are same.

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean
Male Monkey 44 19.00 6.00 0.90
Female Monkey 44 7.00 4.00 0.60

Difference = mu (1) - mu (2)
Estimate for difference: 12.00
95% lower bound for difference: 10.19
T-Test of difference = 0 (vs >): T-Value = 11.04 P-Value = 0.000 DF = 74

The test statistic =11.0 and P-value=0.000

Since P-value<0.05 so we reject null hypothesis at 5% level of significance and conclude that these data provides convincing evidence that the mean percentage of the time spent playing with the police car is greater for male monkeys than for female monkeys.

Assume two data sets come from normal populations.

Test and CI for Two Variances

Method

Null hypothesis Sigma(1) / Sigma(2) = 1
Alternative hypothesis Sigma(1) / Sigma(2) not = 1
Significance level Alpha = 0.05

Statistics

Sample N StDev Variance
Male Monkey 44 2.000 4.000
Female Monkey 44 4.000 16.000

Ratio of standard deviations = 0.500
Ratio of variances = 0.250

Test
Method DF1 DF2 Statistic P-Value
F Test (normal) 43 43 0.25 0.000

Since p-value<0.05 so we can't assume two population variances are same.

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean
Male Monkey 44 10.00 2.00 0.30
Female Monkey 44 19.00 4.00 0.60

Difference = mu (1) - mu (2)
Estimate for difference: -9.000
95% upper bound for difference: -7.874
T-Test of difference = 0 (vs <): T-Value = -13.35 P-Value = 0.000 DF = 63

The test statistic=-13.4 and P-value=0.000. Since P-value<0.05 so we reject null hypothesis at 5% level of significance and conclude that these data provides convincing evidence that the mean percentage of the time spent playing with the doll is greater for female monkeys than for male monkeys.

Assume two data sets come from normal populations.

Test and CI for Two Variances

Method

Null hypothesis Sigma(1) / Sigma(2) = 1
Alternative hypothesis Sigma(1) / Sigma(2) not = 1
Significance level Alpha = 0.05

Statistics

Sample N StDev Variance
Male Monkey 44 5.000 25.000
Female Monkey 44 5.000 25.000

Ratio of standard deviations = 1.000
Ratio of variances = 1.000

Test
Method DF1 DF2 Statistic P-Value
F Test (normal) 43 43 1.00 1.000

Since p-value>0.05 so we assume two population variances are same.

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean
Male Monkey 44 26.00 5.00 0.75
Female Monkey 44 20.00 5.00 0.75

Difference = mu (1) - mu (2)
Estimate for difference: 6.00
95% CI for difference: (3.88, 8.12)
T-Test of difference = 0 (vs not =): T-Value = 5.63 P-Value = 0.000 DF = 86
Both use Pooled StDev = 5.0000

The test statistic=5.6 and P-value=0.000.  Since P-value<0.05 so we reject null hypothesis at 5% level of significance and conclude that these data provides convincing evidence that the mean percentage of the time spent playing with the furry dog is not the same for male and female monkeys.