Worksheet 2A: Describing data Part One: The love gene a study in search of roman

Question

Worksheet 2A: Describing data Part One: The love gene a study in search of romance Studies of monogamous prairie voles have identified a single gene, that seems to play a key role in pair bonding. The the data provided here comes from Lim et 2004. In this study, gene was experimentally enhanced in 10 male voles and their pair bonding behavior was compared to a control group. Enhanced V1A grou Control Group 100,97, 96,93,89, 98,96,88,77,74,70, percentage of time spent 8884776761 huddling with female mate 52,47 40,35,29 13,6 work on these calculations at home. You can use a calculator for this section. 1. Calculate the mean value for each group Mean percentage huddling time for the control group Mean percentage huddling time for the experimental group 2. Calculate the median value for each group Median percentage huddling time for the control group Median percentage huddling time for the experimental group 3. Which of these estimates best represents the central tendency for the control group Why? maaltan Which of these estimates best represents the central tendency for the experimental To mean ble the olat nat stem Why? 4. Calculate the standard deviation for your sample. Use your lecture notes. SD. 12.12 29. X controls SS exp 85. Control STG 121 1319 44 I Iu00 1024 72 107 93,41

Explanation / Answer

Question 5

The SD for control group is given as 30.57818898 while the SD for experimental group is given as 13.09622507.

From these statistics we deduced that the variation in the control group is more than the variation in the experimental group.

Question 6

The formula for standard error is given as below:

Standard error = SE = Standard deviation / sqrt(n)

Standard error for the control group

Standard error = 30.57818898 / sqrt(13) = 8.480863712

Standard error for the experimental group

Standard error = 13.09622507/sqrt(10) = 4.141389997

From these statistics we deduced that the margin of error for the sampling distribution of the experimental group would be less than the control group.

Question 7

Here, we have to find out the 95% confidence intervals for the population mean for control group.

The confidence interval formula is given as below:

Confidence interval = Xbar -/+ t*SE

Where, SE = SD/sqrt(n)

Data

Sample Standard Deviation

30.57818898

Sample Mean

55.76923077

Sample Size

13

Confidence Level

95%

Intermediate Calculations

Standard Error of the Mean

8.480863712

Degrees of Freedom

12

t Value

2.1788

Interval Half Width

18.4782

Confidence Interval

Interval Lower Limit

37.29

Interval Upper Limit

74.25

Now, we have to find out the 95% confidence interval for the population mean for experimental group.

Data

Sample Standard Deviation

13.09622507

Sample Mean

85.2

Sample Size

10

Confidence Level

95%

Intermediate Calculations

Standard Error of the Mean

4.141389997

Degrees of Freedom

9

t Value

2.2622

Interval Half Width

9.3685

Confidence Interval

Interval Lower Limit

75.83

Interval Upper Limit

94.57

From these confidence intervals, it is observed that the width of the confidence interval for control group is more than the experimental group.

Question 8

The above data and confidence intervals do not support the scientist’s hypothesis that the VIA gene plays a role in pair bonding as we find out there is significant difference in the above two intervals.

Data

Sample Standard Deviation

30.57818898

Sample Mean

55.76923077

Sample Size

13

Confidence Level

95%

Intermediate Calculations

Standard Error of the Mean

8.480863712

Degrees of Freedom

12

t Value

2.1788

Interval Half Width

18.4782

Confidence Interval

Interval Lower Limit

37.29

Interval Upper Limit

74.25

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