Question

The following data shows self reported heights and measured heights (in inches) for 8 randomly selected teenage girls. Is there sufficient evidence, at a 0.05 significance levelto sugest that there is a difference between self reported and measured height? Reported 53,64, 61, 66, 64, 65, 68, 63 Measured 58.1 62.7 61.1 64.8 63.2 66.4 67.6 63.5

Answer 1

Since the same set of girls are taken as sample data , we use a matched pairs t test.

The mean of the difference = 0.43, and the standard deviation of the difference S_d = 2.0947

Let be the difference in scores of the 2 populations.

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The Hypothesis:

H0: = 0 : There is no difference between the self reported and measured heights.

Ha: 0: There is a difference between the self reported and measured heights.

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The Test Statistic: Since sample size is small, and population std. deviation is unknown, we use the students t test.

The p value: (2 tailed) at t = 0.58, degrees of freedom = n -1 = 8 - 1 = 7 is 0.5801

The Critical values: (2 tailed) for = 0.05, df = 7 are +2.365 and -2.365

The Decision Rule:   If t observed is > t critical or if t observed is < -t critical, Then reject H0.

Also if P value is < , Then Reject H0.

The Decision:   Since t observed (0.58) lies between the 2 critical values We Fail to Reject H0.

Also since P value (0.5801) is > (0.05) , We fail to Reject H0.

The Conclusion: There isn't sufficient evidence at the 95% significance level to support the claim that there is a difference between the self reported and measured heights.

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Calculation for the mean and standard deviation:

Mean = Sum of observation / Total Observations

Standard deviation = SQRT(Variance)

Variance = Sum Of Squares (SS) / n - 1,

where SS = SUM(X - Mean)².

#	Difference	Mean	(X-Mean)2
1	5.1	0.43	21.8089
2	-1.3	0.43	2.9929
3	0.1	0.43	0.1089
4	-1.2	0.43	2.6569
5	-0.8	0.43	1.5129
6	1.4	0.43	0.9409
7	-0.4	0.43	0.6889
8	0.5	0.43	0.0049
Total	3.4	SS	30.7152

n	8
Sum	3.4
Mean	0.43
SS	30.7152
Variance	4.3879
Std Dev	2.0947