Compare the heights of people in Team 1 and Team 2. We assume heights are normal

Question

Compare the heights of people in Team 1 and Team 2. We assume heights are normally distributed, and we consider the sample sizes involved in this problem to be small. A random sample of 15 individuals from Team 1, we have sample average of 68 and sample standard deviation of 3. From a random sample of 11 individuals from Team 2, we have sample average of 70 and sample standard deviation of 4.

i. Construct a two-sided 90% confidence interval for the difference in mean heights of people at Team 1 and Team 2.

ii. The Team leader of Team 2 claims that the mean height of people in Team 2 are significantly greater than the people in Team 1. Perform an appropriate seven step hypothesis test, find the P-value and draw a conclusion for the Team leader’s claim.

iii. Suppose the Team leader 2 believes that the height variances of the two populations (camp 1 and 2) are equal, where the variance is 9 in-sq for both camps. Compute the pooled variance, the test statistic, and the p–value for the same hypothesis test in ii.

Explanation / Answer

Team - 1

n1 = 15, mean1 = 68, sd1 = 3

Team - 2

n1 = 11, mean1 = 70, sd1 = 4

(i) two-sided 90% confidence interval for the difference in mean heights

Calculation

Pooled Variance
s2p = (SS1 + SS2) / (df1 + df2) = 25 / 24 = 1.04

Standard Error
s(M1 - M2) = ((s2p/n1) + (s2p/n2)) = ((1.04/15) + (1.04/11)) = 0.41

Confidence Interval
1 - 2 = (M1 - M2) ± ts(M1 - M2) = 2 ± (1.71 * 0.41) = 2 ± 0.69

1 - 2 = (M1 - M2) = 2, 90% CI [1.31, 2.69].

You can be 90% confident that the difference between your two population means (1 - 2) lies between 1.31 and 2.69.

(ii)

State the hypotheses.

Null hypothesis: 1< 2
Alternative hypothesis: 1 < 2

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.10 (90%CI). Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s12/n1) + (s22/n2)]
SE = sqrt[(32/15) + (42/11)] = 1.4333686

DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (32/15 + 42/11)2 / { [ (32 / 15)2 / (14) ] + [ (42 / 11)2 / (10) ] }
DF = 4.221157 / { 0.0257142857 + 0.2115702479 }= 17.789

t = [ (x1 - x2) ] / SE = [(68 - 70) ] / 1.4333686 = -1.395

where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is the size of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

We use the t Distribution Calculator to find P(t < -1.395)

The P-Value is 0.089996.
The result is significant at p < 0.10.

Interpret results. Since the P-value (0.089996) is less than the significance level (0.10), we cannot accept the null hypothesis.

Conclusion. Reject null hypothesis. We have sufficient evidence that the mean height of people in Team 2 are significantly greater than the people in Team 1.

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