X9 Total Flight Time

N = 306

SD = 5836.1251

SE = SD/n

5836.1251/306

SE = 333.629

(Software package -SPSS Calculations)

	Statistic	Std. Error
X9 = Total Flight Time	Mean	6185.339	333.6291
95% Confidence Interval for Mean	Lower Bound	5528.833
Upper Bound	6841.845
5% Trimmed Mean	5526.826
Median	4355.000
Variance	34060356.278
Std. Deviation	5836.1251
Minimum	11.7
Maximum	35000.0
Range	34988.3
Interquartile Range	6425.0
Skewness	1.758	.139
Kurtosis	3.390	.278

1. Using a critical value of 1.96 and the standard error you calculated above, constructthe 95% confidence interval for the population mean.

Answer:

6119.426	Lower
6251.252	Upper

(By Hand Calculation)

SE	t*
33.6291	1.96	65.91304
Population Mean			95%CI
6185.339	-65.91304	6119.426	Lower
6185.339	65.91304	6251.252	Upper

Questions:

1. Using a critical value of 1.96 and the standard error you calculated above, constructthe 95% confidence interval for the population mean.

2. Comparethe95%CIyouconstructedtowhatyourstatisticalsoftware packageprovides.

3. Interpret the 95% CI in the context of the given research setting (use the CI provided byyoursoftwarepackage).

4.Examinethe shapeof the distributionof participants' totalflight time.

5. Whatisthe shapeof this distribution?

6. Perform an outlier analysis using Jackknife distances and remove the outliers. Does theshape of the distribution change?

7. Compare the SEand 95% CI of this modified distribution where the outliers have beenremoved to the respective SEand 95% CI of the initial distribution where the outliers werepresent. Describe the impact the outliers are having on the SEand the 95% CI. Whichdistribution do you think provides a better estimation of the population's mean flight time?Why?

8. Although the sample datain both the presence and absence of outliersare not normallydistributed,wouldtheinterpretationof the95% CIstill be meaningful? Explain.

9. Suppose the data were from a sample of convenience instead of a random sample. Wouldtheinterpretation of the95% CIstill bemeaningful?Explain.