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X9 Total Flight Time N = 306 SD = 5836.1251 SE = SD/n 5836.1251/306 SE = 333.629 Statistic Std. Error X9 = Total Flight Time
X9 Total Flight Time
N = 306
SD = 5836.1251
SE = SD/n
5836.1251/306
SE = 333.629
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 |
- Interpret SEin the context of the given research setting .
- Using a critical value of 1.96 and the standard error of 333.629 constructthe 95% confidence interval for the population mean.
- Comparethe95%CIyouconstructedtowhatyourstatisticalsoftware packageprovides.
- Interpret the 95% CI in the context of the given research setting (use the CI provided byyoursoftwarepackage).
- Examinethe shapeof the distributionof participants' totalflight time.
- Whatisthe shapeof this distribution?
- Perform an outlier analysis using Jackknife distances and remove the outliers. Does theshape of the distribution change?
- 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?
- Although the sample datain both the presence and absence of outliersare not normallydistributed,wouldtheinterpretationof the95% CIstill be meaningful? Explain.
- Suppose the data were from a sample of convenience instead of a random sample. Wouldtheinterpretation of the95% CIstill bemeaningful?Explain.
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