Problem 11-18 (modified). NASA scientists obtained the followingdata from airborne radiometer scans using the ESTAR method ofaperature synthesis.Soil Moisture Percentages: 19,16,15,16.5,15,14Source: D. M. LeVine, A. J. Griffis, C. T. Swift, and T. J.Jackson, "ESTAR: A Synthetic Aperature Microwave Radiometer forRemote Sensing Applications," Proceedings of the IEEE,December 1994, p. 1795.These data represent sample measurements taken over the 220-240degree K brightness temperature range. The theoretical model statesthat the mean volumetric soil moisture percentage should be 13.5%.Use the confidence interval approach to conclude whether the sampledata support or refute the null hypothesis that the mean ESTARmeasurements equal the theoretical level, using a .05 level ofsignificance.Lower 95% confidence limit = 1.Upper 95% confidence limit = 2.

Respuesta :

Answer:

95% confidence interval is 15.92±1.67. Since lower confidence limit 14.25% is higher than the theoretical level 13.5%, we can reject the null hypothesis that the mean ESTAR measurements equal the theoretical level, at 0.05 significance.

Step-by-step explanation:

Confidence Interval can be calculated using M±ME where

  • M is the sample mean soil moisture percentage
  • ME is the margin of error from the mean

margin of error (ME) from the mean can be calculated using the formula

ME=[tex]\frac{t*s}{\sqrt{N} }[/tex] where

  • t is the corresponding statistic in the 95% confidence level
  • s is the standard deviation of the sample
  • N is the sample size (6)

Mean of the sample is=[tex]\frac{19+16+15+16.5+15+14}{6} [/tex] ≈ 15.92

Standard deviation of the sample is [tex]\sqrt{\frac{(19-15.92)^2+(16-15.92)^2+(15-15.92)^2+(16.5-15.92)^2+(15-15.92)^2+(14-15.92)^2}{6} }[/tex]≈1.59

two tailed t value of 95% confidence level in 5 degrees of freedom is 2.571

Then ME=[tex]\frac{2.571*1.59}{\sqrt{6} }[/tex] ≈ 1.67

95% confidence interval is then 15.92±1.67

Lower 95% confidence limit =14.25%

Upper 95% confidence limit= 17.59%