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4.4722 10.10 According to a paper by Sokal and Hunter published in the Annals of the Entomological Society of America in 1955, wing lengths were measured for a sample of 100 houseflies. The data were used to produce a 95% confidence interval; lengths were recorded in millimeters. Variable N Mean St Dev SE Mean 95.0% CI length 100 4.5500 0.3920 0.0392 ( 4.4722, 4.6278) a. Report the sample mean wing length. 4661278 b. Find the center point of the 95% confidence interval. c. Explain why your answers to parts (a) and (b) are equal. 1556 d. Show that the interval is approximately equal to sample mean plus or minus 2 standard errors, where a standard error is the standard deviation divided by the square root of the sample size. e. Which one of these is the correct interpretation of the interval? 1. There is a 95% probability that we produce an interval that contains population mean wing length. 2. There is a 95% probability that we produce an interval that contains sample mean wing length. 3. The probability is 95% that population mean wing length falls in this interval. 4. The probability is 95% that sample mean wing length falls in this interval. f. Standardize the sample mean, if population mean equals 4.50. g. Use the result of part (f) to argue that 4.50 is a plausible value for the population mean wing length.

Answer 1

a) Sample mean = 4.55

b) Center point of 95% confidence interval:

(Lower + Upper limit)/2

= (4.4722 + 4.6278)/2 = 4.55

c) because sample mean is the center of confidence interval. Margin of error is added and subtracted from mean to get the upper and lower limit.

d) Lower limit = 4.55 - 2*0.0392 = 4.4716

Upper limit = 4.55 + 2*0.0392 = 4.6284

Which is approximately equal to the confidence interval

e) Answer 3. The probability is 95% that population mean wing length falls in this interval.