Help with C, D, and E 1. Understand how to estimate the value of the Mean for a

Question

Help with C, D, and E

1. Understand how to estimate the value of the Mean for a quantitative variable and a Proportion for a qualitative variable. 2. Know how to find a point estimate and a margin of error 3. Know when to use the T-distribution versus the Z-distribution. Problem (1). The FAA wants to estimate the average flight times (in minutes) from Albuquerque, NM to Dallas Texas, for flights with American Airlines. He randomly selects nine flights between the two cities and obtains the data shown. Assume that the population Standard Deviation (a) equals 8 minutes. Answer 195 110, 1103-1111-191- 1 100 106 a)X=103.44 Use the data to compute a point estimate for the population mean flight time between Albuquerque and Dallas on an American Airlines flight. (a) a Miers Because the sample size is small, we must verify that the flight time is approximately normally distributed and the sample does not contain any outliers or is skewed Construct your own box-plot and calculate the skew-measure (b) c) (c) Construct a 95% confidence interval for the flight time. Interpret this interval Moran of error: (d) Construct a 90% confidence Intervafor the flight time. Interpret this interval ) 107.33) decreased (size decreases) (e) What happened to the width of the interval when the level of confidence decreased?

Explanation / Answer

(a) Here mean flight time x = 103.4444 (as formula you have written is correct.)

(b) There are no outliers here/

(c) Here standard error of the mean se0 = /sqrt(n) = 8/ sqrt(9) = 8/3 minutes

Margin of error = Critica test statistic * Standard error of the mean

Critica test statistic = Z(95%) = +-1.96

Margin of error = 1.96 * 8/3 = 5.2267 minutes

95% confidence interval = sample mean +- margin of error = 103.4444 +- 5.2267 = (98.22 min, 108.67 min)

(d) For 90% confidence interval

Critical test statistic Z = +- 1.645

Margin of error = 1.645 * 8/3 = 4.3867 minutes

95% confidence interval = sample mean +- margin of error = 103.4444 +- 4.3867 = (99.06 min, 107.83 min)

(e) If level of confidence interval will decrease then the width of confidence interval will decrease.

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