Question

A real estate agent wants to estimate the average age of those buying 35 30 52 48 37 investment property in his area. He randomly selects 15 of his clients who 48 50 41 44 43 purchased an investment property and obtains the data shown. Use this information to answer the following questions. 58 43 56 45 40 Click the icon to view a boxplot and a normal probability plot. The buyer ages are approximately normally distributed and the sample does not contain any outliers. Construct a 95% confidence interval for the mean age for all the real estate agents clients who purchased investment property. I0 (Use ascending order. Round to one decimal place as needed.) Interpret this interval. 0 A. We are 95% confident that the interval actually does not contain the true value of the mean B. We are 95% confident that the interval actually does contain the true value of the mean. ° C. There is a 95% chance that the true value of the mean will fall in the interval 0 D. There is a 95% chance that the true value of the mean will not fall in the interval. Use the normal probability plot and boxplot to assist in verifying the model requirements. O A. The requirements are not met since all the data lie within the bounds of the normal probability plot and the boxplot does not reveal any outliers. O B. The requirements are not met since not all the data lie within the bounds of the normal probability plot and the boxplot does reveals outliers. ° C. The requirements are met since all the data lie within the bounds of the normal probability plot and the boxplot does not reveal any outie s O D. The requirements are met since not all the data lie within the bounds of the normal probability plot and the boxplot does reveals outliers
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Answer #1

(a) The computation table (x-7 35 30 52 -9.66667 14.6667 93.44451 215.1112 53.77773 11.11109 58.77783 11.11109 28.44441 13.44447 0.444449 2.777789 7.33333 3.33333 -7.66667 37 3.33333 5.33333 -3.66667 0.66667 1.66667 13.33333 1.66667 11.33333 0.33333 -4.66667 50 58 2.777789 128.4444 0.111109 21.77781 56 40 X 670 Σ(x,-ZF-819 3333 i-1 i-1 The formula for calculating the value of sample mean is 35 + 30 +… + 40 670 44.66667 X = 44 66667 The formula for calculating the value of standard deviation is (819.3333) 15-1 819.3333 14 58.52381 7.650086 Og ~ 7.650086

b) The sample average is X -44.66667 The standard deviation is s= 7.650086 The sample size is n 15 The degrees of freedom is df n-1 -15-1 = 14 The level of significance is α=0.05 The table values are cacul ated by using the excel function tinv(probability, deg rees of freedom) ti(0.05,14) 2.144787 x-1 2.144787 The 95% confidence interval for the population mean is cal2,2-1 2620 7.650086 → 44.66667-2144787 μ 44.66667 + 2、144787 44.66667- 2.144787 2.144787 3.872983 SHS44.66667 +2.144787 3.872983 44. 66667-2144787 (1.975244) μ 44. 66667 + 2. 144787 (1.975244) 44.66667- 4.236476 SHS 44.66667 +4.236476 40.43019 SuS48.90314 Interpretati on: We can be 95% confident that the true population mean is between 40.4302 and 48.9031

c) The step-by-step Instructions to obtain Normal probability Plot using Minitab 18: * Store the variable (X) in the Column C1 of the Minitab18 work sheet. *Choose the Graph -Probability plot Single. Select the variable (X) in the variable box xClick ok. The Normal Probability Plot:

Probability Plot of X Probability Plot ofX Normal-95% CI 95 90 Mean 44.67 StDev 7.650 15 0.116 P-Value 0987 AD 80 70 60 50 40 30 20 Gridline 10 20 30 40 50 60 70

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(d) The step-by-step instructions to obtain a boxplot output using Minitab 18 software: Store the variable _X in the minitab worksheet * Choose Graph ->Boxplot ->One Y -Simple * Select the Graph variables in the box. *Click ok The boxplo t using Minitab 18:

Boxplot of x Boxplot of X 30 35 40 45 50 60

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Interpretation:

The box plot and normal probability plots reveals that the requirements are met since all the data lie within the normal probability plot and the box plot does not reveal any outliers. There is not much departures from normality. The distribution is approximately normal based on the normal probability plot in the above part (c).

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