n

Mean

St. dev.

LQ

Median

UQ

Athenia

135

1756

69

1706

1754

1803

Olympic

150

1733

101

1670

1731

1803

(b) Let X1 be the cost of a two week stay with Athenia in September, and let X2 be the cost of a two week stay with Olympic in September. Assume that both X1 and X2 follow a Normal distribution, with means and standard deviations as given in the table above.

Let A be upper quartile cost of a two week stay with Athenia. For a randomly chosen customer with Athenia and a randomly chosen customer with Olympic, find P(X1 > A) and P(X2 > A), respectively. Comment.

P(X1>A)

P(X1>1803) = 1-P (Z<1803-1755.9/69) = 1- P (Z<0.6819) = 1-0.7517 = 0.2483

P(X2>A)

P(X2>1803) = 1-P (Z<1803-1733.3/100.996) = 1-P (Z<0.69) = 1- 0.7549 = 0.2451

The probability for the cost of a two week stay with Athenia in September that is higher than £1803 is higher than that of the Olympic. So, it can be seen that Olympic has a higher probability to offer cheap luxury accommodation.

(c) Using your answers to parts (a) and (b), compare and contrast the costs of renting a luxury apartment with these two companies.

To compare Athenia and Olympic, we can see that for the lower quartile range for Athenia (£1706) is higher than that of Olympic (£1671). As well as for the mean and median, Olympics’ are cheaper than Athenias’. Also, from (b), we can see that the probability for getting higher than £1803, Athenia’s one is higher than Olympic’s. So, these show the cost of renting a luxury apartment with Olympic is cheaper than Athenia.

(d) Now compare the costs between the two companies more formally by performing an appropriate hypothesis test. Remember to clearly state your null and alternative hypotheses, and make sure you explain your findings clearly, in plain English.

[You may use Minitab to perform this hypothesis test, if you wish, although full marks can also be achieved by working through this question by hand]

Athenia:

Null and alternate hypotheses:

H0: µ = £ 1756

H1: µ ≠ £1756

2 (a) Produce a scatterplot of these data, and comment. [You may use Minitab for your plot, if you wish]

(b) Calculate Sx1y, Sx1x1 and Syy. You may use the following summaries:

¯x1 = 6353.214 ¯y = 6290.714

∑x²1 = 582613061 Ʃ y² = 589703770 Ʃx1y = 578290105

Sx1y = 578290105-14x6353.214x6290.714 = 18762573.43

Sx1x1 = 582613061 -14x6353.214x6353.214

= 17526467.18

Syy = 589703770- 14x6290.714x6290.714 = 35680613.18

2 (c) Use your answers to part (b) to estimate the simple linear regression equation Y = β0 + β1X1 + ǫ, and superimpose your regression line on the scatterplot in part (a).

β1= Sx1y/Sx1x1= 18762573.43/17526467.18 = 1.07 β0 = ˉy- β1ˉx1 = 6290.714 – 1.0705x6353.214 = -510.58 ≈ -511 Thus, the regression equation is: Y = -511+ 1.07X1 + ǫ

(d) Enter the data from the table above into two separate columns of a Minitab worksheet. Now use Minitab to perform a simple linear regression of Y on X1 to confirm your answer to part (c). In your solutions,