Question

A student desires to know if two-bedroom apartments for rent near the BRCC campus in Baton Rouge are significantly different, on average, than one-bedroom apartments for rent near the BRCC campus in Baton Rouge. The student collects data for a random sample of ten (10) advertisements for each of one-bedroom and two-bedroom apartments for rent near the BRCC campus in Baton Rouge. The data below shows the advertised rents (in dollars per month) for the selected apartments.

1 Bedroom
550
695
640
580
595
600
495
675
595
525
2 Bedroom
645
725
580
750
575
635
695
650
655
575

Set up and carry out an appropriate test to determine whether there is convincing evidence that drinking coffee improves memory. Be sure to provide the type of test used, the correct hypotheses, the test statistic, the P-value, assumptions necessary (if any), and the conclusions in context. Assume unequal variances in the populations.
Then, provide and interpret a 99% confidence interval for the hypotheses that you test in (a). What conclusion(s) do you make in context?

Answer 1

Null hypothesis : There is no significant difference between the advertised rent of one bedroom apartments and two bedroom apartments.

i.e, The Advertised rent of one bedroom apartment () = Advertised rent of two bedroom apartment .

Alternative hypothesis : There is significant difference between the advertised rent of one bedroom apartments and the two bedroom apartments.

i.e,

Rent of one bedroom apartment (X)	Rent of two bedroom apartment (Y)
550 695 640 580 595 600 495 675 595 525	645 725 580 750 575 635 695 650 655 575	2025 10000 2025 225 0 25 10000 6400 0 4900	12.25 5852.25 4692.25 10302.25 5402.25 182.25 2162.25 2.25 42.25 5402.25
		= 3955.55	= 3783.61

Sample Mean of X = 5950/10 = 595

Sample Mean of Y = 6485/10 = 648.5

Sample standard deviation of X = Sample standard deviation of Y =

Assuming unequal population variance, pooled standard deviation =

= 62.21

Test statistic is given by -

= -1.92

P value = P(|t| > 1.92) = Area under the t curve with 18 degrees of freedom on the left side of t = -1.92 and the right side of t = 1.92

= 0.071

Can be obtained using the formula =TDIST(1.92,18,2) in excel

Since, P value (0.071) > 0.01( level of significance), we may not reject the null hypothesis at 0.01 level of significance. So, we conclude that there is no significant difference between the advertised rent of one bedroom apartments and two bedroom apartments.