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Yuliya22 [10]
2 years ago
4

Monthly rent paid by undergraduates and graduate students. Undergraduate Student Rents (n = 10) 760 770 890 660 730 790 790 690

1,060 680 Graduate Student Rents (n = 12) 1,080 920 930 880 720 920 740 830 960 880 860 870 Click here for the Excel Data File (a) Construct a 99 percent confidence interval for the difference of mean monthly rent paid by undergraduates and graduate students, using the assumption of unequal variances. (Use Minitab. Round your final answers to 3 decimal places.) The 99% confidence interval is from to (b) What do you conclude? We cannot conclude there is a significant difference in means for undergraduate and graduate rent. We can conclude there is a significant difference in means for undergraduate and graduate rent.
Mathematics
1 answer:
Vinvika [58]2 years ago
6 0

Answer:

(a) 99% Confidence interval =  [ -230.11 , 29.11 ]

(b) We cannot conclude there is a significant difference in means for undergraduate and graduate rent.

Step-by-step explanation:

We are given the Monthly rent paid by undergraduates and graduate students.

Undergraduate Student Rents (n = 10) : 760, 770, 890, 660, 730, 790, 790, 690, 1,060, 680

Graduate Student Rents (n = 12) : 1,080, 920, 930, 880, 720, 920, 740, 830, 960, 880, 860, 870

Firstly let X_1bar = Sample mean of Undergraduate Student Rents

                          = Sum of all rent values ÷ n = 782

s_1^{2} = variance of Undergraduate Student Rents = \frac{\sum (X_1-X_1bar)^{2} }{n-1} = 14018

X_2bar = Sample mean of Graduate Student Rents = 882.5

s_2^{2} = variance of Graduate Student Rents = \frac{\sum (X_2-X_2bar)^{2} }{n-1} = 9111.4

The pivotal quantity used here for confidence interval is;

                         \frac{(X_1bar -X_2bar) - (\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } } ~ t_n__1+n_2-2

where,

P(-2.845 < t_2_0 < 2.845) = 0.99

P(-2.845 < \frac{(X_1bar -X_2bar) - (\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } } < 2.845) = 0.99

P(-2.845*s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}<(X_1bar -X_2bar) - (\mu_1-\mu_2)<2.845*s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} ) = 0.99

P((X_1bar -X_2bar) - 2.845*s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} < (\mu_1-\mu_2) < (X_1bar -X_2bar) + 2.845*s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} ) = 0.99

99% Confidence interval for (\mu_1-\mu_2) =

[ (X_1bar -X_2bar) - 2.845*s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} , (X_1bar -X_2bar) + 2.845*s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} ]

[(782 - 882.5) - 2.845*106.4\sqrt{\frac{1}{10}+\frac{1}{12} , (782 - 882.5) + 2.845*106.4\sqrt{\frac{1}{10}+\frac{1}{12} ]

 =  [ -230.11 , 29.11 ]

(b) After seeing the 99% confidence interval for the difference of mean monthly rent paid by undergraduates and graduate students, we cannot conclude that there is a significant difference in means for undergraduate and graduate rent because in the above interval 0 lies in between them .

                 

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