Mathematics, 27.02.2020 21:55 Flameking1223

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.

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Answer from: sanafarghal

(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* $s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}$ < $(\mu_1-\mu_2)$ < + 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* $s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}$ , + 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 .

Answer from: Quest

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Answer from: Quest

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Monthly rent paid by undergraduates and graduate students. Undergraduate Student Rents (n = 10) 760...

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