SOLUTION: Use the following n=10 sample data to construct a 90% confidence interval of the population mean μ. Assume a simple random sample was taken the population SD is unknown, and

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Question 864078: Use the following n=10 sample data to construct a 90% confidence interval of the population mean μ. Assume a simple random sample was taken the population SD is unknown, and the population is normally distributed.

86 135 250 357 447 553 646 769 830 1135
Found 2 solutions by ewatrrr, sheville151:
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!


Hi

Calculator and/0r use of Electronic Worksheets Absolutely necessary

x	m	(x-µ)	(x-µ)^2

86	520.8	-434.80	189051.040

135	520.8	-385.80	148841.640

250	520.8	-270.80	73332.640

357	520.8	-163.80	26830.440

447	520.8	-73.80	5446.440

553	520.8	32.20	1036.840

646	520.8	125.20	15675.040

769	520.8	248.20	61603.240

830	520.8	309.20	95604.640

1135	520.8	614.20	377241.640 Sum

5208			994663.600 sum/9 = V

			110518.178 sqrt V

		SD	332.443

ME = 1.645*332.4/sqrt(10) = 173

CI: 521-173-ME < u < 521 + 173   (348, 694) not listed; B the closest match 

Answer by sheville151(1)   (Show Source): You can put this solution on YOUR website!
 Hi



Calculator and/0r use of Electronic Worksheets Absolutely necessary



x	µ	(x-µ)	(x-µ)^2

86	520.8	-434.8	189051.04

135	520.8	-385.8	148841.64

250	520.8	-270.8	73332.64

357	520.8	-163.8	26830.44

447	520.8	-73.8	5446.44

553	520.8	32.2	1036.84

646	520.8	125.2	15675.04

769	520.8	248.2	61603.24

830	520.8	309.2	95604.64

1135	520.8	614.2	377241.640 Sum



5208			994663.600 sum/9 = V



			110518.178 sqrt V



		SD	332.443



Degrees of Freedom = n - 1 --> 10 - 1 = 9 + 2 tailed .10 (90% interval)



9.10 on t value table = 1.833



ME = 1.833*332.4/sqrt(10) = 192.69



CI: 520.8 - 192.69 < μ < 520.8 + 192.69   (328, 714)

 

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