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) About Me  (Show Source):
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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) About Me  (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)