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sample of 50 items sel. The mean value of the sampled items was found to = $542.50 with...

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j1195059 | Student, Undergraduate | (Level 2) eNoter

Posted November 10, 2010 at 8:28 AM via web

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sample of 50 items sel. The mean value of the sampled items was found to = $542.50 with a sample stand. dev. of $120.56.

(a) what is point estimator for pupulation mean. 9b)construct a 95% confidence interval for the pop. mean. (interpret) (C) construct a 90% confidence interval for the population mean.

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neela | High School Teacher | (Level 3) Valedictorian

Posted November 10, 2010 at 12:59 PM (Answer #1)

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a) The point estimator of the poulation mean M is the sample mean m itself.Therefore, $542.50 is the estimated value of the population mean by point estimation.

b)

If the sample of size n has a standrd devition of s, then the sample mean ha the standard deviation of s/sqrt(n).

We know that z = |M-m|/(s/sqrtn) is a standard normal variate.

Therefore probability that the estimated value of M is within  limits is 95% confidetial limits  is  as good as P(z <= |(M-m)|/ (s/sqrtn)  <  t ) =  95% = 0.95.

Therefore from tables t = 2.74.

Threfore |(M-542.5)|/(120.56/sqrt50) < 2.74.

 542.5 - 2.74(120.56/sqrt50) < = M  < = 542+ 2.74 (120/sqrt50)

542.50-46.72 <=  M <= 542.5+46.72

496.78 <=  M <= 589.22.

Therefore the population mean leis between 496.78  and 589.78 with 95% confidence.

c)

Similar to the b.

P(z = |M-m|/(s/sqrtn) < 0.90

P(z = {|M-542.5|/120.52/sqrt50) < x } = 0.90

x = 1.645 from normal distribution tables.

Therefore  542.5- (1.645)(120.52/sqrt50) < M < 542.5 + (1.645)(120.52/sqrt50)

 542.5 -28.04 <= M < = 542.5 +28.04

514 .46 <=  M <= 570.54.

Therefore the 90% confidence interval for the population mean M is 514.46 to 570.54.

 

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