An electric scale gives a reading equal to the true weight plus a random error t
ID: 3362206 • Letter: A
Question
An electric scale gives a reading equal to the true weight plus a random error that is normally distributed with rnean 0 and variance 2 mg2 Suppose that the results of five successive weighings of the same object are as follows: 3.142, 3.163, 3.155, 3.150, 3.141. Assume that 2- 0.01. (a) Calculate a 95% confidence interval for , the true weight of the object. (b) Calculate the minimum sample size required for a 90% confidence interval for to have width no greater than 0.1 mg. (c) Calculate the sample standard deviation for the observed data. d) A researcher believes that the known variance assumption is wrong Use the sample standard deviation to construct (i) an appropriate 95% confidence interval and (ii) a 90% upper confidence bound for .Explanation / Answer
The mean of the five successive weights = (3.142 + 3.163 + 3.155 + 3.150 + 3.141)/5 = 3.1502
A) At 95% cinfidence interval the critical value is 1.96
Cinfidence interval is
mean +/- z* * SD/sqrt (n )
= 3.1502 +/- 1.96 * 0.1/sqrt(5)
= 3.1502 +/- 0.0877
= 3.0625, 3.2379
B) at 90% cinfidence interval the critical value is 1.645
Margin of error = 0.1
Or, z* * SD/sqrt (n ) = 0.1
Or, 1.645 * 0.1/sqrt(n) = 0.1
Or, sqrt(n) = 1.645
Or, n = 2.7 = 3
D) i) sample standard deviation (s) = sqrt(((3.142 - 3.1502)^2 + (3.163 - 3.1502)^2 + (3.155 - 3.1502)^2 + (3.150 - 3.1502)^2 + (3.141 - 3.1502)^2)/4) = 0.0092
At 4 degrees of freedom and 95% cinfidence interval the critical value is 2.776
Confidence interval is
Mean +/- t* * s/sqrt (5)
= 3.1502 +/- 2.776 * 0.0092/sqrt(5)
= 3.1502 +/- 0.0114
= 3.1388, 3.1616
ii) with 4 degrees of freedom and 90% cinfidence interval the critical value is 2.132
The upper Confidence bound is
Mean + t* * s/sqrt (5)
= 3.1502 + 2.132 * 0.0092/sqrt(5)
= 3.1502 + 0.0088
= 3.159
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.