Someone helps? how to solve this problem. Thank u very very much.?

Question

The Nevada Highway Commission wants to estimate the mean weight of commercial vehicles traveling on a particular section of interstate highway. An inspector takes a sample of 64 randomly selected trucks and weighs them. The resulting mean weight is 14.7 tons with a standard deviation of 4.8 tons. Construct a 80, 85, and 90% confidence interval estimate for the true mean weight and interpret your answer in the words of the problem.

Answer 1

ANSWER: 80% CONFIDENCE INTERVAL = [13.9,  15.5],

85% CONFIDENCE INTERVAL = [13.8, 15.6],

90% CONFIDENCE INTERVAL = [13.7, 15.7]

Why???

SMALL SAMPLE, CONFIDENCE INTERVAL, NORMAL POPULATION DISTRIBUTION

x-bar = Sample mean [14.7]

s = Sample standard deviation [4.8]

n = Number of samples [64]

df = degrees of freedom [64 - 1 = 63]

For confidence level of 80%, two-sided interval ("look-up" from Table) "t critical value" is (approx) 1.30

Resulting Confidence Interval for "true mean":

x-bar +/- (t critical value) * s/SQRT(n) = 14.7 +/- 1.30 * 4.8/SQRT(64) = [13.9,  15.5]

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For confidence level of 85%, two-sided interval ("look-up" from Table) "t critical value" is (approx) 1.46

Resulting Confidence Interval for "true mean":

x-bar +/- (t critical value) * s/SQRT(n) = 14.7 +/- 1.46 * 4.8/SQRT(64) = [13.8,  15.6]

- - - - - - - - - - - - - - - - - - - - - - - - - -

For confidence level of 90%, two-sided interval ("look-up" from Table) "t critical value" is (approx) 1.67

Resulting Confidence Interval for "true mean":

x-bar +/- (t critical value) * s/SQRT(n) = 14.7 +/- 1.67 * 4.8/SQRT(64) = [13.7,  15.7]

Answer 2

Number of sample = n = 64

Mean = m = 14.7

Standard deviation = S = 4.8

Standard deviation of the mean "Sm":

Sm = S/[square-root(n)] = 4.8 / [square-root(64)] = 0.6

Consider n=64 is a small sample size, use t-distribution.

Use a t-distribution table or Excel, with degree of freedom = 64-1=63.

For 90% confidence level: a = 0.05

t0.05(63) = 1.669

confidence interval = m-t*Sm ~ m+t*Sm

= 14.7-1.669*0.6 ~ 14.7+1.669*0.6

= 13.70 ~ 15.70

do the 80 & 85% cases in the same way by using a = 0.1 & 0.125, respectively.

Note: if using Excel for the t values, use the equations:

=TINV(0.1,63) for 90% confidence level

=TINV(0.2,63) for 80% confidence level

=TINV(0.15,63) for 85% confidence level

since Excel always gives 2-tailed values for t-distribution.