The weights of packages of grass seed distributed by a company have a normal dis
ID: 3219522 • Letter: T
Question
The weights of packages of grass seed distributed by a company have a normal distribution with mean mu. and variance sigma^2. A random sample of 10 packages of grass seed is taken and the weights (in decagrams) are: x_1 = 46.4 x_2 = 46.1 x_3 = 45.8 x_4 = 47.0 x_5 = 46.1 x_6 = 45.9 x_7 = 45.8 x_8 = 46.9 x_9 = 45.2 x_10 = 46.0 (a) Give (to the neatest hundredth) an unbiased point estimate for mu. (b) Give (to the nearest thousandth) an unbiased point estimate for sigma^2. (c) Find a 97% confidence interval for the mean weight mu of all such packages of grass seed distributed by this company. (d) Find a 95% confidence interval for the variance sigma^2 of all such packages of grass seed distributed by this company.Explanation / Answer
We know that the sample mean( X) is the unbiased estimator of population mean (µ .
X = 1/n in xi = 461.2/10 = 46.12
Unbiased point estimate for µ = 46.12
Sample standard deviation(s) is the unbiased estimator for population standard deviation ( )
s = [1/(n-1) in (xi - X )2]
s = (2.576/9) = 0.286222 = 0.535
Sample variance = s2 = 0.5352 = 0.286
Unbiased point estimate for 2 = 0.286
97% confidence interval for µ = X ± t/2 (s/n)
= 0.03, d.f = 9 t0.03/2 = invnorm( 0.97 + (1-0.97)/2 ) = 2.1701
46.12 ± 2.1701*(0.535/3)
97% confidence interval for µ = (46.5070, 45.732)
95% confidence interval for 2 = (n-1)s2 / 21-(/2) < 2 < (n-1)s2 / 2(/2)
= 1-0.95 = 0.05
20.975 = 21.920 , 20.025 = 3.816 ( from chi square table)
9*0.286/21.920 < 2 < 9*0.286/3.816
0.1174 < 2 < 0.6745
95% confidence interval for variance = (0.6754, 0.1174)
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