(a) To find a point estimate for [tex]\(\mu\)[/tex], the true mean sodium content for hot dogs, we first sum the sodium contents of all the hot dogs and then divide by the number of hot dogs to find the mean.
The sodium contents are:
[tex]$
430, 480, 340, 350, 280, 560, 520
$[/tex]
To find the mean:
[tex]\[
\mu = \frac{430 + 480 + 340 + 350 + 280 + 560 + 520}{7}
\][/tex]
Calculating the sum:
[tex]\[
430 + 480 + 340 + 350 + 280 + 560 + 520 = 2960
\][/tex]
Now, divide by the number of data points (7):
[tex]\[
\mu = \frac{2960}{7} = 422.8571
\][/tex]
Therefore, the point estimate of [tex]\(\mu\)[/tex] is:
[tex]\[
\mu = 422.8571
\][/tex]
(b) To find a point estimate for [tex]\(\sigma^2\)[/tex], the variance of sodium content, we use the formula for the sample variance:
[tex]\[
\sigma^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i - \mu)^2
\][/tex]
where [tex]\( x_i \)[/tex] are the individual data points, [tex]\( \mu \)[/tex] is the mean sodium content, and [tex]\( n \)[/tex] is the number of data points.
We have already found [tex]\(\mu = 422.8571\)[/tex].
Now, let's calculate each squared deviation from the mean and sum them up:
[tex]\[
(430 - 422.8571)^2 = 51.3060
\][/tex]
[tex]\[
(480 - 422.8571)^2 = 3260.3060
\][/tex]
[tex]\[
(340 - 422.8571)^2 = 6883.0204
\][/tex]
[tex]\[
(350 - 422.8571)^2 = 5328.7347
\][/tex]
[tex]\[
(280 - 422.8571)^2 = 20516.7347
\][/tex]
[tex]\[
(560 - 422.8571)^2 = 18702.8776
\][/tex]
[tex]\[
(520 - 422.8571)^2 = 9425.8776
\][/tex]
Summing these squared deviations:
[tex]\[
51.3060 + 3260.3060 + 6883.0204 + 5328.7347 + 20516.7347 + 18702.8776 + 9425.8776 = 64169.8570
\][/tex]
Now, divide this sum by [tex]\( n-1 = 6 \)[/tex]:
[tex]\[
\sigma^2 = \frac{64169.8570}{6} = 10690.4762
\][/tex]
So, the point estimate of [tex]\(\sigma^2\)[/tex] is:
[tex]\[
\sigma^2 = 10690.4762
\][/tex]
(c) To find a point estimate for [tex]\(\sigma\)[/tex], the standard deviation of sodium content, take the square root of the variance [tex]\(\sigma^2\)[/tex]:
[tex]\[
\sigma = \sqrt{10690.4762} = 103.3948
\][/tex]
Therefore, the point estimate of [tex]\(\sigma\)[/tex] is:
[tex]\[
\sigma = 103.3948
\][/tex]
Is the statistic you used to produce your estimate of [tex]\(\sigma\)[/tex] unbiased?
Yes, the sample standard deviation is an unbiased estimator of the population standard deviation when we use [tex]\( n-1 \)[/tex] in the variance calculation formula. So the statistic used is indeed unbiased.
