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Given below are the sodium contents (in mg) for seven brands of hot dogs rated as "very good."

[tex]\[ 430 \quad 480 \quad 340 \quad 350 \quad 280 \quad 560 \quad 520 \][/tex]

(a) Use the given data to produce a point estimate of [tex]\(\mu\)[/tex], the true mean sodium content for hot dogs. (Round your answer to four decimal places.)

[tex]\[ \mu = 422.8571 \][/tex]

(b) Use the given data to produce a point estimate of [tex]\(\sigma^2\)[/tex], the variance of sodium content for hot dogs. (Round your answer to four decimal places.)

[tex]\[ \sigma^2 = \boxed{} \][/tex]

(c) Use the given data to produce an estimate of [tex]\(\sigma\)[/tex], the standard deviation of sodium content. (Round your answer to four decimal places.)

[tex]\[ \sigma = \boxed{} \][/tex]

Is the statistic you used to produce your estimate of [tex]\(\sigma\)[/tex] unbiased? (Hint: See the discussion following Example 9.3.)

- Yes
- No

Sagot :

GinnyAnsweridn
(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.
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