Mathematics, 19.02.2020 06:01 avaryan84

1. A food safety guideline is that the mercury in fish should be below 1 part per million (ppm). Listed below are the amounts of mercury (ppm) found in tuna sushi sampled at different stores in a major city. Construct a 99% confidence interval estimate of the mean amount of mercury in the population. Does it appear that there is too much mercury in tuna sushi?a. 0.58.b. 0.82.c. 0.10.d. 0.98e. 1.27 f. 0.56.g. 0.96.2. What is the confidence interval estimate of the population mean μ? Does it appear that there is too much mercury in tuna sushi? A. No, because it is possible that the mean is not greater than 1 ppm. Also, at least one of the sample values is less than 1 ppm, so at least some of the fish are safe. B. No, because it is not possible that the mean is greater than 1 ppm. Also, at least one of the sample values is less than 1 ppm, so at least some of the fish are safe. C. Yes, because it is possible that the mean is not greater than 1 ppm. Also, at least one of the sample values exceeds 1 ppm, so at least some of the fish have safe.

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Answer from: jholland03

Using the t-distribution, we get that:

The 99% confidence interval estimate of the mean amount of mercury in the population is (0.22 ppm, 1.28 ppm).A. No, because it is possible that the mean is not greater than 1 ppm. Also, at least one of the sample values is less than 1 ppm, so at least some of the fish are safe.

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The observations are: {0.58, 0.82, 0.1, 0.98, 1.27, 0.56, 0.96}

Before building the confidence interval, we need to find the sample mean and the sample standard deviation.

$\overline{x} = \frac{0.58 + 0.82 + 0.1 + 0.98 + 1.27 + 0.56 + 0.96}{7} = 0.75$

$s = \sqrt{\frac{(0.58 - 0.75)^2 + (0.82 - 0.75)^2 + (0.1 - 0.75)^2 + (0.98 - 0.75)^2 + (1.27 - 0.75)^2 + (0.56 - 0.75)^2 + (0.96- 0.75)^2}{6}} = 0.3785$

Now, we have the standard deviation for the sample, thus the t-distribution is used.

The first step to solve this problem is finding how many degrees of freedom,which is the sample size subtracted by 1. So

df = 7 - 1 = 6

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 6 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$ . So we have T = 3.71.

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 3.71\frac{0.3785}{\sqrt{7}} = 0.53$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 0.75 - 0.53 = 0.22 ppm

The upper end of the interval is the sample mean added to M. So it is 0.75 + 0.53 = 1.28 ppm.

The 99% confidence interval estimate of the mean amount of mercury in the population is (0.22 ppm, 1.28 ppm).

From this, the correct option for the interpretation is:

A. No, because it is possible that the mean is not greater than 1 ppm. Also, at least one of the sample values is less than 1 ppm, so at least some of the fish are safe.

A similar problem is given at link

1. A food safety guideline is that the mercury in fish should be below 1 part per million (ppm). Lis

Answer from: studentstudies9816

The confidence interval estimate of the population mean is :

(0.61 ppm, 0.90 ppm)

The correct option is (A).

Step-by-step explanation:

The amounts of mercury (ppm) found in tuna sushi sampled at different stores in a major city are:

S = {0.58, 0.82, 0.10, 0.98, 1.27, 0.56, 0.96}

A $(100-\alpha )\%$ confidence interval for the population mean (μ) is an interval estimate of the true value of the mean. This interval has a $(100-\alpha )\%$ probability of consisting the true value of mean.

⇒ Since the population standard deviation is not provided we will use the t-distribution to construct the 99% confidence interval for mean.

⇒ The formula for confidence interval for the population mean is:

$\bar x\pm t_{\alpha/2,(n-1)}\times \frac{s}{\sqrt{n}}$

Here,

$\bar x$ = sample mean

s = sample standard deviation

n = sample size

$t_{\alpha/2,(n-1)}$ = critical value.

The degrees of freedom for the critical value is, (n - 1) = 7 - 1 = 6.

The significance level is: $\alpha =1-Confidence\ level=1-0.99=0.01$

The critical value is:

$t_{\alpha/2,(n-1)}=t_{0.01/2, 7}=t_{0.05,7}=3.143$

**Use the t-table for the critical value.

Compute the sample mean and sample standard deviation as follows:

$\bar x=\frac{1}{7}(0.58+ 0.82+ 0.10+ 0.98+ 1.27+ 0.56+ 0.96) =0.753$

$\int\limits^a_b {x} \, dx s=\sqrt{\frac{\sum (x{i}-\bar x)^{2}}{n-1} } =\sqrt{\frac{1}{6} \times 0.859743} =0.379$

The 99% confidence interval for μ is:

$x^{2} CI=0.753\pm 3.143\times\frac{0.379}{\sqrt{7}} \\=0.753\pm0.143\\=(0.61, 0.896)\\\approx(0.61, 0.90)$

The confidence interval estimate of the population mean is:

(0.61 ppm, 0.90 ppm)

The upper and lower limit of the 99% confidence interval indicates that the true mean value is less than 1 ppm. This implies that there is not too much mercury in tuna sushi

Because it is possible that the mean is not greater than 1 ppm. Also, at least one of the sample values is less than 1 ppm, so at least some of the fish are safe.

Thus, the correct option is (A).

Answer from: Quest

about 5.6 days

step-by-step explanation:

we can fill the given information into the decay equation and solve for the value of t at which half the sample remains.

$\dfrac{11}{2} =11e^{-0.1247t} \qquad\text{decay equation with values filled in}\\\\\dfrac{1}{2}=e^{-0.1247t} \qquad\text{divide by 11}\\\\-\ln{(2)}=-0.1247t \qquad\text{take the natural log}\\\\\dfrac{\ln{(2)}}{0.1247}=t\approx 5.559\approx 5.6\qquad\text{divide by -0.1247}$

the half-life is about 5.6 days.

Answer from: Quest

Suredude why not right

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