Question

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.

Answer 1

Answer:

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).