Question

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 98% confidence interval estimate of the mean amount of mercury in the population. 0.57 0.71 0.10 0.93 1.24 0.55 0.82 what is the confidence interval estimate of the population mean μ? □ ppm < μ < □ ppm (round to three decimal places as needed.)

Explanation:

Step1: Calculate the sample mean ($\bar{x}$)

First, sum up all the data points: $0.57 + 0.71 + 0.10 + 0.93 + 1.24 + 0.55 + 0.82 = 4.92$
There are $n = 7$ data points. So, the sample mean is $\bar{x}=\frac{4.92}{7}\approx0.7029$

Step2: Calculate the sample standard deviation ($s$)

First, find the squared differences from the mean:

• $(0.57 - 0.7029)^2\approx0.0177$
• $(0.71 - 0.7029)^2\approx0.0001$
• $(0.10 - 0.7029)^2\approx0.3635$
• $(0.93 - 0.7029)^2\approx0.0516$
• $(1.24 - 0.7029)^2\approx0.2885$
• $(0.55 - 0.7029)^2\approx0.0234$
• $(0.82 - 0.7029)^2\approx0.0137$

Sum these squared differences: $0.0177 + 0.0001 + 0.3635 + 0.0516 + 0.2885 + 0.0234 + 0.0137 = 0.7585$

The sample variance is $s^2=\frac{0.7585}{n - 1}=\frac{0.7585}{6}\approx0.1264$
So, the sample standard deviation is $s=\sqrt{0.1264}\approx0.3555$

Step3: Determine the t - critical value

For a 98% confidence interval and $n - 1 = 6$ degrees of freedom, the t - critical value ($t_{\alpha/2}$) from the t - distribution table is approximately $t_{0.01,6}=3.143$

Step4: Calculate the margin of error ($E$)

The formula for the margin of error is $E = t_{\alpha/2}\times\frac{s}{\sqrt{n}}$
Substitute the values: $E = 3.143\times\frac{0.3555}{\sqrt{7}}\approx3.143\times\frac{0.3555}{2.6458}\approx3.143\times0.1344\approx0.422$

Step5: Calculate the confidence interval

The confidence interval is $\bar{x}-E < \mu < \bar{x}+E$
Substitute the values: $0.7029 - 0.422 < \mu < 0.7029 + 0.422$
Which simplifies to $0.281 < \mu < 1.125$ (rounded to three decimal places)

Answer:

$0.281$ ppm $<\mu < 1.125$ ppm