Respuesta :
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 [tex](100-\alpha )\%[/tex] confidence interval for the population mean (μ) is an interval estimate of the true value of the mean. This interval has a [tex](100-\alpha )\%[/tex] 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:
[tex]\bar x\pm t_{\alpha/2,(n-1)}\times \frac{s}{\sqrt{n}}[/tex]
Here,
[tex]\bar x[/tex] = sample mean
s = sample standard deviation
n = sample size
[tex]t_{\alpha/2,(n-1)}[/tex] = critical value.
The degrees of freedom for the critical value is, (n - 1) = 7 - 1 = 6.
The significance level is: [tex]\alpha =1-Confidence\ level=1-0.99=0.01[/tex]
The critical value is:
[tex]t_{\alpha/2,(n-1)}=t_{0.01/2, 7}=t_{0.05,7}=3.143[/tex]
**Use the t-table for the critical value.
Compute the sample mean and sample standard deviation as follows:
[tex]\bar x=\frac{1}{7}(0.58+ 0.82+ 0.10+ 0.98+ 1.27+ 0.56+ 0.96) =0.753[/tex]
[tex]\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[/tex]
The 99% confidence interval for μ is:
[tex]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)[/tex]
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).
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.
[tex]\overline{x} = \frac{0.58 + 0.82 + 0.1 + 0.98 + 1.27 + 0.56 + 0.96}{7} = 0.75[/tex]
[tex]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[/tex]
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 [tex]1 - \frac{1 - 0.99}{2} = 0.995[/tex]. So we have T = 3.71.
The margin of error is:
[tex]M = T\frac{s}{\sqrt{n}} = 3.71\frac{0.3785}{\sqrt{7}} = 0.53[/tex]
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 https://brainly.com/question/24232455
