Thank you for visiting Caviar is an expensive delicacy so companies that package it pay very close attention to the amount of product in their tins Suppose a company. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
We are given a simple random sample (SRS) of 20 tins, with a sample mean of
[tex]$$\overline{x} = 99.8$$[/tex]
grams and a sample standard deviation of
[tex]$$s = 0.9$$[/tex]
grams. Since the sample size is small ([tex]$n=20$[/tex]) and the population standard deviation is unknown, we use the [tex]$t$[/tex]-distribution to construct a 95% confidence interval for the true mean amount of caviar per tin.
The steps are as follows:
1. Determine the Degrees of Freedom:
The degrees of freedom (df) is given by
[tex]$$\text{df} = n - 1 = 20 - 1 = 19.$$[/tex]
2. Find the Critical [tex]$t$[/tex] Value:
For a 95% confidence interval, we need the [tex]$t$[/tex]-value corresponding to a cumulative probability of [tex]$0.975$[/tex] (since 95% confidence level leaves 2.5% in each tail). For 19 degrees of freedom, this critical value is approximately
[tex]$$t^ \approx 2.093.$$[/tex]
3. Calculate the Standard Error (SE):
The standard error of the mean is
[tex]$$\text{SE} = \frac{s}{\sqrt{n}} = \frac{0.9}{\sqrt{20}}.$$[/tex]
4. Compute the Margin of Error (ME):
Multiply the critical [tex]$t$[/tex] value by the standard error:
[tex]$$\text{ME} = t^ \cdot \text{SE} = 2.093 \cdot \frac{0.9}{\sqrt{20}}.$$[/tex]
5. Form the Confidence Interval:
The 95% confidence interval for the mean is given by:
[tex]$$\overline{x} \pm \text{ME} = 99.8 \pm 2.093\left(\frac{0.9}{\sqrt{20}}\right).$$[/tex]
Thus, the correct answer is:
(B) [tex]$$99.8 \pm 2.093\left(\frac{0.9}{\sqrt{20}}\right).$$[/tex]
[tex]$$\overline{x} = 99.8$$[/tex]
grams and a sample standard deviation of
[tex]$$s = 0.9$$[/tex]
grams. Since the sample size is small ([tex]$n=20$[/tex]) and the population standard deviation is unknown, we use the [tex]$t$[/tex]-distribution to construct a 95% confidence interval for the true mean amount of caviar per tin.
The steps are as follows:
1. Determine the Degrees of Freedom:
The degrees of freedom (df) is given by
[tex]$$\text{df} = n - 1 = 20 - 1 = 19.$$[/tex]
2. Find the Critical [tex]$t$[/tex] Value:
For a 95% confidence interval, we need the [tex]$t$[/tex]-value corresponding to a cumulative probability of [tex]$0.975$[/tex] (since 95% confidence level leaves 2.5% in each tail). For 19 degrees of freedom, this critical value is approximately
[tex]$$t^ \approx 2.093.$$[/tex]
3. Calculate the Standard Error (SE):
The standard error of the mean is
[tex]$$\text{SE} = \frac{s}{\sqrt{n}} = \frac{0.9}{\sqrt{20}}.$$[/tex]
4. Compute the Margin of Error (ME):
Multiply the critical [tex]$t$[/tex] value by the standard error:
[tex]$$\text{ME} = t^ \cdot \text{SE} = 2.093 \cdot \frac{0.9}{\sqrt{20}}.$$[/tex]
5. Form the Confidence Interval:
The 95% confidence interval for the mean is given by:
[tex]$$\overline{x} \pm \text{ME} = 99.8 \pm 2.093\left(\frac{0.9}{\sqrt{20}}\right).$$[/tex]
Thus, the correct answer is:
(B) [tex]$$99.8 \pm 2.093\left(\frac{0.9}{\sqrt{20}}\right).$$[/tex]
Thank you for reading the article Caviar is an expensive delicacy so companies that package it pay very close attention to the amount of product in their tins Suppose a company. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
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