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]