91Ó°ÊÓ

The data in the Table are the result of a random survey of $39$national flags (with replacement between picks) from various countries.

$\begin{array}{|cc|}\hline \mathit{X}& \text{Freq.}\\ 1& 1\\ 2& 7\\ 3& 18\\ 4& 7\\ 5& 6\\ \hline\end{array}$

If $\stackrel{¯}{X}$and $s$are the mean and the standard deviation of the random sample from a normal distribution with unknown variance $100(1-\mathrm{Î\pm })\%$confidence interval is given by

$\stackrel{¯}{x}-t_{\frac{\mathrm{Î\pm }}{2},n-1}\frac{s}{\sqrt{n}}≤\mathrm{μ}≤\stackrel{¯}{x}+t_{\frac{\mathrm{Î\pm }}{2},n-1}\frac{s}{\sqrt{n}}$

where $t_{\frac{\mathrm{Î\pm }}{2},n-1}$is the upper $100\frac{\mathrm{Î\pm }}{2}$percentage point of the $t$distribution with $n-1$degrees of freedom.

The simple mean is

localid="1650538313491" $$\begin{gathered} \stackrel{¯}{x}=\frac{1}{n}\underset{i=1}{\overset{n}{∑}}{x}_i \\ =\frac{1×1+7×2+18×3+7×4+6×5}{39} \\ =3.26 \end{gathered}$$

And standard deviation is

localid="1650538384998" $$\begin{gathered} s={\left(\frac{{∑}_{i=1}^n{\left(x_i-\stackrel{¯}{x}\right)}^2}{n-1}\right)}^{\frac{1}{2}} \\ ={\left(\frac{{∑}_{i=1}^{39}{\left(x_i-3.26\right)}^2}{38}\right)}^{\frac{1}{2}} \\ =1.02 \end{gathered}$$

Now, we need to find a localid="1650538417579" $95\%\mathrm{Cl}$on the population mean, then

localid="1650538452918" $$\begin{gathered} \frac{\mathrm{Î\pm }}{2}=\frac{1-0.95}{2} \\ =0.025 \\ ⇒t_{\frac{\mathrm{Î\pm }}{2},n-1} \\ =t_{0.025,38} \\ =2.03 \end{gathered}$$

We used a probability table for the Student's $t$-distribution to find the value of $t$. The table gives $t$-scores that correspond to degrees of freedom (row) and the confidence level (column). The $t$-score is found where the row and column intersect in the table.

$7.8-2.03\frac{6.22}{\sqrt{5}}≤\mathrm{μ}≤7.8+2.03\frac{6.22}{\sqrt{5}}$

$7.8-0.33≤\mathrm{μ}≤7.8+0.33$

Therefore,

Lower limit is$2.92$.

The data in the Table are the result of a random survey of $39$national flags (with replacement between picks) from various countries.

$\begin{array}{|cc|}\hline \mathit{X}& \text{Freq.}\\ 1& 1\\ 2& 7\\ 3& 18\\ 4& 7\\ 5& 6\\ \hline\end{array}$

If $\stackrel{¯}{X}$and $s$are the mean and the standard deviation of the random sample from a normal distribution with unknown variance $100(1-\mathrm{Î\pm })\%$confidence interval is given by

$\stackrel{¯}{x}-t_{\frac{\mathrm{Î\pm }}{2},n-1}\frac{s}{\sqrt{n}}≤\mathrm{μ}≤\stackrel{¯}{x}+t_{\frac{\mathrm{Î\pm }}{2},n-1}\frac{s}{\sqrt{n}}$

where $t_{\frac{\mathrm{Î\pm }}{2},n-1}$is the upper $100\frac{\mathrm{Î\pm }}{2}$percentage point of the $t$distribution with $n-1$degrees of freedom.

The simple mean is

localid="1650538592407" $$\begin{gathered} \stackrel{¯}{x}=\frac{1}{n}\underset{i=1}{\overset{n}{∑}}{x}_i \\ =\frac{1×1+7×2+18×3+7×4+6×5}{39} \\ =3.26 \end{gathered}$$

And standard deviation is

localid="1650538944116" $$\begin{gathered} s={\left(\frac{{∑}_{i=1}^n{\left(x_i-\stackrel{¯}{x}\right)}^2}{n-1}\right)}^{\frac{1}{2}} \\ ={\left(\frac{{∑}_{i=1}^{39}{\left(x_i-3.26\right)}^2}{38}\right)}^{\frac{1}{2}} \\ =1.02 \end{gathered}$$

Now, we need to find a $95\%\mathrm{Cl}$on the population mean, then

$\frac{\mathrm{Î\pm }}{2}=\frac{1-0.95}{2}=0.025⇒t_{\frac{\mathrm{Î\pm }}{2},n-1}=t_{0.025,38}=2.03$

We used a probability table for the Student's $t$-distribution to find the value of $t$. The table gives $t$-scores that correspond to degrees of freedom (row) and the confidence level (column). The $t$-score is found where the row and column intersect in the table.

$7.8-2.03\frac{6.22}{\sqrt{5}}≤\mathrm{μ}≤7.8+2.03\frac{6.22}{\sqrt{5}}$

$7.8-0.33≤\mathrm{μ}≤7.8+0.33$

Therefore,

Upper limit is$3.59$.

The data in the Table are the result of a random survey of $39$national flags (with replacement between picks) from various countries.

$\begin{array}{|cc|}\hline \mathit{X}& \text{Freq.}\\ 1& 1\\ 2& 7\\ 3& 18\\ 4& 7\\ 5& 6\\ \hline\end{array}$

If $\stackrel{¯}{X}$and $s$are the mean and the standard deviation of the random sample from a normal distribution with unknown variance $100(1-\mathrm{Î\pm })\%$confidence interval is given by

$\stackrel{¯}{x}-t_{\frac{\mathrm{Î\pm }}{2},n-1}\frac{s}{\sqrt{n}}≤\mathrm{μ}≤\stackrel{¯}{x}+t_{\frac{\mathrm{Î\pm }}{2},n-1}\frac{s}{\sqrt{n}}$

where $t_{\frac{\mathrm{Î\pm }}{2},n-1}$is the upper $100\frac{\mathrm{Î\pm }}{2}$percentage point of the $t$distribution with $n-1$degrees of freedom.

The simple mean is

localid="1650539042896" $$\begin{gathered} \stackrel{¯}{x}=\frac{1}{n}\underset{i=1}{\overset{n}{∑}}{x}_i \\ =\frac{1×1+7×2+18×3+7×4+6×5}{39} \\ =3.26 \end{gathered}$$

And standard deviation is

localid="1650539055897" $$\begin{gathered} s={\left(\frac{{∑}_{i=1}^n{\left(x_i-\stackrel{¯}{x}\right)}^2}{n-1}\right)}^{\frac{1}{2}} \\ ={\left(\frac{{∑}_{i=1}^{39}{\left(x_i-3.26\right)}^2}{38}\right)}^{\frac{1}{2}} \\ =1.02 \end{gathered}$$

Now, we need to find a $95\%\mathrm{Cl}$on the population mean, then

$\frac{\mathrm{Î\pm }}{2}=\frac{1-0.95}{2}=0.025⇒t_{\frac{\mathrm{Î\pm }}{2},n-1}=t_{0.025,38}=2.03$

We used a probability table for the Student's $t$-distribution to find the value of $t$. The table gives $t$-scores that correspond to degrees of freedom (row) and the confidence level (column). The $t$-score is found where the row and column intersect in the table.

$7.8-2.03\frac{6.22}{\sqrt{5}}≤\mathrm{μ}≤7.8+2.03\frac{6.22}{\sqrt{5}}$

$7.8-0.33≤\mathrm{μ}≤7.8+0.33$

Therefore,

Error bound is $0.33$.