91影视

For each part, values of the sample mean $\left(\stackrel{炉}{x}\right)$, sample standard deviation (s) and degrees of freedom (n) are given.

Given $\stackrel{炉}{x}=21,s=2.5,n=50$

The 90% confidence interval can be calculated using the formula,

$\sqrt{\frac{\left(n-1\right){\mathbf{s}}^{\mathbf{2}}}{{\mathbf{蠂}}_{\frac{\mathbf{伪}}{\mathbf{2}}}^{\mathbf{2}}}}\mathbf{猢�}\mathit{蟽}\mathbf{猢�}\sqrt{\frac{\left(n-1\right){\mathbf{s}}^{\mathbf{2}}}{{\mathbf{蠂}}_{\left(1-\frac{伪}{2}\right)}^{\mathbf{2}}}}$

From the table values, at the 0.10 level of significance and at 49 degrees of freedom, the value for ${蠂}_{\frac{伪}{2}}^2$is 66.3387, and the value for ${蠂}_{\left(1-\frac{伪}{2}\right)}^2$is 33.9303.

Substitute the values to get the required confidence interval.

$$\begin{gathered} \left(\sqrt{\frac{\left(50-1\right){2.5}^2}{66.3387}}猢�蟽猢�\sqrt{\frac{\left(50-1\right){2.5}^2}{33.9303}}\right)=\sqrt{4.6165}猢�蟽猢�\sqrt{9.0259} \\ =2.1486猢�蟽猢�3.0043 \end{gathered}$$

Therefore, the 90% confidence interval for $蟽$is $\left(2.1486,3.0043\right)$.

Given $\stackrel{炉}{x}=1.3,s=0.02,n=15$

The 90% confidence interval can be calculated using the formula,

$\sqrt{\frac{\left(n-1\right){\mathbf{s}}^{\mathbf{2}}}{{\mathbf{蠂}}_{\frac{\mathbf{伪}}{\mathbf{2}}}^{\mathbf{2}}}}\mathbf{猢�}\mathit{蟽}\mathbf{猢�}\sqrt{\frac{\left(n-1\right){\mathbf{s}}^{\mathbf{2}}}{{\mathbf{蠂}}_{\left(1-\frac{伪}{2}\right)}^{\mathbf{2}}}}$

From the table values, at the 0.10 level of significance and at 14 degrees of freedom, the value for ${蠂}_{\frac{伪}{2}}^2$is 23.6848, and the value for ${蠂}_{\left(1-\frac{伪}{2}\right)}^2$is 6.5706.

Substitute the values to get the required confidence interval.

$$\begin{gathered} \left(\sqrt{\frac{\left(15-1\right){0.02}^2}{23.6848}}猢�蟽猢�\sqrt{\frac{\left(14-1\right){0.02}^2}{6.5706}}\right)=\sqrt{0.00024}猢�蟽猢�\sqrt{0.00085} \\ =0.0155猢�蟽猢�0.0292 \end{gathered}$$

Therefore, the 90% confidence interval for $蟽$is$\left(0.0155,0.0292\right)$.

Given$\stackrel{炉}{x}=167,s=31.6,n=22$

The 90% confidence interval can be calculated using the formula,

role="math" localid="1668660233980" $\sqrt{\frac{\left(n-1\right)s^2}{{蠂}_{\frac{伪}{2}}^2}}猢�蟽猢�\sqrt{\frac{\left(n-1\right)s^2}{{蠂}_{\left(1-\frac{伪}{2}\right)}^2}}$

From the table values, at the 0.10 level of significance and at 21 degrees of freedom, the value for ${蠂}_{\frac{伪}{2}}^2$is 32.6706, and the value for ${蠂}_{\left(1-\frac{伪}{2}\right)}^2$is 11.5913.

Substitute the values to get the required confidence interval.

$$\begin{gathered} \left(\sqrt{\frac{\left(22-1\right){31.6}^2}{32.6706}}猢�蟽猢�\sqrt{\frac{\left(22-1\right){31.6}^2}{11.5913}}\right)=\sqrt{641.85}猢�蟽猢�\sqrt{1809.09} \\ =25.3348猢�蟽猢�42.5334 \end{gathered}$$

Therefore, the 90% confidence interval for $蟽$is $\left(25.3348,42.5334\right)$.

Given$\stackrel{炉}{x}=9.4,s=1.5,n=5$

The 90% confidence interval can be calculated using the formula,

$\sqrt{\frac{\left(n-1\right){\mathbf{s}}^{\mathbf{2}}}{{\mathbf{蠂}}_{\frac{\mathbf{伪}}{\mathbf{2}}}^{\mathbf{2}}}}\mathbf{猢�}\mathit{蟽}\mathbf{猢�}\sqrt{\frac{\left(n-1\right){\mathbf{s}}^{\mathbf{2}}}{{\mathbf{蠂}}_{\left(1-\frac{伪}{2}\right)}^{\mathbf{2}}}}$

From the table values, at the 0.10 level of significance and at 4 degrees of freedom, the value for${蠂}_{\frac{伪}{2}}^2$ is 9.4877, and the value for ${蠂}_{\left(1-\frac{伪}{2}\right)}^2$is 0.7107.

Substitute the values to get the required confidence interval.

$$\begin{gathered} \left(\sqrt{\frac{\left(5-1\right){1.5}^2}{9.4877}}猢�蟽猢�\sqrt{\frac{\left(5-1\right){1.5}^2}{0.7107}}\right)=\sqrt{0.9486}猢�蟽猢�\sqrt{12.6632} \\ =0.9740猢�蟽猢�3.5585 \end{gathered}$$

Therefore, the 90% confidence interval for $蟽$is $\left(0.9740,3.5585\right)$.