Chapter 10, Problem 10.79SE

(a)

To determine

To find: 95% confidence interval for the mean diameter of the species.

Answer to Problem 10.79SE

A 95% confidence interval gives the upper and lower limits for $\mu$ is $(169.121,199.879)$

Explanation of Solution

Given:

Specimen	Diameter	Height	D/H
OSU 36651	185	78	2.37
OSU 36652	194	65	2.98
OSU 36653	173	77	2.25
OSU 36654	200	76	2.63
OSU 36655	179	72	2.49
OSU 36656	213	76	2.80
OSU 36657	134	75	1.79
OSU 36658	191	77	2.48
OSU 36659	177	69	2.57
OSU 36660	199	65	3.06
$\overline{X}$ :	184.5	73	2.54
S:	21.5	5	.37

Calculation:

For the Diameter of the Species:

The sample mean and standard deviation for the recorded data are,$\overline{x}=184.5$

  $s=21.5$

If you choose a 5% level of significance ( $\alpha =0.05$ ), the critical value can be found from the values of t from Table 4 of Appendix I. With $df=n-1=9$ the critical value is $t_{\alpha /2}=t_{0.05/2}=2.262$

A 95% confidence interval gives the upper and lower limits for $\mu$ as,$\overline{x}\pm t_{\alpha /2}\left(\frac{s}{\sqrt{n}}\right)$

  $=>184.5\pm 2.262\left(\frac{21.5}{\sqrt{10}}\right)$

  $=>184.5\pm 15.379$

  $=>(169.121,199.879)$

Thus, you can estimate with 95% confidence that the average diameter lies between $169.121$ and $199.879$ .

Conclusion: Thus, a 95% confidence interval gives the upper and lower limits for $\mu$ as $(169.121,199.879)$ .

(b)

To determine

To find: a 95% confidence interval for the mean height of the species.

Answer to Problem 10.79SE

A 95% confidence interval gives the upper and lower limits for $\mu$ is $(69.423,76.577)$ .

Explanation of Solution

Given:

Specimen	Diameter	Height	D/H
OSU 36651	185	78	2.37
OSU 36652	194	65	2.98
OSU 36653	173	77	2.25
OSU 36654	200	76	2.63
OSU 36655	179	72	2.49
OSU 36656	213	76	2.80
OSU 36657	134	75	1.79
OSU 36658	191	77	2.48
OSU 36659	177	69	2.57
OSU 36660	199	65	3.06
$\overline{X}$ :	184.5	73	2.54
S:	21.5	5	.37

Calculation:

For the Height of the Species:

The sample mean and standard deviation for the recorded data are,$\overline{x}=73$

  $s=5$

A 95% confidence interval gives the upper and lower limits for $\mu$ as,$\overline{x}\pm t_{\alpha /2}\left(\frac{s}{\sqrt{n}}\right)$

  $=>73\pm 2.262\left(\frac{5}{\sqrt{10}}\right)$

  $=>73\pm 3.577$

  $=>(69.423,76.577)$

Thus, you can estimate with 95% confidence that the average height lies between $69.423$ and $76.577$ .

Conclusion: Thus, a 95% confidence interval gives the upper and lower limits for $\mu$ as $(69.423,76.577)$ .

(c)

To determine

To find: a 95% confidence interval for the mean ratio of diameter to height.

Answer to Problem 10.79SE

A 95% confidence interval for the mean ratio of diameter to height is $(2.275,2.805)$ .

Explanation of Solution

Given:

Specimen	Diameter	Height	D/H
OSU 36651	185	78	2.37
OSU 36652	194	65	2.98
OSU 36653	173	77	2.25
OSU 36654	200	76	2.63
OSU 36655	179	72	2.49
OSU 36656	213	76	2.80
OSU 36657	134	75	1.79
OSU 36658	191	77	2.48
OSU 36659	177	69	2.57
OSU 36660	199	65	3.06
$\overline{X}$ :	184.5	73	2.54
S:	21.5	5	.37

Calculation:

For the ratio of diameter to height of the species:

The sample mean and standard deviation for the recorded data are,$\overline{x}=2.54$

  $s=0.37$

A 95% confidence interval gives the upper and lower limits for $\mu$ as,$\overline{x}\pm t_{\alpha /2}\left(\frac{s}{\sqrt{n}}\right)$

  $=>2.54\pm 2.262\left(\frac{0.37}{\sqrt{10}}\right)$

  $=>2.54\pm 0.265$

  $=>(2.275,2.805)$

Conclusion: Thus, a 95% confidence interval for the mean ratio of diameter to height is $(2.275,2.805)$ .

(d)

To determine

To compare:the intervals found in parts a, b, and c and explain whether the average of the ratios is same as the ratio of the average diameter to average height

Answer to Problem 10.79SE

The average of the ratios is same as the ratio of the average diameter to average height.

Explanation of Solution

Calculation:

The width of the interval for the mean diameter of the species is wider than the other two intervals because of its larger standard deviation.

  $\frac{\text{Average Diameter}}{\text{Average Height}}=\frac{184.5}{73}=2.53$

  $\text{Average of }\frac{\text{Diameter}}{\text{Height}}=2.54$

Since, the difference is very low we can assume that both are equal.

Conclusion: Thus, the average of the ratios is same as the ratio of the average diameter to average height.