What is the optimal time for a scuba diver to be on the bottom of the ocean?

permaneceerc 2021-02-12 Answered

What is the optimal time for a scuba diver to be on the bottom of the ocean? That depends on the depth of the dive. The U.S. Navy has done a lot of research on this topic. The Navy defines the "optimal time" to be the time at each depth for the best balance between length of work period and decompression time after surfacing. Let $x=$ depth of dive in meters, and let $y=$ optimal time in hours. A random sample of divers gave the following data.
$\begin{array}{|cccccccc|}\hline x& 13.1& 23.3& 31.2& 38.3& 51.3& 20.5& 22.7\\ y& 2.78& 2.18& 1.48& 1.03& 0.75& 2.38& 2.20\\ \hline\end{array}$
(a)
Find $\Sigma x,\Sigma y,\Sigma {x}^{2},\Sigma {y}^{2},\Sigma xy,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}r$. (Round r to three decimal places.)
$\Sigma x=$
$\Sigma y=$
$\Sigma {x}^{2}=$
$\Sigma {y}^{2}=$
$\Sigma xy=$
$r=$
(b)
Use a $1\mathrm{%}$ level of significance to test the claim that $\rho <0$. (Round your answers to two decimal places.)
$t=$
critical $t=$
Conclusion
Reject the null hypothesis. There is sufficient evidence that $\rho <0$.Reject the null hypothesis. There is insufficient evidence that $\rho <0$.
Fail to reject the null hypothesis. There is sufficient evidence that $\rho <0$.Fail to reject the null hypothesis. There is insufficient evidence that $\rho <0.$
(c)
Find ${S}_{e},a,\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}b$. (Round your answers to four decimal places.)
${S}_{e}=$
$a=$
$b=$

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Expert Answer

tafzijdeq
Answered 2021-02-13 Author has 92 answers

Step 1
(a)
The value of $$\displaystyle\sum{x}$$ is,
$$\displaystyle\sum{x}={13.1}+{23.3}+{31.2}+{38.5}+{51.3}+{20.5}+{22.7}={200.6}$$
Thus, the value of $$\displaystyle\sum{x}\ {i}{s}\ {200.6}$$
The value of $$\displaystyle\sum{y}$$ is,
$$\displaystyle\sum{y}={2.78}+{2.18}+{1.48}+{1.03}+{0.75}+{2.38}+{2.2}={12.8}$$
Thus, the value of $$\displaystyle\sum{y}\ {i}{s}\ {12.8}$$
The value of $$\displaystyle\sum{x}^{2}$$ is,
$$\displaystyle\sum{x}^{2}={13.1}^{2}+{23.3}^{2}+{31.2}^{2}+{38.5}^{2}+{51.3}^{2}+{20.5}^{2}+{22.7}^{2}$$
$$\displaystyle={171.61}+{542.89}+{973.44}+{1482.25}+{2631.69}+{420.25}+{515.29}$$
$$\displaystyle={6737.42}$$ Thus, the value of $$\displaystyle\sum{x}^{2}\ {i}{s}\ {6737.42}$$
Thus, the value of $$\displaystyle\sum{y}^{2}$$ is,
$$\displaystyle\sum{y}^{2}={2.78}^{2}+{2.18}^{2}+{1.48}^{2}+{1.03}^{2}+{0.75}^{2}+{2.38}^{2}+{2.2}^{2}$$
$$\displaystyle={7.7284}+{4.7524}+{2.1904}+{1.0609}+{0.5625}+{5.6644}+{4.84}$$
$$\displaystyle={26.7990}$$
Thus, the value of $$\displaystyle\sum{y}^{2}\ {i}{s}\ {26.7990}.$$
The value of $$\displaystyle\sum{x}{y}$$ is,
$$\displaystyle\sum{x}{y}={\left({13.1}\times{2.78}\right)}+{\left({23.3}\times{2.18}\right)}+{\left({31.2}\times{1.48}\right)}+{\left({38.5}\times{1.03}\right)}+{\left({51.3}\times{0.75}\right)}+{\left({20.5}\times{2.38}\right)}+{\left({22.7}\times{2.2}\right)}$$
$$\displaystyle={36.418}+{50.794}+{46.176}+{39.655}+{38.475}+{48.790}+{49.940}$$
$$\displaystyle={310.248}$$
Thus, the value of $$\displaystyle\sum{x}{y}\ {i}{s}\ {310.248}.$$
The value of r is,
$$\displaystyle{r}=\frac{{{n}\sum{x}{y}-{\left(\sum{x}\right)}{\left(\sum{y}\right)}}}{\sqrt{{{\left[{n}\sum{x}^{2}-{\left(\sum{x}\right)}^{2}\right]}{\left[\sum{y}^{2}-{\left(\sum{y}\right)}^{2}\right]}}}}$$
$$\displaystyle=\frac{{{7}{\left({310.248}\right)}-{\left({200.6}\times{12.8}\right)}}}{{\sqrt{{{\left[{\left({7}\times{6737.42}\right)}-{\left({200.6}\right)}^{2}\right]}{\left[{\left({7}\times{26.7990}\right)}-{\left({12.8}\right)}^{2}\right]}}}}}$$
$$\displaystyle=\frac{{-{395.944}}}{{405.4729}}$$
$$\displaystyle=-{0.976}$$
Hence, the correlation coefficient r is $$\displaystyle-{0.976}.$$
Step 2
(b)
The hypothesis is,
Null hypothesis:
$$\displaystyle{H}_{{0}}:\rho={0}$$
Alternative hypothesis:
$$\displaystyle{H}_{{1}}:\rho<{0}$$
The test statistic is,
$$\displaystyle{t}=-{r}\sqrt{{\frac{{{n}-{2}}}{{{1}-{r}^{2}}}}}$$
$$\displaystyle=-{0.976}\sqrt{{\frac{{{7}-{2}}}{{{1}-{\left(-{0.976}\right)}^{2}}}}}$$
$$\displaystyle=\frac{{-{2.1824}}}{{0.2178}}$$
$$\displaystyle=-{10.02}$$
Thus, the test statistic is $$\displaystyle-{10.02}.$$
The degrees of freedom is,
$$\displaystyle{d}{f}={n}-{2}$$
$$\displaystyle={7}-{2}$$
$$\displaystyle={5}$$
The degrees of freedom is 5.
Decision rules for correlation coefficient:
If the test statistic is less than the negative critical value, then reject the null hypothesis.
Computation of critical value:
The critical value of t-distribution for 0.01 level of significance at 5 degrees of freedom can be obtained using the excel formula “=T.INV(0.01,5)”. The critical value is $$\displaystyle-{3.36}.$$
Thus, the critical value is $$\displaystyle-{3.36}.$$
Conclusion:
The test statistic is $$\displaystyle-{10.02}$$ less than critical value $$\displaystyle-{3.36}.$$
Based on the decision rule, reject the null hypothesis.
Reject the null hypothesis. There is sufficient evidence that $$\displaystyle\rho<{0}.$$
Step 3
c)
The value of $$\displaystyle\overline{{x}}$$ is,
$$\displaystyle\overline{{x}}=\frac{{\sum{x}}}{{n}}$$
$$\displaystyle=\frac{200.6}{{7}}$$
$$\displaystyle={28.6571}$$
The value of $$\displaystyle\overline{{y}}$$ is,
$$\displaystyle\overline{{y}}=\frac{{\sum{y}}}{{n}}$$
$$\displaystyle=\frac{12.8}{{7}}$$
$$\displaystyle={1.8286}$$
The value of b is,
$$\displaystyle{b}=\frac{{{n}\sum{x}{y}-{\left(\sum{x}\right)}{\left(\sum{y}\right)}}}{{{n}\sum{x}^{2}-{\left(\sum{x}\right)}^{2}}}$$
$$\displaystyle=\frac{{{7}{\left({310.248}\right)}-{\left({200.6}\times{12.8}\right)}}}{{{\left({7}\times{6737.42}\right)}-{\left({200.6}\right)}^{2}}}$$
$$\displaystyle=-\frac{395.944}{{6921.58}}$$
$$\displaystyle=-{0.0572}$$
Thus, the value of b is $$\displaystyle-{0.0572}.$$
The value of a is,
$$\displaystyle{a}=\overline{{y}}-{b}\overline{{x}}$$
$$\displaystyle={1.8286}-{\left(-{0.0572}\times{28.6571}\right)}$$
$$\displaystyle={1.8286}+{1.6392}$$
$$\displaystyle={3.4678}$$
Thus, the value of a is 3.4678.
The value of $$\displaystyle{S}_{{e}}$$ is,
$$\displaystyle{S}_{{e}}=\sqrt{{\frac{{\sum{y}^{2}-{a}\sum{y}-{b}\sum{x}{y}}}{{{n}-{2}}}}}$$
$$\displaystyle=\sqrt{{\frac{{{26.7990}-{\left({3.4678}\times{12.8}\right)}-{\left(-{0.0572}\times{310.248}\right)}}}{{{7}-{2}}}}}$$
$$\displaystyle=\sqrt{{\frac{{{0.1573456}}}{{5}}}}$$
$$\displaystyle={0.1774}$$
Hence, the value of $$\displaystyle{S}_{{e}}$$ is 0.1776.

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