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

Step 1
(a)
The value of $\sum x$ is,
$\sum x=13.1+23.3+31.2+38.5+51.3+20.5+22.7=200.6$
Thus, the value of $\sum x \mathrm{is} 200.6$
The value of $\sum y$ is,
$\sum y=2.78+2.18+1.48+1.03+0.75+2.38+2.2=12.8$
Thus, the value of $\sum y \mathrm{is} 12.8$
The value of $\sum x^2$ is,
$\sum x^2={13.1}^2+{23.3}^2+{31.2}^2+{38.5}^2+{51.3}^2+{20.5}^2+{22.7}^2$
$=171.61+542.89+973.44+1482.25+2631.69+420.25+515.29$
$=6737.42$ Thus, the value of $\sum x^2 \mathrm{is} 6737.42$
Thus, the value of $\sum y^2$ is,
$\sum y^2={2.78}^2+{2.18}^2+{1.48}^2+{1.03}^2+{0.75}^2+{2.38}^2+{2.2}^2$
$=7.7284+4.7524+2.1904+1.0609+0.5625+5.6644+4.84$
$=26.7990$
Thus, the value of $\sum y^2 \mathrm{is} 26.7990.$
The value of $\sum xy$ is,
$\sum xy=\left(13.1\times 2.78\right)+\left(23.3\times 2.18\right)+\left(31.2\times 1.48\right)+\left(38.5\times 1.03\right)+\left(51.3\times 0.75\right)+\left(20.5\times 2.38\right)+\left(22.7\times 2.2\right)$
$=36.418+50.794+46.176+39.655+38.475+48.790+49.940$

$=310.248$
Thus, the value of $\sum xy \mathrm{is} 310.248.$
The value of $r$ is,
$r=\frac{n\sum xy-\left(\sum x\right)\left(\sum y\right)}{\sqrt{\left[n\sum x^2-{\left(\sum x\right)}^2\right]\left[\sum y^2-{\left(\sum y\right)}^2\right]}}$
$=\frac{7\left(310.248\right)-\left(200.6\times 12.8\right)}{\sqrt{\left[\left(7\times 6737.42\right)-{\left(200.6\right)}^2\right]\left[\left(7\times 26.7990\right)-{\left(12.8\right)}^2\right]}}$

$=\frac{-395.944}{405.4729}$
$=-0.976$

Hence, the correlation coefficient $r$ is $-0.976.$
Step 2
(b)
The hypothesis is,
Null hypothesis:
$H_0:\rho =0$Alternative hypothesis:
$H_1:\rho <0$The test statistic is,
$t=-r\sqrt{\frac{n-2}{1-r^2}}$

$=-0.976\sqrt{\frac{7-2}{1-{\left(-0.976\right)}^2}}$
$=\frac{-2.1824}{0.2178}$

$=-10.02$
Thus, the test statistic is $-10.02.$
The degrees of freedom is,
$df=n-2$
$=7-2$
$=5$
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 $-3.36.$
Thus, the critical value is $-3.36.$
Conclusion:
The test statistic is $-10.02$ less than critical value $-3.36.$
Based on the decision rule, reject the null hypothesis.
Reject the null hypothesis. There is sufficient evidence that $\rho <0.$
Step 3
c)
The value of $\overline{x}$ is,
$\overline{x}=\frac{\sum x}{n}$
$=\frac{200.6}{7}$
$=28.6571$
The value of  $\overline{y}$ is,
$\overline{y}=\frac{\sum y}{n}$
$=\frac{12.8}{7}$
$=1.8286$
The value of b is,
$b=\frac{n\sum xy-\left(\sum x\right)\left(\sum y\right)}{n\sum x^2-{\left(\sum x\right)}^2}$
$=\frac{7\left(310.248\right)-\left(200.6\times 12.8\right)}{\left(7\times 6737.42\right)-{\left(200.6\right)}^2}$
$=-\frac{395.944}{6921.58}$
$=-0.0572$
Thus, the value of b is $-0.0572.$
The value of a is,
$a=\overline{y}-b\overline{x}$
$=1.8286-\left(-0.0572\times 28.6571\right)$
$=1.8286+1.6392$
$=3.4678$
Thus, the value of a is 3.4678.
The value of $S_e$is,
$S_e=\sqrt{\frac{\sum y^2-a\sum y-b\sum xy}{n-2}}$
$=\sqrt{\frac{26.7990-\left(3.4678\times 12.8\right)-\left(-0.0572\times 310.248\right)}{7-2}}$
$=\sqrt{\frac{0.1573456}{5}}$
$=0.1774$
Hence, the value of $S_e$ is $0.1776.$