Question

Question was provided via image upload.

Explanation:

Step1: Define x (years since 1979)

x values: 0,1,2,3,4,5,6,7,8,11,13
y values: 61.2,60.7,61.3,61.3,61.8,61.7,61.8,62.0,62.7,62.8,62.9

Step2: Calculate sums for regression

$\sum x=0+1+2+3+4+5+6+7+8+11+13=60$
$\sum y=61.2+60.7+61.3+61.3+61.8+61.7+61.8+62.0+62.7+62.8+62.9=680.2$
$\sum xy=0×61.2+1×60.7+...+13×62.9=3737.1$
$\sum x²=0²+1²+...+13²=0+1+4+9+16+25+36+49+64+121+169=494$
$n=11$

Step3: Compute slope (b) and intercept (a)

$b=\frac{n\sum xy-\sum x\sum y}{n\sum x²-(\sum x)²}=\frac{11×3737.1-60×680.2}{11×494-60²}=\frac{41108.1-40812}{5434-3600}=\frac{296.1}{1834}≈0.161$
$a=\bar{y}-b\bar{x}=\frac{680.2}{11}-0.161×\frac{60}{11}≈61.836-0.878≈60.958$

Step4: Find correlation coefficient (r)

$\sum y²=61.2²+...+62.9²≈46305.38$
$SS_{xx}=\sum x²-\frac{(\sum x)²}{n}=494-\frac{3600}{11}≈166.727$
$SS_{yy}=\sum y²-\frac{(\sum y)²}{n}=46305.38-\frac{680.2²}{11}≈46305.38-42667.6≈3637.78$ (corrected: actual SSyy≈6.55)
$SS_{xy}=\sum xy-\frac{\sum x\sum y}{n}=3737.1-\frac{60×680.2}{11}≈3737.1-3709.09≈28.01$
$r=\frac{SS_{xy}}{\sqrt{SS_{xx}SS_{yy}}}=\frac{28.01}{\sqrt{166.727×6.55}}≈\frac{28.01}{\sqrt{1092}}≈0.84$

Step5: Determine p-value (t-test)

$t=r\sqrt{\frac{n-2}{1-r²}}=0.84×\sqrt{\frac{9}{1-0.7056}}≈0.84×\sqrt{30.56}≈0.84×5.53≈4.64$
df=9, one-tailed p-value≈0.0005 (two-tailed≈0.001)

Step6: Predict 1991 (x=12)

$y=60.958+0.161×12≈60.958+1.932≈62.89$

Step7: Predict 2050 (x=71)

$y=60.958+0.161×71≈60.958+11.431≈72.39$

Answer:

(a) $H_0:
ho = 0$ (no linear relationship), $H_1:
ho ≠ 0$ (significant linear relationship)
(b) $y = 60.96 + 0.16x$
(c) $r ≈ 0.84$
(d) $p$-value ≈ 0.001
(e) Reject $H_0$; p-value < 0.05
(f) There is a significant linear relationship between year and percent at α=0.05
(g) $r²≈0.7056$; ~70.6% of y variation is explained by x
(h) ≈62.9%
(i) ≈72.4%; inaccurate (extrapolation beyond data range)
(j) Percent increases by ~0.16% per year since 1979