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H_0:σ_men^2=σ_women^2
H_1: σ_men^2>σ_women^2
Assuming both the men and the women follow a normal distribution we can say that the test statistic f is:
f=(s_men^2)/(s_(women  )^2 )=0.8362
n_men=25
n_women=21
f_c=2.859

Since f is less than f_c we can say that we cannot reject the null hypothesis and we cannot accept the alternative.


Using MATLAB, we compute the ANOVA for the data
brick =

   21.8000   21.7000   21.9000   21.9000
   21.9000   21.4000   21.8000   21.7000
   21.7000   21.5000   21.8000   21.8000
   21.6000   21.5000   21.6000   21.7000
   21.7000       NaN   21.5000   21.6000
   21.5000       NaN       NaN   21.8000
   21.8000       NaN       NaN       NaN

>> anova1(brick)

ans =

    0.0827


	Source 	SS 	   df 	    MS 		F	 Prob>F
	Groups 	0.13911    3 	   0.04637 	2.62 	0.0827
	Error 		0.31907    18      0.01773
	Total 		0.45818     21

As we can see, the f value is 2.62, which is smaller than our critical f value of 3.16 therefore we cannot reject the null hypothesis so we can’t show that the firing temperature affects the density of the brick.


    H_0:σ_before^2-σ_after^2=3
H_1:σ_before^2-σ_after^2>3


> temp = c(22.9,24.0,22.9,23.0,20.5,26.2,25.8,26.1,26.9,22.8,27.0,26.1,26.2,26.6)
> turbidity=c(125,118,103,105,26,90,99,100,105,55,267,286,235,265)
> plot(temp,turbidity)
> fit <- lm(turbidity~temp)
> fit

Call:
lm(formula = turbidity ~ temp)

Coefficients:
(Intercept)         temp
    -510.71        26.31


> fit$residuals
         1          2          3          4          5          6
 33.252837  -2.686318  11.252837  10.622005  -2.607188 -88.564628
         7          8          9         10         11         12
-69.041299 -75.933796 -91.980454 -34.116330  67.388714 110.066204
        13         14
 56.435372  75.912043

c and d.
> summary(fit)

Call:
lm(formula = turbidity ~ temp)

Residuals:
    Min      1Q  Median      3Q     Max
-91.980 -60.310   4.007  50.640 110.066

Coefficients:
            Estimate Std. Error t value Pr(>|t|)
(Intercept) -510.713    228.196  -2.238   0.0450 *
temp          26.308      9.178   2.867   0.0142 *
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 67.68 on 12 degrees of freedom
Multiple R-squared:  0.4064,	Adjusted R-squared:  0.357
F-statistic: 8.217 on 1 and 12 DF,  p-value: 0.01418


Multiple R-squared:  0.4064,	Adjusted R-squared:  0.357


Both the slope and the intercept have p-values from the t-test that are smaller than 0.05, therefore it is likely that they are a good fit for the data we have.


Linear regression seems like it will work pretty well here since it looks like we have a line.


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[Workspace loaded from ~/.RData]

> blood = c(1,0,1,2,5,1,4,6,2,3,5,4,6,8,4,5,7,9,7,6)
> sound = c(60,63,65,70,70,70,80,90,80,80,85,89,90,90,90,90,94,100,100,100)
> plot(blood,sound)
Error in plot.new() : figure margins too large
> plot(sound,blood)
> plot(blood,sound)
> fit <- lm(blood~sound)
> fit

Call:
lm(formula = blood ~ sound)

Coefficients:
(Intercept)        sound
   -10.1315       0.1743

> summary(fit)

Call:
lm(formula = blood ~ sound)

Residuals:
    Min      1Q  Median      3Q     Max
-1.8120 -0.9040 -0.1333  0.5023  2.9310

Coefficients:
             Estimate Std. Error t value Pr(>|t|)
(Intercept) -10.13154    1.99490  -5.079 7.83e-05 ***
sound         0.17429    0.02383   7.314 8.57e-07 ***
---
Signif. codes:
0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.318 on 18 degrees of freedom
Multiple R-squared:  0.7483,	Adjusted R-squared:  0.7343
F-statistic:  53.5 on 1 and 18 DF,  p-value: 8.567e-07

> 85* 0.17429 -10.13154
[1] 4.68311