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In a test of the ability of a certain polymer to remove toxic wastes from water, experiments where conducted at three different temperatures. The data below give the percentages of impurities that where removed by the polymer in 17 independent attempts. Fewer samples were taken when the temperature was high since these experiments where more expensive.

low = [31,36,34,37,39,34,33]

medium = [36,31,39,32,36,30,25]

high = [29,24,28]

Assume the samples are drawn from normal distributions with unknown but equal variances and with means $μ_{low}$, $μ_{medium}$ and $μ_{high}$

(a) With a significance level of 95% percent, test the null hypothesis that the temperature has no effect on the ability to remove toxic wastes from the water, i.e. the null hypothesis $H_0$ : $μ_{low}$ = $μ_{medium}$ = $μ_{high}$. What are the values $SS_W. SS_B$ and the test statistics?

I have ran a one way ANOVA though MATLABs anova1() function on the date which give me this: ANOVA1

From this i read that: $SS_W.$ = 188,286 $SS_B$ = 129,832 and the test statistics = 4,83

Further the p-value is 2,5% and thereby lower than 5% so we can reject with 95% confidence that temperature has no effect on the polymers ability to remove toxic waste.

This is my first time running an ANOVA-test. Is this the right way to interpret the results?

(b) Do not consider the measurements performed with the high temperature in the following question. Compute the interval that, with 90% percent confidence, will contain the difference $μ_{low}$ - $μ_{medium}$ ?

No sure on how to approach this one?

Maybe: Lower limit = $μ - z_{1-(0.1/2)} δ_μ$ Upper limit = $μ + z_{1-(0.1/2)} δ_μ$

$z_{1-(0.1/2)} =z_.95 = ? $ and $δ_μ = \text{standard devivation} /\sqrt{14} = ?$

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    $\begingroup$ For (a) Perhaps see this Answer to a Question where design is 'balanced' (all three groups have the same sample size). For (b) do a two-sample t confidence interval. (I am not sure about the notation $\delta_\mu.$) See my Answer below. $\endgroup$
    – BruceET
    Aug 14 '17 at 18:21
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Your interpretation in (a) seems fine. Once you reject $H_0$ that all population means are equal, the next task is to investigate what the pattern of differences may be.

Specifically, part (b) investigates whether $\mu_L = \mu_M$ by looking at a CI for the difference $\mu_L - \mu_M.$ The assumption of the ANOVA in (a) is that all three populations have the same variance. Still assuming the Low and Medium populations have the same variance, a 95% confidence interval (CI) based on a two-sample "pooled" t procedure is $(-2.296, 6.581),$ as computed in R statistical software below. (You can verify this by hand using a formula in your text, or by using MATLAB.)

Because the CI contains $0,$ we conclude that a $0$ difference (i.e. equal means) is believable. Also, from the output below, the P-value for a two-sample pooled t test is about $0.31 > 0.05,$ so we cannot reject $H_0: \mu_L = \mu_M.$

Low = c(31,36,34,37,39,34,33)
Med = c(36,31,39,32,36,30,25)

t.test(Low, Med, var.eq=T)

        Two Sample t-test

data:  Low and Med
t = 1.0519, df = 12, p-value = 0.3136
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-2.295540  6.581254
sample estimates:
mean of x mean of y 
 34.85714  32.71429 

If you were do do all three possible comparisons of pairs of means from the original data, you would have three comparisons. If each comparison is made at the 5% level, then there is some doubt what the overall error probability is. There are several kinds of multiple comparison procedures for dealing with situations. Some names are 'Fisher LSD', 'Tukey HSD', and 'Bonferroni'. I suppose this may be the next topic in your course.

The Bonferroni method is to do each of the three tests at level $\alpha = .05/3 = .017.$ (Or each of three CIs at confidence level 98.3%.) This ensures that the overall error probability is not greater than 5%.

Note: If you are comparing levels Med and High it would be best to use a Welch ('separate-variances') two-sample procedure because failure of the equal-variance assumption can be especially dangerous when group sample sizes are unequal. To get the Welch CI in R, omit the argument var.eq to obtain the 95% CI $(0.164, 11.265),$ which does not contain $0;$ the P-value for testing $H_0: \mu_M = \mu_H$ against the two sided alternative is $0.045 < .05,$ so you could reject at the 5% level.

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  • $\begingroup$ Just a quick note, you made the calculations for a 95% CI. b) asked for a s 90%. I've calculated a 95% CI to (-2,2955, 6,5813) and a 90% CI ( -1.4878, 5.7735) . $\endgroup$
    – Daniel
    Aug 15 '17 at 8:24
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    $\begingroup$ 90% Welch CI for Med vs. High is as you say. $\endgroup$
    – BruceET
    Aug 15 '17 at 18:23

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