Question

Vitamin C (ascorbate) boosts the human immune system and is effective in preventing a variety of illnesses. In a study by E. Cameron and L. Pauling, published as the paper "Supplemental Ascorbate in the Supportive Treatment of Cancer; Reevaluation of Prolongation of Survival Times in Terminal Human Cancer", patients in advanced stages of cancer were given a vitamin C supplement. Patient were grouped according to the organ affected by cancer; stomach, bronchus, colon, ovary or breast. The study yielded the survival times, given on the WeissStats site.

a. Obtain individual normal probability plots and the standard deviations of the sample.

b. Perform a residual analysis

c. use your results from part (a) and (b) to decide whether conducting a one-way ANOVA test on the data is reasonable. If so, also do parts (d) and (e).

d. use a one-way ANOVA test to decide, at the\(5%\) significance level whether, the data provide sufficient evidence to conclude that a difference exists among the means of the populations from which the samples were taken.

e. Interpret your results from part (d).

Short answer

Part a.

Part b.

Part c.

Part d. The null hypothesis is rejected that mean data provided a sufficient evidence to support the claim for the means of the population from which the sample were drawn are not all the same.

Step-by-step solution

Part a. Step 1. Given information

The sample data is given into the paper.

Part a. Step 2. Calculation

Let’s take the random sample of length \(120\).

Draw a normal probability plot using function “normplot” in MATLAB.

Program:

Query:

• First, we have defined the random samples.
• Then generate the normal probability plot.

Part b. Step 1. Calculation

Let’s take the random sample of length \(120\).

Then calculate the residual using relation

\(residual = data -fit\)

Then, draw a normal probability plot using function “normplot” in MATLAB.

Program:

Query:

• First, we have defined the random samples.

• Then generate the normal probability plot of the residuals.

Part c. Step 1. Calculation

Calculate the SST, SSTR and SSE using given relation

\(SST=\sum x^{2}-\frac{(\sum x)^{2}}{n}\)

\(SST=1561154-\frac{(4228)^{2}}{12}=7.25\times 10^{4}\)

\(SSTR=\frac{\sum (x_{i})^{2}}{n_{i}}-\frac{\sum (x)^{2}}{n}\)

\(SSTR=\frac{1828^{2}}{4}+\frac{1225^{2}}{4}+\frac{1175^{2}}{4} -\frac{(4228)^{2}}{12}=6.20\times 10^{4}\)

\(SSE=SST-SSTR=5.25\times 10^{3}\)

Then,

\(df_{T}=k-1=3-1=2\)

\(df_{E}=n-k=12-2=10\)

\(MSTR=\frac{SSTR}{df_{T}}=\frac{6.60\times 10^{4}}{3}=4.30\times 10^{4}\)

\(MSE=\frac{SSE}{df_{E}}=\frac{5.45\times 10^{3}}{10}=0.545\times 10^{3}\)

\(F=\frac{MSTR}{MSE}=\frac{3.30\times 10^{4}}{0.545\times 10^{3}}\approx 59.61\times 10^{3}\)

Then make an ANOVA table.

Part d. Step 1. Calculation

The \(p-\)value is the probability value which obtaining by the test statistics, or a value more extreme. The \(P-\)value is the number in the row title of the \(F-\)distribution table which containing \(F-\)value in the row \(dfd=df_{E}=7\) and in the column \(dfn=df_{T}=2\)

So, the \(p-\)value lie between

\(0.025<P<0.050\)

And if the \(p-\)value is less than significance level then it will reject the null hypothesis.

\(P<0.05\Rightarrow\) Reject \(H_{0}\)