Q. 13.71 A study by E. Anionwu et al. pub... [FREE SOLUTION]

Chapter 13: Q. 13.71 (page 546)

A study by E. Anionwu et al. published as the paper "Sickle Cell Disease in a British Urban Community", measured the steady-state hemoglobin levels of patients with three different types of sickle cell disease: HB SS, HB ST and HB SC. The data are presented 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

Expert verified

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鈥檚 take the random sample of length \(120\).

Draw a normal probability plot using function 鈥渘ormplot鈥� in MATLAB.

Program:

Query:

First, we have defined the random samples.

Then generate the normal probability plot.

Part b. Step 1. Calculation

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

Then calculate the residual using relation

\(residual = data -fit\)

Then, draw a normal probability plot using function 鈥渘ormplot鈥� 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

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

\(SST=0.4569-\frac{(3.2130)^{2}}{25}=0.0440\)

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

\(SSTR=\frac{0.0901^{2}}{5}+\frac{0.1558^{2}}{10}+\frac{0.1439^{2}}{6}+\frac{0.0672^{2}}{4} -\frac{(3.2130)^{2}}{25}=0.0023\)

\(SSE=SST-SSTR=0.0417\)

Then,

\(df_{T}=k-1=4-1=3\)

\(df_{E}=n-k=25-4=21\)

\(MSTR=\frac{SSTR}{df_{T}}=\frac{0.0023}{3}=0.00077\)

\(MSE=\frac{SSE}{df_{E}}=\frac{0.0417}{21}=0.001986\)

\(F=\frac{MSTR}{MSE}=\frac{0.00077}{0.001986}\approx 0.3877\)

\(F=\frac{MST}{MSE}=\frac{12}{2.2857}\approx 5.25\)

After calculating these values put all into the table and get 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 \(H_{0}\)