 For each of the following test, outline a research scenario lunnatican4 2021-12-09 Answered
For each of the following test, outline a research scenario and use any hypothetical data to show how the test is conducted
i. Analysis of variance
ii. Chi-square for goodness-of-fit test
iii. Chi-square test for independence

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i.Analysis of Variance (ANOVA)
Analysis of variance (ANOVA) is used to determine whether there are any statistically significant differences between the means of two or more independent unrelated groups.
A Research scenario for the Analysis of Variance ANOVA is described below:
A clinical trial is run to compare weight loss programs. Three popular weight loss programs are considered (Low calorie, Low fat and Low carbohydrate). A total of 15 patients agree to participate in the study and are randomly assigned to one of the three diets. The data on the weight loss of the 15 patients under different weight loss program are given below:
$\begin{array}{|c|c|}\hline Low \ Calorie & Low \ fat & Low \ carbohydrate \\ \hline 2 & 2 & 2 \\ \hline 6 & 5 & 1 \\ \hline 5 & 6 & 5 \\ \hline 1 & 3 & 2 \\ \hline 4 & 2 & 6 \\ \hline \end{array}$
Is there a statistically significant difference in the mean weight loss among the three diets?
Step 1 - Setting of hypothesis and determination of level of significance
$$\displaystyle{H}_{{{0}}}:\mu_{{{1}}}=\mu_{{{2}}}=\mu_{{{3}}}$$
$$\displaystyle{H}_{{{1}}}:$$ All means are not equal
Level of significance a=0.05
Step 2 - Selection of appropriate test statistics
$$\displaystyle{F}={\frac{{{M}{S}{B}}}{{{M}{S}{E}}}}$$
Step 3 - Setting up of Decision rule:
Degrees of freedom.
$$\displaystyle{d}{f}_{{{1}}}={k}-{1}$$
=3-1
= 2
$$\displaystyle{d}{f}_{{{2}}}={N}-{k}$$
=15-3
=12
The critical value of F-statistics for degrees of freedoms 2,12 at 0.05 level of significance is 3.885
Therefore, Decision rule: Reject $$\displaystyle{H}_{{{0}}}:{\quad\text{if}\quad}\ {F}{>}{3.885}$$
Step 4 - computation of test statistics:
$\begin{array}{|c|c|}\hline & Low \ Calorie & Low \ fat & Low \ carbohydrate \\ \hline n & 5 & 5 & 5 \\ \hline Group \ mean & 3.6 & 3.6 & 3.2 \\ \hline \end{array}$
The overvall mean is obtained as $$\displaystyle\overline{{{X}}}={3.46}$$
Therefore,
$$\displaystyle{S}{S}{B}=\sum{n}_{{{j}}},{\left(\overline{{{X}_{{{j}}}}}-\overline{{{X}}}\right)}^{{{2}}}$$
$$\displaystyle={5}{\left({3.6}-{3.46}\right)}^{{{2}}}+{5}{\left({3.6}-{3.46}\right)}^{{{2}}}+{5}{\left({3.2}-{3.46}\right)}^{{{2}}}$$
=0.534
$$\displaystyle{S}{S}{E}=\sum\sum{\left({X}-\overline{{{X}}}_{{{j}}}\right)}^{{{2}}}$$
=(17.2+13.2+18.8)
= 49.2
The required ANOVA table is given as follows:
$\begin{array}{|c|c|}\hline Sources \ of \ variation & Sum \ of \ squares (SS) & Degrees \ of \ freedom(df) & Means \ Square \ (MS) & F \\ \hline Between \ treatment & 0.534 & 3-1=2 & 0.534/2=0.267 & 0.267/4.1=0.065 \\ \hline Error & 49.2 & 15-3=12 & 49.2/12=4.1 & \\ \hline Total & 49.734 & 15-1=14 & & \\ \hline \end{array}$
Conclusion:
Since $$\displaystyle{F}{\left({0.065}\right)}{<}{3.885}$$</span>, Therefore, there is no evidence to reject the null hypothesis and we can conclude that there is no statistically significant difference in the mean weight loss among the three diets.
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ii.Chi-square for goodness of fit test.
The Chi-square for goodness of fit test is used to test the significant difference between the theory and experiment.
A Research scenario for the Chi-square for goodness of fit test
is described below
Toy company prints baseball cards. The company claims 30% of the cards are rookies,60% veterans but not all stars and 10% are veteran all stars. Suppose a random sample of 100 cards has 50 rookies,45 veterans and 5 all-stars. Is this consistent with the company claim. Significance level 0.05
Step1 - Hypothesis to be tested are as follows:
$$\displaystyle{H}_{{{0}}}$$: The proportion of rookies, veterans and all-stars is 30%,60% and 10% respectively
$$\displaystyle{H}_{{{1}}}$$: At least one of the proportions in the null hypothesis is false
Step 2 - The test statistics is given as follows:
$$\displaystyle{x}^{{{2}}}=\sum{\left[{\frac{{{\left({O}_{{{r},{c}}}-{E}_{{{r},{c}}}\right)}^{{{2}}}}}{{{E}_{{{r},{c}}}}}}\right]}$$
Step 3: Analysis of sample data
Degrees of freedom (df)=k-1=3-1=2
Expected values are calculated as follows:
$$\displaystyle{E}_{{{i}}}={n}\times{p}_{{{i}}}$$
$$\displaystyle{E}_{{{1}}}={100}\times{0.30}={30}$$
$$\displaystyle{E}_{{{2}}}={100}\times{0.60}={60}$$
$$\displaystyle{E}_{{{3}}}={100}\times{0.10}={10}$$
Step 3 - The test statistic is calculated as follow:
$$\displaystyle{x}^{{{2}}}={\frac{{{\left({50}-{30}\right)}^{{{2}}}}}{{{30}}}}+{\frac{{{\left({45}-{60}\right)}^{{{2}}}}}{{{60}}}}+{\frac{{{\left({5}-{10}\right)}^{{{2}}}}}{{{10}}}}$$
=19.58
Step 4: From chi-distribution table the critical value of statistics at 0.05 level of significance for 2 df is 5.991
Decision Rule: Reject $$\displaystyle{H}_{{{0}}},{\quad\text{if}\quad}\ {x}^{{{2}}}{>}{5.991}$$
Conclusion:
Since $$\displaystyle{x}^{{{2}}}{\left({19.58}\right)}{>}{5.991}$$
Therefore, there is sufficient evidence to reject the null hypothesis and conclude that the result is not consistent with the company claim.