In order to examine the differences in ages of health inspectors among five coun

Question

In order to examine the differences in ages of health inspectors among five counties, a Health Department statistician took random samples of six inspectors' ages in each county. The data are listed below. Use alpha = 0.05 for one-way ANOVA test in Excel. State the null and alternative hypothesis. What is the p-value? State your conclusion. Use Fisher's LSD method. What is the critical value? Which means differ? Use Bonferroni's adjustment method. What is the critical value? Which means differ? Use Tukey's method. What is the critical value? Winch means differ?

Explanation / Answer

There are 5 ages of inspectors among five school districts.

Here we have to test hypothesis that,

H0 : mu1 = mu2 = mu3 = mu4 = mu5

H1 : Atleast one of the mean is differ.

Assume alpha = level of significance = 0.05

Steps of ANOVA in EXCEL is :

Enter data in EXCEL sheet --> Data --> Data Analysis --> Anova: Single Factor --> ok --> Input Range : select all data --> Grouped by : columns --> Alpha : 0.05 --> Output Range : select one empty cell --> ok

Output is,

Here test statistic F = 11.1124

P-value = 0.0000256

P-value < alpha

Reject H0 at 5% level of significance.

Conclusion : Atleast one of the mean is differ.

We use Fisher's LSD method for checking which means are differ.

In Fisher LSD the test of hypothesis is,

H0 : Two means are equal.

H1 : Two means are not equal.

LSD we can find by using formula,

LSD = tc*sqrt(MSW*(1/n1 + 1/n2) )

tc = critical value from the t-distribution table
MSW = mean square within, obtained from the results of your ANOVA test
n = number of scores used to calculate the means

First we have to find critical value.

The critical value we can find by using EXCEL.

=TINV(probability, deg_freedom)

where probability = 0.05

deg_freedom = degrees of freedom for within groups in ANOVA = 25

tc = 2.060

LSD = 2.060*sqrt(40.9933*(1/6 + 1/6)) = 7.6149

y1bar = mean of column1 = 44.33

y2bar = mean of column2 = 46.33

y3bar = mean of column3 = 29.83

y4bar = mean of column4 = 46.00

y5bar = mean of column5 = 53.67

y1bar - y2bar = 44.33 - 46.33 = -2.00

If | y1bar - y2bar | >= LSD

then reject H0 at 5% level of significance.

Here  | y1bar - y2bar | < LSD

Accept H0 at 5% level of significance.

Conclusion : Column1 mean and column2 mean are equal.

| y2bar - y3bar | = | 46.33-29.83 | = 16.50

| y2bar - y3bar | > LSD

Reject H0 at 5% level of significance.

Conclusion : Column2 mean and column3 mean are not equal.

| y3bar - y4bar | = | 29.83 - 46 | = 16.17

| y3bar - y4bar | > LSD

Reject H0 at 5% level of significance.

Conclusion : Column3 mean and column4 mean are not equal.

| y4bar - y5bar | = | 46 - 53.67 | = 7.67

| y4bar - y5bar | > LSD

Reject H0 at 5% level of significance.

Conclusion : Column4 mean and column5 mean are not equal.

Mean for column2, column3, column4 and column5 are not same.

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