Data https://drive.google.com/file/d/0B99gErxWiB1acUpyWHNJRUQ0S00/view Question

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Data

https://drive.google.com/file/d/0B99gErxWiB1acUpyWHNJRUQ0S00/view

Question 22 In order to verify if the proportions of houses (having furniture) differ between houses (having different amount of bathroom), the statistician Andrew Handerson believes that it is necessary to test formally whether the amount of bathroom in a house and whether it has furniture, are independent. 1. Make the null hypothesis, the distribution type of test statistic and the approximate distribution of the test statistic for carrying out this test. Provide a cross-tabulation of the data of these houses by these two factors. (It is useful to generate this summary table by using computer.) 2. Perform a formal test of whether the proportions of houses having furniture varies with the amount of bathrooms, that is, whether having furniture and number of bathrooms is independent of each other. Using computer may be a useful way to solve this problem. It is required to indicate the table of expected frequencies, the observed value and relevant DF for the distribution of the test statistic, the resulting p-value. 3. Demonstrate the calculation of the expected value for the table cell identified the house has 1 bathroom and no furniture. Demonstrate the calculation for the contribution to the observed test statistic from the cell identified the house has 1 bathroom and has furniture. 4. Analyse the p-value for this test

Explanation / Answer

1. Null Hypothesis : H0 : "the amout of bathroom in houses" and "whether it has furniture or not" is independent in nature.

ALternative Hypothesis : Ha : "the amout of bathroom in houses" and "whether it has furniture is not" is dependent in nature

Distribution type of test statistic is binomial distriubtion and approximate distibution of test statistic to carry this test is Chi - square distribution.

Observed Table

2. Expected frequency

I will take one example

Expected frquency for No furniture and 1 bathroom = Total number of 1 bathroom * total number of no furnitire/ Total number of sample = 16 * 71/103 = 11.03

similarly we will calculate the Chi - square test

Degree of freedom = 2 * 1 = 2

3. Calculation is demonstrated above.

Now chi - square statistic contribution X2 (1 bathroom - no furniture) = (O - E)2 /E = (11 - 11.03)2 / 11.03 = 0.0000816

similarly, the chi - square contribution table with formula

X2 = (o - E)2 /E

so CHi - square statistic

X2 = 3.195

where dF = 2 and alpha = 0.05

so critical value of X2 cr  = 5.99

so here X2 < X2 cr so we shall reject the null hypothesis and can conclude that "the amout of bathroom in houses" and "whether it has furniture or not" is independent in nature.

4. p - value = CHITEST (Observed, Expected) = CHIINV (X2 > 3.195; 2) = 0.202

Observed Table Have Furniture Yes No Total Number of Bathrooms 1 5 11 16 2 23 40 63 3 4 20 24 Total 32 71 103

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