Mathematics A) Spreadsheet Data Analysis B) State Essay

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Mathematics

a) Spreadsheet Data Analysis

b) State hypotheses:

Ho: Type of soft drink and resident are independent. (null hypothesis)

Ha: Type of soft drink and resident are not independent. (alternate hypothesis)

c) Conduct chi-square test of independence at 0.05 significance level: = 0.05

d) Calculate degrees of freedom: Number of rows = 2, Number of columns = 2

Degrees of freedom (df) = (Number of rows - 1) x (Number of columns - 1) = (2-1)(2-1) = 1

e) Calculate the expected frequencies from actual data:

f) Calculate the chi-square statistic using both actual and expected frequency tables:

In excel, the cursor was placed in cell M17 and the formula of chi test typed as:

=CHITEST (actual frequency range, expected frequency range)

=CHITEST (L4:M5, L13:M14), as shown in the formula bar above

The result gives, the calculated chi-square probability, p = 1.2203 x 10-10

g) Look up the Chi-Square value, ?2 from chi-square distribution tables for df = 1 and ? = 0.05

2 = 3.

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84 from tables

h) Now we need to check whether or not the calculated chi-square probability, p is less than or equal to ? = 0.05 OR the Chi-Square value, ?2 from chi-square distribution tables is equal to or greater than ? = 0.05. If p OR ?2, we reject the null hypothesis and conclude that at the 0.05 significance level, there is a relationship between the row variables and column variables. Since p = 1.2203 x 10-10 < ? = 0.05 (OR ?2 = 3.84 > ? = 0.05) we conclude that there is a relationship between type of soft drink and resident response.

2. a) Forbes Data

Based on the statistical report and values highlighted, we see that there is a.....

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