The table below gives the frequency with which a sample of women drink tea and coffee (Arab, Lenore et al. "Gender differences in tea, coffee, and cognitive decline in the elderly: the Cardiovascular Health Study." Journal of Alzheimer's disease : JAD vol. 27,3 (2011): 553-66. doi:10.3233/JAD-2011-110431). Do women drink tea and coffee with the same frequency at the alpha = 0.01 significance level?

Freq	Tea	Coffee
< 5 / year	580	1019
5 - 10 / year	289	169
1 - 3 / month	503	178
1 - 4 / week	618	208
> 5 / week	742	1148

• Enter the data into an array.
• Calculate the chi-square statistic and p-value
• Write an if statement to determine if the null hypothesis that women drink tea and coffee with the same frequency is rejected or not.

Ex: If the data for men is used,

Freq	Tea	Coffee
< 5 / year	551	637
5 - 10 / year	244	139
1 - 3 / month	387	152
1 - 4 / week	452	225
> 5 / week	443	915

the output is:

377.6187409281143 1.9030487979418984e-80 The hypothesis that tea and coffee are drunk with the same frequency is rejected.

Template:

# Import the necessary modules

# Construct a contingency table

# Calculate the test statistic and p-value chi2, p, df, ex = # Code for calculating test statistic and p-value print(chi2) print(p)

# Determine if null hypothesis is rejected or not if # Write appropriate if statement print("The hypothesis that tea and coffee are drunk with the same frequency is rejected.") else: print("The hypothesis that tea and coffee are drunk with the same frequency is not rejected.")

Reference:

https://learn.zybooks.com/zybook/Applications_of_Business_Analytics_II_-_61113/chapter/7/section/10