Week7 Data Dive - Hypothesis Testing
Woods
2026-03-02
Create Groups & Define Variables
Hypothesis 1 Grouping
Group A: Coffees roasted in Taiwan
Group B: Coffees roasted in all other countries
Main Variable: ‘rating’ (continuous)
Hypothesis 2 Grouping
Main Variable: ‘high_rating’ (binary)
Group A vs Group B: ‘high_price’ vs ‘low_price’ based on median split of ‘100g_USD’
coffee_clean <- coffee_clean %>%
mutate(
# Hypothesis 1 grouping: Taiwan vs Non-Taiwan
country_group = if_else(loc_country == "Taiwan", "Taiwan", "Non-Taiwan"),
# Hypothesis 2: define high/low price and high/low rating
high_price = if_else(`100g_USD` > median(`100g_USD`, na.rm = TRUE),
"High price", "Low price"),
high_rating = if_else(rating >= 93, 1, 0) # binary success
)Median rating = 93; Found in last weeks data dive
Hypothesis 1: Neyman-Pearson Framework
Main variable: ‘rating’
Group A: Taiwan
Group B: Non-Taiwan
1. Null/Alternative Hypothesis
Null Hypothesis (H0): The mean rating of Taiwan roasted coffees equals the mean rating of non-Taiwan roasted coffees.
\[ ^μTaiwan = ^μNon-Taiwan \]
Alternative Hypothesis (H1): The mean rating of Taiwan roasted coffees differs from that of non-Taiwan roasted coffees.
\[ ^μTaiwan \neq ^μNon-Taiwan \]
Interpretation: Do coffees roasted in Taiwan differ in mean rating from non-Taiwan roasted coffees?
2. Intentional Desgin
\(\alpha\) = 0.05 (5% false positive risk)
Power = 0.80 (20% false negative risk)
Minimum practicality meaningful difference = 1.00 rating point
- 1.00 is small enough but not too small, and I did not want to go higher than a full rating point since the reviews are all between 84-98, despite the rating scale being in between 0-100 technically
3. Sample Size Calculation
power_calc <- power.t.test(
delta = 1, # minimum difference in means we care about
sd = 2, # assumed SD of rating
sig.level = 0.05, # alpha
power = 0.80,
type = "two.sample",
alternative = "two.sided"
)
power_calc##
## Two-sample t test power calculation
##
## n = 63.76576
## delta = 1
## sd = 2
## sig.level = 0.05
## power = 0.8
## alternative = two.sided
##
## NOTE: n is number in *each* groupcoffee_clean %>% count(country_group)## # A tibble: 2 × 2
## country_group n
## <chr> <int>
## 1 Non-Taiwan 1531
## 2 Taiwan 549Sample Size Conclusion: There are hundreds of observations for variable A and B, so I have enough data to test based on my design and effect size! This is far more than the required minimum as n =63.77, which means that I need about 64 Taiwan and 64 non-Taiwan roasted coffee observations to reliably detect a 1 rating point difference with 80% power, and my actual sample size is far larger.
4. Perform Test
t_test_res <- t.test(
rating ~ country_group,
data = coffee_clean,
var.equal = FALSE # Welch t-test
)
t_test_res##
## Welch Two Sample t-test
##
## data: rating by country_group
## t = -6.7582, df = 1183.2, p-value = 2.189e-11
## alternative hypothesis: true difference in means between group Non-Taiwan and group Taiwan is not equal to 0
## 95 percent confidence interval:
## -0.6099728 -0.3354930
## sample estimates:
## mean in group Non-Taiwan mean in group Taiwan
## 92.98628 93.45902Test Summary Insights: The mean rating of coffees roasted in Taiwan is +0.47 rating points higher than that of non-Taiwan roasted coffees. After flipping my confidence interval to interpret Taiwan to non-Taiwan, my 95% confidence interval is: [0.335, 0.610], which is quite narrow, meaning it has high precision.
5. Visualization
ggplot(coffee_clean, aes(x = country_group, y = rating, fill = country_group)) +
geom_boxplot(alpha = 0.7) +
geom_jitter(width = 0.15, alpha = 0.4) +
labs(
title = "Coffee Ratings: Taiwan vs Non-Taiwan",
x = "Country group",
y = "Rating"
) +
theme_minimal() +
theme(legend.position = "none")
6. Conclusion
Reject H0: There is extremely strong evidence that Taiwan roasted coffees have higher mean ratings than non-Taiwan roasted coffees. Despite a somewhat small effect size at 0.47 rating points, it is nonetheless consistent, precise, and statistically convincing due to a large sample size. Overall, the effect is small but quite real and detectable due to the large dataset. The visualization strongly supports this evidence depicting a visibly higher mean rating for coffees roasted in Taiwan, with very few coffees receiving below a 90/100 rating, where as non-Taiwan roasted coffees have a much more significant variance with many sub 90/100 ratings!
Hypothesis 2: Fisher’s Significance Testing
Main Variable: ‘high_rating’ (binary: 1 if rating \(\geq\) 93; 0 if rating < 93)
Group A/B: ‘high_price’ & ‘low_price’ (“High price” vs “Low price”)
High price = above median price
Low price = below median price
1. Null/Alternative Hypothesis
Null Hypothesis (H0): Proportion of high ratings is the same in both price groups
\[ P_{High price} = P_{Low price} \]
Alternative Hypothesis (H1): The proportions differ
\[ P_{High price} \neq P_{Low price} \]
Interpretation: Are high price coffees more likely to receive high ratings?
2. 2x2 Table
tab_price_rating <- table(coffee_clean$high_price, coffee_clean$high_rating)
tab_price_rating##
## 0 1
## High price 220 815
## Low price 393 652High price group: \[ P_1 = \frac{815}{1035} = 0.787 \]
Low price group: \[ P_2 = \frac{652}{1045} = 0.624 \]
Difference in proportions:\[ P_1 - P_2 = 0.163 \]
There is a 16.3 percentage point difference.
3. Perform Fisher / Two-proportion Test
prop_res <- prop.test(
x = c(
sum(coffee_clean$high_rating[coffee_clean$high_price == "High price"], na.rm = TRUE),
sum(coffee_clean$high_rating[coffee_clean$high_price == "Low price"], na.rm = TRUE)
),
n = c(
sum(coffee_clean$high_price == "High price", na.rm = TRUE),
sum(coffee_clean$high_price == "Low price", na.rm = TRUE)
),
alternative = "two.sided",
correct = TRUE
)
prop_res##
## 2-sample test for equality of proportions with continuity correction
##
## data: c(sum(coffee_clean$high_rating[coffee_clean$high_price == "High price"], na.rm = TRUE), sum(coffee_clean$high_rating[coffee_clean$high_price == "Low price"], na.rm = TRUE)) out of c(sum(coffee_clean$high_price == "High price", na.rm = TRUE), sum(coffee_clean$high_price == "Low price", na.rm = TRUE))
## X-squared = 66.104, df = 1, p-value = 4.277e-16
## alternative hypothesis: two.sided
## 95 percent confidence interval:
## 0.1240346 0.2029977
## sample estimates:
## prop 1 prop 2
## 0.7874396 0.6239234Fisher / Two-proportion Test Insights: The 95% confidence interval is [0.124, 0.203], which tells us that high price coffees are 12.4% to 20.3% more likely to receive a rating \[\geq\] 93 than low price coffees. Additionally, the p-value is far below 0.05, suggesting that we should reject \[H_0\] , since there is strong evidence that the probability of receiving high ratings differs significantly between coffees priced above the median and coffees priced below the median.
4. Visualization
prop_df <- coffee_clean %>%
group_by(high_price) %>%
summarise(
prop_high = mean(high_rating, na.rm = TRUE),
n = n()
)
ggplot(prop_df, aes(x = high_price, y = prop_high, fill = high_price)) +
geom_col(alpha = 0.8) +
scale_fill_manual(values = c("High price" = "green",
"Low price" = "lightgreen")) +
geom_text(aes(label = round(prop_high, 2)), vjust = -0.5) +
ylim(0, 1) +
labs(
title = "Proportion of High Ratings (≥ 93) by Price Group",
x = "Price group",
y = "Proportion high rating"
) +
theme_minimal()
5. Conclusion
Reject H0: The Two-Proportion and Fisher Test strongly support that higher prices have a much higher chance of receiving a high rating compared to lower priced coffees. The visual supports and clearly depicts these statistically significant results that the correlation between low priced coffee ratings and high priced coffee ratings is certainly not proportional. The bigger picture insight we gain is the justification of higher priced coffees, as they consistently live up to their value, receiving much higher ratings than lower priced coffees. It is important for consumers to know that they are receiving premium coffee when they buy higher priced coffees. Ultimately, this supports the notion that premium pricing aligns with higher quality coffee far more often than not.