##Problem 1 We conduct a chi-square goodness-of-fit to examine if the two alleles (R and X) are not equally likely. Hypotheses H0: PR = PX Ha: PR ≠ PX
#Chi-Square Goodness-of-Fit Test for ACTN3 Alleles
## Observed counts
observed <- c(R = 244, X = 192)
# Null values
theoritical_prop <- rep(1/2, 2)\(H_0\):\(p_R\) = \(p_X\) = 1/2 \(H_a\): at least on \(p_i\) \(\neq\) 1/2
# Expected values
expected_values <- theoritical_prop*sum(observed)
expected_values## [1] 218 218#Chi-Square Test
test_result <- chisq.test(observed)
test_result##
## Chi-squared test for given probabilities
##
## data: observed
## X-squared = 6.2018, df = 1, p-value = 0.01276**Interpretation* -The test statistic, X-squared = 6.2018, df = 1, p-value = 0.01276. -We reject the null hypothesis since the p-value is less than 0.05. -It implies there is sufficient statistically significant evidence that the two ACTN3 allele options (R and X) are not equally likely. -Conclusion: The alleles R and X are not equally likely.
The dataset imported R-studio.
#Create contingency table
tableVG <- table(NutritionStudy$VitaminUse, NutritionStudy$Sex)
tableVG ##
## Female Male
## No 87 24
## Occasional 77 5
## Regular 109 13\(H_0\) : Vitamin use is not associated with Gender \(H_a\) : Vitamin use is associated with Gender
#Chi-square test of association
chisq.test(tableVG)##
## Pearson's Chi-squared test
##
## data: tableVG
## X-squared = 11.071, df = 2, p-value = 0.003944-The test statistic, X-squared = 11.071, df = 1, -The p-value = 0.003944. -Since the p-value is less than 0.05, we reject the null hypothesis. - This implies there is enough evidence of significant association between vitamin use and gender. -Conclusion: Female are statistically more likely to take vitamins compared to male.
Is there a significant difference in gill rates by Calcium Level? Hypotheses H0: µL = µM = µH Ha: µL ≠ µM ≠ µH
#LOading FishGills3 dataset
setwd("C:/Users/user/Downloads")
Gill_rate <- read_csv("FishGills3.csv")## Rows: 90 Columns: 2
## ── Column specification ────────────────────────────────────────────────────────
## Delimiter: ","
## chr (1): Calcium
## dbl (1): GillRate
##
## ℹ Use `spec()` to retrieve the full column specification for this data.
## ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.head(Gill_rate)## # A tibble: 6 × 2
## Calcium GillRate
## <chr> <dbl>
## 1 Low 55
## 2 Low 63
## 3 Low 78
## 4 Low 85
## 5 Low 65
## 6 Low 98Hypothesis
\(H_0\): \(\mu_L\) = \(\mu_M\) = \(\mu_H\)
\(H_a\): not all \(\mu_i\) are equal
# Perform One-way ANOVA model
fit=aov(GillRate ~ Calcium, data = Gill_rate)
# summary results
summary(fit)## Df Sum Sq Mean Sq F value Pr(>F)
## Calcium 2 2037 1018.6 4.648 0.0121 *
## Residuals 87 19064 219.1
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1interpretation -The test statistic, F (2, 87) = 4.648, p = 0.0121. -The null hypothesis is rejected since the p-value is less than 0.05. -Hence, there is sufficient evidence to support the claim. -Conclusion: There is a statistically significant difference in gill rate based on the calcium level of the water.