Statistics Essay for You

Q1
The investigator wants to approximate which treatments between A and B is effective in decreasing cholesterol levels. The study uses the hypotheses tests detailed below
H0 – The effect of treatment A in reducing cholesterol levels is equal to the effect of treatment B. That is µA - µB = 0
HA – The effect of treatment A in reducing cholesterol levels is not equal to the effect of treatment B. That is µA - µB ≠ 0.
Q2
The paired samples t-test is utilized in the estimation of the statistical significance of the difference in the means. The pooled t-test is used because the statistical test to establish if the variance were equal is insignificant, that is 0.6235 > 0.05 . The alternative hypothesis that the variances are not equal is accepted.
Q3
The point estimate of the value of the deviation of the mean cholesterol levels is calculated to be 27.265.
Q4
The p-value is computed to be is 0.0024 and because the value is lower than 0.05, the null hypothesis is therefore rejected.
Q5
The p-value indicate that the test is significant and as a consequence the null hypothesis is rejected and it is deduced that no difference exists betwixt the average of the treatments.
Q6
(9.87, 44.67) is the 95% CI for the difference in the treatments. There is a 0.95 chance that the difference between the treatments mean are between these limits.
Q7

The scatterplot of the API against ELL is presented below

Q8
The correlation coefficient is calculated to be 0.768 which can be interpreted to mean that there is a strong negative correlation between the API and ELL. The assumptions for the analysis have been fulfilled as there is a correlation between API and ELL and both variables fall under the category of ratio or interval data.
Q9
The R2 value is observed to be 0.589 and this is interpreted to be proportion of deviation of the variable API that can be defined by the variable ELL.

Q10

The regression standardized equation is
API = -4.396(ELL) + 785.89
The hypothesis that the slope is equal to zero is rejected and the hypothesis that the slope is not equal to zero is accepted because the 95% confidence interval indicates the limits of the slope o be -4.75 and -4.034
The gradient is negative which indicates that ELL increases with a decrease in API. The 95% CI of the gradient is computed to be (-4.758, -4.034)
The estimated API for a value of ELL of 75 is -4.396(75)+785 = 455.3. therefore, the value of API is 455.3
Most of the variables used in the analysis are in the form of categorical data and as a consequence their use in forecasting the API is limited. Variables like parent's educaion which could have been used in the forecasting are eliminated because of missing data. The correlation between the variable API and the percent of students in the first year is weak. Therefore, the scatterplot of the variable API and the variable ELL was the only variables considered.