Question

2. Consider a study comparing is the length of time (in days) for recovery. The medications were randomly assigned to the patients. In group 1, the ni = 15 patients were given medication 1. In group 2, the n2 = 18 patients were given medication 2. We will use a simple linear regression model to analyse the recovery time according to the medication. We import the data with R and display a few two medications for severe bladder infections. The response y rows > head (bladder) time group 20 1 1 12 2 1 3 17 1 4 21 1 5 18 1 16 6 1 The levels for the group variable were coded as 1 and 2, which are numeric. So we will use (b) We fit a simple linear regression model and display a summary of the fit. We are also displaying confidence interval for the coefficients > summary(mod) Call bladder lm (formula = time group, data Residuals: Median Min 1Q 3Q Маx -5.3889 -2.1333 -0.3889 1.8667 6.6111 Coefficients: Estimate Std. Error t value Pr(> |t|) (Intercept)16.1333 group2 0.7704 20.941 2e-16 *** 4.2556 1.0432 4.079 0.000293 0 ***0.001 ** 0.01 * 0.05 . 0.1 Signif. codes: 1 Residual standard error: 2.984 on 31 degrees of freedom Multiple R-squared: 0.3493, 0.3283 Adjusted R-squared: p-value: 0.0002928 F-statistic: 16.64 on 1 and 31 DF, > confint (mod) 2.5% 97.5 % (Intercept) 14.562018 17.704649 2.127984 6.383127 group2 (i) Give an estimate of the mean recovery time for patients on medication 1 (ii) Give an estimate of the mean recovery time for patients on medication 2 (iii) Give an estimate the common population variance o2. (iv) Give the value of the coefficient of determination and interpret it within the context of the study (iv) Test for the equality of the mean recovery time. Formulate the hypotheses. Give the value of the t-test statistic, and write down the p-value symbolically. Give the value of the p-value and the conclusion at a = 5%. (v) Give a 95% confidence for B1. Based on this interval, which medication is best?

Answer 1

(i) raw data is needed

(ii)raw data is needed

(iii) MSE(here residual standard error) is the estimate of the common population variance=2.984

(iv) coefficient of determination=R-square=0.3493

(iv) null hypothesis H0:$\mu$group1=$\mu$group2

alternate hypothesis Ha: $\mu$group1 $\ne$$\mu$group2

statistic t=4.079

p-value=0.000293

Reject H0 , as the p-value is less than alpha=0.05

(v) (1-alpha)*100% confidence interval $\beta$1=$\hat{\beta}$1​​​​±t(alpha/2,error df)*SE($\hat{\beta}$1)

95% confidence interval =4.2556±t(0.05/2, n-1)*1.0432=4.2556±2.04*1.0432=4.2556±2.1281=(2.1275,6.3837)

given is (2.1280, 6.3831) this is due to decimal place approximation error