Statistics II20785 marksNumerical: confidence interval for proportionEstimation
A survey was conducted among 70 students studying B.Sc. CSIT in some colleges randomly. Among them, 50 students secured more than 80% marks in statistics. Compute 99% and 95% confidence intervals for the population proportion of students who secured more than 80% marks in subject statistics, and com
A survey was conducted among 70 students studying B.Sc. CSIT in some colleges randomly. Among them, 50 students secured more than 80% marks in statistics. Compute 99% and 95% confidence intervals for the population proportion of students who secured more than 80% marks in subject statistics, and comment on the results. [5]
Parameter Value ------------------ Sample size $n$ 70 Students scoring 80% ($X$) 50 Sample proportion $p = X/n$ $50/70 = 0.7143$ $q = 1 - p$ $0.2857$ $Z$ for 99% $2.576$ $Z$ for 95% $1.96$ All required data present. $$S.E.(p) = \sqrt{\frac{pq}{n}} = \sqrt{\frac{0.7143 \times 0.2857}{70}}$$ $$= \s...