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.

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Suresh Chand · Accepted Answer

Solution: Sample size (n) = 70 Proportion of student secured more than 80% marks in statistics is p = (frac{x}{n}) = (frac{50}{70}) = 0.71 Hence an estimate of standard error of P is S.E. (p) = (sqrt{frac{pq}{n}}) = (sqrt{frac{0.71 times 0.29}{70}}) = 0.034 Now, At 95% level of confidence interval for P i.e. (p enspace pm enspace z_{alpha / 2} sqrt{frac{pq}{n}}) = 0.71 ± 1.96 x 0.034 = (0.77664, 0.64336) Hence, At 95% level of confidence interval, Between 77.66% to 64.34% students scored more than 80% in statistics. At 99% level of confidence interval for P i.e. (p enspace pm enspace z_{alpha / 2} sqrt{frac{pq}{n}}) = 0.71 ± 2.58 x 0.034 = (0.79772, 0.62228) Hence, At 99% level of confidence interval, Between 79.77% to 62.23% students scored more than 80% in statistics.

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