Suppose that the weights (in kgs) of six months old babies, monitored at a healthcare facility, have $ N(\mu, \sigma^2) $ distribution, where $ \mu \in \mathbb{R} $ and $ \sigma 0 $ are unknown parameters. Let $ X_1, X_2, \ldots, X_9 $ be a random sample of the weights of such babies. Let $ \overline{X} = \frac{1}{9} \sum_{i=1}^{9} X_i $, $ S = \sqrt{\frac{1}{8} \sum_{i=1}^{9} (X_i - \overline{X})^2} $ and let a 95% confidence interval for $ \mu $ based on $ t $-distribution be of the form $ (\overline{X} - h(S), \overline{X} + h(S)) $, for an appropriate function $ h $ of random variable $ S $. If the observed values of $ \overline{X} $ and $ S^2 $ are 9 and 9.5, respectively, then the width of the confidence interval is equal to __________ (round off to 2 decimal places) (You may use $ t_{9,0.025} = 2.262, t_{8,0.025} = 2.306, t_{9,0.05} = 1.833, t_{8,0.05} = 1.86 $).

Question

Suppose that the weights (in kgs) of six months old babies, monitored at a healthcare facility, have $ N(\mu, \sigma^2) $ distribution, where $ \mu \in \mathbb{R} $ and $ \sigma > 0 $ are unknown parameters. Let $ X_1, X_2, \ldots, X_9 $ be a random sample of the weights of such babies. Let $ \overline{X} = \frac{1}{9} \sum_{i=1}^{9} X_i $, $ S = \sqrt{\frac{1}{8} \sum_{i=1}^{9} (X_i - \overline{X})^2} $ and let a 95% confidence interval for $ \mu $ based on $ t $-distribution be of the form $ (\overline{X} - h(S), \overline{X} + h(S)) $, for an appropriate function $ h $ of random variable $ S $. If the observed values of $ \overline{X} $ and $ S^2 $ are 9 and 9.5, respectively, then the width of the confidence interval is equal to __________ (round off to 2 decimal places) (You may use $ t_{9,0.025} = 2.262, t_{8,0.025} = 2.306, t_{9,0.05} = 1.833, t_{8,0.05} = 1.86 $).

cdquestions Admin · Accepted Answer

Step 1: Identify the formula for $ h(S) $. For a 95\% confidence interval based on the $ t $-distribution, $ h(S) $ is given by: $$ h(S) = t_{\alpha/2, \, n-1} \cdot \frac{S}{\sqrt{n}}, $$ where: $ t_{\alpha/2, \, n-1} $ is the critical value of the $ t $-distribution for a 95\% confidence level with $ n-1 $ degrees of freedom, $ S $ is the sample standard deviation, $ n $ is the sample size. Step 2: Calculate $ S $ from the given variance. The sample variance is given as $ S^2 = 9.5 $. Thus, the sample standard deviation is: $$ S = \sqrt{S^2} = \sqrt{9.5} \approx 3.08. $$ Step 3: Determine the value of $ h(S) $. The sample size is $ n = 9 $, so the degrees of freedom are $ n-1 = 8 $. For a 95\% confidence level, the critical value is $ t_{0.025, 8} = 2.306 $. Substituting the values: $$ h(S) = t_{0.025, 8} \cdot \frac{S}{\sqrt{n}} = 2.306 \cdot \frac{3.08}{\sqrt{9}} = 2.306 \cdot \frac{3.08}{3}. $$ Simplify: $$ h(S) \approx 2.306 \cdot 1.0267 \approx 2.37. $$ Step 4: Calculate the width of the confidence interval. The width of the confidence interval is: $$ 2 \cdot h(S) \approx 2 \cdot 2.37 = 4.74. $$ Conclusion: The width of the confidence interval is: $$ \boxed{4.74}. $$

Correct Answer: 4.7 - 4.8

Solution and Explanation

Top Questions on Multivariate Distributions

Questions Asked in IIT JAM MS exam