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 \( t_{9,0.025} = 2.262, t_{8,0.025} = 2.306, t_{9,0.05} = 1.833, t_{8,0.05} = 1.86 \)).
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 \( t_{9,0.025} = 2.262, t_{8,0.025} = 2.306, t_{9,0.05} = 1.833, t_{8,0.05} = 1.86 \)).
Updated On: Oct 1, 2024
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Correct Answer: 4.7 - 4.8
Solution and Explanation
The correct Answer is : 4.70 -4.80(Approx.)
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