section6sol

# section6sol - Stat 104 Section 6 Handout(Solutions Marion...

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Stat 104, Section 6 Handout (Solutions) Marion, Thursday, 4pm, SC107 Key Concepts 2-sample t -based confidence intervals for the difference of two means/ two proportions Hypothesis Testing Review 1. The weight of an adult swan is normally distributed with a mean of 30 pounds and a standard deviation of 10 pounds. A farmer randomly selected 36 swans and loaded them into his truck. What is the probability that this flock of swans weights more than 1000 pounds? The wrong approach to this problem is to find the average weight and standard deviation of a flock of swans and then calculate the Z-score from that. We have no information about a flock of swans and how that weight is distributed; we are told that individual swan weights are normally distributed. If a flock of swans needs to weigh less than 1000 pounds, then an average weight of the swans needs to weigh 1000/36 pounds. If the true weight of an adult swan is , then the average weight of a sample of 36 is . We need to find the P(>1000/36). P(>1000/36) = P( ) = P (Z>-1.33) = .908 2. The weights of boxes of cookies produced by a certain manufacturer approximately follow a normal distribution with a mean of 202 g and a standard deviation of 3 g. a. Between what values do the middle 95% of the weights of boxes of cookies lie? From a normal distribution, we know that 95% of the observations lie within 1.96 standard deviation from the mean. Therefore: μ ± 1.96*σ = 202 ± 1.96*(3) = (196.12, 207.88) b. If the manufacturer stamps 206 g on all the boxes, what percent of boxes of cookies are overweight?

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## This note was uploaded on 03/27/2012 for the course STATS 104 taught by Professor Michaelparzen during the Fall '11 term at Harvard.

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section6sol - Stat 104 Section 6 Handout(Solutions Marion...

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