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anonymous
 one year ago
The results of a fitness trial is a random variable X which is normally distributed with mean µ and standard deviation 2.4. A researcher uses the results from a random sample of 90 trials to calculate a 98% confidence interval for µ. What is the width of this interval ?
anonymous
 one year ago
The results of a fitness trial is a random variable X which is normally distributed with mean µ and standard deviation 2.4. A researcher uses the results from a random sample of 90 trials to calculate a 98% confidence interval for µ. What is the width of this interval ?

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phi
 one year ago
Best ResponseYou've already chosen the best response.1if you have a random sample of 90 trials and find the average of them, you will get mu \( \pm\) some standard deviation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how to find the average ?

phi
 one year ago
Best ResponseYou've already chosen the best response.1the std dev would be the population std dev / sqrt(size of sample) \[ \sigma= \frac{\sigma_{pop}} { \sqrt{90} } =\frac{2.4} { \sqrt{90} } \]

phi
 one year ago
Best ResponseYou've already chosen the best response.1you don't find the average... you are told it is \( \mu\) what they want are the lower and upper bounds x so that \[ \mu \pm x\] represents 90% of the area under the curve.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we dont have to find mu ?

phi
 one year ago
Best ResponseYou've already chosen the best response.1no. find the number of std deviations so that you get 98% of the area dw:1438259659150:dw

phi
 one year ago
Best ResponseYou've already chosen the best response.1the tails use 2% , 1% on each end

phi
 one year ago
Best ResponseYou've already chosen the best response.1after you get the value for x (# of std deviations) you calculate the interval as \( 2 \cdot x\cdot \sigma\) where \( \sigma= \frac{2.4}{\sqrt{90}} \)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but we do not have sample mean.. how to calculate ?

phi
 one year ago
Best ResponseYou've already chosen the best response.1they are asking for What is the width of this interval ? the "middle" of the interval is \( \mu\) the upper bound of the interval is \( \mu+x\) the lower bound is \( \mu  x \) the width of this interval is upper bound  lower bound which is \( \mu+x  ( \mux) \) = 2x so you need to find x

phi
 one year ago
Best ResponseYou've already chosen the best response.1x is the # of std dev so that you get 98% of the area times sigma (which you calculate from the info they gave us)
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