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Mathematics 17 Online
OpenStudy (anonymous):

In a study of 205 adults, the mean heart rate was 75 beats per minute. Assume the population of heart rates is known to be approximately normal with a standard deviation of 8 beats per minute. What is the 95% confidence interval for the mean beats per minute?

OpenStudy (anonymous):

a) 73.9-76.1 b)73.7-76.1 c)73.9-76.3 d)70.9-73.3

OpenStudy (anonymous):

@mathstudent55 @kirbykirby

OpenStudy (anonymous):

@Hero @amistre64 @dan815

OpenStudy (amistre64):

\[\hat p\pm z(SE)\]

OpenStudy (amistre64):

in this case p hat is the sample mean z is related to 95% about the mean and SE is as it always is: sigma/sqrt(n)

OpenStudy (anonymous):

okay, so what would be the next step? I am very confused. @amistre64

OpenStudy (amistre64):

well, everything is defined for you, might need to look up the z score that relates to the 95% stuff tho

OpenStudy (anonymous):

would the answer be C?

OpenStudy (amistre64):

x bar is normally used instead of phat for means ..... but its the same run down im not sure what the answer is, i havent worked the problem. show me how you arrive at C

OpenStudy (anonymous):

I plugged the values given in the question to the equation you listed

OpenStudy (amistre64):

let me verify it then

OpenStudy (amistre64):

the wolfs gone wacky .... im thinking 1.96 for the z score, 75-1.96(8/sqrt(205)) but then 75 is not the midpoint of C so it cant be it.

OpenStudy (anonymous):

with that formula ^ I got -41.46?

OpenStudy (anonymous):

@amistre64

OpenStudy (amistre64):

i got 73.9, but not C

OpenStudy (anonymous):

so A. @amistre64

OpenStudy (amistre64):

i get A out of it, assuming 1.96 is the proper z value ... wich im like 96% sure of lol

OpenStudy (amistre64):

i gotta run so good luck

OpenStudy (anonymous):

thanks

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