# Let X be the number of hours slept per night by an adult. Assume that X is nor
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# Let X be the number of hours slept per night by an adult. Assume that X is normally distributed about a mean value with standard deviation = 0.5 hours. Suppose that 400 adults are surveyed, and are found to have slept an average of 6.5 hours the preceding night. What is the 90% confidence interval for the mean of X? #Referring to question above, suppose we want to estimate the number of hours slept with a margin of error of at most 10 minutes (1/6 hour), with confidence of 95%. How many people must we survey?Explanation / Answer
# Let X be the number of hours slept per night by an adult. Assume that X is normally distributed about a mean value with standard deviation = 0.5 hours. Suppose that 400 adults are surveyed, and are found to have slept an average of 6.5 hours the preceding night. What is the 90% confidence interval for the mean of X
alpha / 2 = 0.05
Z= 1.64 from tables of normal distribution
I: 6.5 +/- 1.64 * 0.5 / srqt 400
6.5 +/- 0.041
6.459 < miu < 6.541
#Referring to question above, suppose we want to estimate the number of hours slept with a margin of error of at most 10 minutes (1/6 hour), with confidence of 95%. How many people must we survey?
n = ( z * sigma / error ) ^2
alpha / 2 = 0.025 Z= 1.96
n = ( 1.96 * 0.5 / (1/6) ) ^2
n = 34.5744
n = 35
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