A study of the ability of individuals to walk in a straight line reported the ac

Question

A study of the ability of individuals to walk in a straight line reported the accompanying data on cadence (strides per second) for a sample of n = 20 randomly selected healthy men. 0.93 0.85 0.92 0.95 0.93 0.88 1.00 0.92 0.85 0.81 0.76 0.93 0.93 1.01 0.93 1.06 1.09 0.96 0.81 0.98 A normal probability plot glves substantial support to the assumption that the population distribution of cadence is approximately normal. A descriptive summary of the data from Minitab follows. an StDev SEMean 20 0.9250 0.93000.9250 0.0828 0.0185 Variable cadence Variable cadence Mean Median TMean Min 0.7600 1.0900 0.8650 0.9700 Max 91 93 (a) Calculate and interpret a 95% confidence interval for population mean cadence. (Round your answers to four decimal places.) strides per second Interpret this interval with 95% confidence, the value of the true mean cadence of all such men falls above the confidence with 95% confidence, the value of the true mean cadence of all such men falls below the confidence with 95% confidence, the value of the true mean cadence of all such men falls inside the confidence (b) Calculate and interpret a 95% prediction interval for the cadence of a single individual randomly selected from this population. (Round your answers to four decimal places.) strides per second Interpret this interval. If this bound is calculated sample after sample, in the long run, 95% of these bounds will fail to capture a future individual value of cadence for a healthy man. If this bound is calculated sample after sample, in the long run, 95% of these bounds will capture a future individual value of cadence for a healthy man. If this bound is calculated once, there is a 95% chance that these bounds will capture a future individual value of cadence for a healthy man. If this bound is calculated once, there is a 5% chance that these bounds will capture a future individual value of cadence for a healthy man. (e) Calculate an interval that includes at least 99% of the cadences in the population distribution using a confidence level of 95%. (Round your answers to four decimal places.) X strides per second Interpret this interval.

Explanation / Answer

Answer

A.a) Given data from Minitab output

Mean=0.9250

Standard deviation(SD)=0.0828

Z-value from table=1.96

n=20

Confidence interval for population mean

=mean±Z*SD/sqrt(n)

Confidence interval is 0.9250-1.96*0.0828/sqrt(20) and 0.9250+1.96*0.0828/sqrt(20)

Solving, we get

Confidence interval=(0.8887,0.9613)

A.b) formula for prediction interval is given as

=mean± Z*SD*Sqrt(1+1/n)

Prediction interval is 0.9250-1.96*0.0828*sqrt(1+1/20) and 0.9250+1.96*0.0828*sqrt(1+1/20)

Solving, we get

Prediction interval=(0.8401, 1.009)

A.c) for interval to include 99% of cadences in the population

We have confidence interval as mean±3*SD

Confidence interval=0.9250-3*0.0828 and 0.9250+3*0.0828

Hence, the confidence interval is (0.6766,1.1734)

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