A study of the ability of individuals to walk in a straight line reported the ac
ID: 3358460 • Letter: A
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
0.79 0.93 0.93 1.05 0.93 1.06 1.06 0.96 0.81 0.99
A normal probability plot gives substantial support to the assumption that the population distribution of cadence is approximately normal. A descriptive summary of the data from Minitab follows.
Variable N Mean Median TrMean StDev SEMean
cadence 20 0.9275 0.9300 0.9278 0.0801 0.0179
Variable Min Max Q1 Q3
cadence 0.7900 1.0600 0.8650 0.9750
(c) 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.)
Explanation / Answer
Solution:
From the given information
Sample size n = 20,
Sample mean x = 0.9275
Standard deviation s = 0.0801
c) We have to determine an interval that includes at least 99% of the cadences in the population distribution using a confidential level of 95%.
The tolerance interval that includes at least 99% of the cadeness in the population distribution distribution using a confidence level of 95%.
Using Tolerance Critical values for Normal Population distributions table, for a confidence level of 95% with n = 20, a teo-sided tolerance interval capturing at least 99% of the cadeness in the population sampled uses the tolerance critical value is 3.615
Thus, the required interval is given by
{ x ± (TOlerance critical value) * s} = { 0.9275 ± (3.615) * 0.0801}
= { 0.9275 ± 0.2896 }
= {0.6379, 1.2171}
That is, we can be 95% confidence that the interval (0.6379, 1.2171) will include at least 99% of the cadeness in the population.
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