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 0.86 1.00 0.92 0.85 0.810.78 0.93 0.93 1.02 0.93 1.06 1.06 0.96 0.81 0.98Assume the population distribution of cadence is approximately normal. A descriptive summary of the data is as follows:VariablecadenceVariablecadenceN20Mean0.9240Min0.7800Median0.9300Max1.0600St Dev0.0792Q10.8550Q30.9700(a) Calculate and interpret a 95% confidence interval for population mean cadence. (Round your answers to four decimal places.)strides per secondInterpret this interval.With 95% confidence, the value of the true mean cadence of all such men falls within the confidence interval.The probability the true mean cadence is within this interval is 0.95.With 95% confidence, the value of the true mean cadence of all such men falls above the confidence interval.With 95% confidence, the value of the true mean cadence of all such men falls below the confidence interval.

The sample size of healthy men selected using random selection, Provided data on cadence:The…

Answered: A study of the ability of individuals…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Suppose students' exam scores are normally distributed with mean u = 65 and standard deviation O = =10 What proportion of the students taking the test score at most 80? O a) 0.07 O b) 0.95 O c) 0.14 d) 0.93 e) 0.99
If you have these information's, draw the histogram Mid-Point frequency Cumulative (0.0,0.1] (0.1,0.2] (0.2,0.3] (0.3,0.4] (0.4,0.5] (0.5,0.6] (0.6,0.7) (0.7,0.8] (0.8,0.9] (0.9,1.0] (1.0,1.1] (1.1,1.2] (1.2,1.3] (1.3,1.4] (1.4,1.5] (1.5,1.6] (1.6,1.7] (1.7,1.8] (1.8,1.9] (1.9,2.0] 0.05 1 1 0.15 35 36 0.25 230 266 0.35 609 875 0.45 1029 1904 0.55 1226 3130 0.65 1247 4377 0.75 1090 5467 0.85 858 6325 0.95 596 6921 1.05 426 7347 1.15 300 7647 1.25 167 7814 1.35 96 7910 1.45 37 7947 1.55 26 7973 1.65 13 7986 1.75 8 7994 1.85 4 7998 1.95 2 8000
A random sample of 100,000 credit sales in a department store showed an average sale of $81.25. From past data, it is known that the standard deviation of the population is $10.00. Determine the standard error of the mean. O 0.0001 O 0.0100 O 0.0316 O 10.00
A researcher found that the two groups (n=25 in each group) they tested were significantly different from each other with a t-value of 2.91, which has a corresponding p-value of p=.06. Report the findings in APA number format.
A sample of 314 patients between 38 and 82 were given a drug intended to reduce their level of cholesterol. The results were the mean reduction in cholesterol of 0.94 millimole per liter. Assume the population standard deviation is σ = 0.18. Construct a 98% C.I.
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Referring to Table 2, the mean square for the factor filling time is____ 1). 0.9030 2). 0.4234 3). 0.4515 4). 0.2117
1:37 PM Wed Apr 21 * 72% 88 Q Stats Exam •.. Untitled Notebook (1) Stats Exam 37. The manager of Green Valley medical center claims that the median number of patients seen by doctors who work at the center is 85. To test this claim, 20 days are randomly selected and the number of patients seen is recorded as shown. At a=0.1, test the claim. (Use sign test) 86 82 90 88 85 85 94 93 69 61 89 90 91 94 78 85 71 77 69 77

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