Practice of Statistics in the Life Sciences
Practice of Statistics in the Life Sciences
4th Edition
ISBN: 9781319013370
Author: Brigitte Baldi, David S. Moore
Publisher: W. H. Freeman
Question
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Chapter 17, Problem 17.15AYK

(a)

To determine

To Explain: the confidence interval on the basis provided information.

(a)

Expert Solution
Check Mark

Answer to Problem 17.15AYK

Confidence interval is 0.8176 and 1.3278

Explanation of Solution

Given:

Practice of Statistics in the Life Sciences, Chapter 17, Problem 17.15AYK

On the basis of given output of confidence interval is 0.8176 and 1.3278 which implies that there are 95% of surety that the mean difference in tremor amplitude with ACT device off and ACT device on for the patients diagnosed with essential tremor.

(b)

To determine

To find: the mean and standard deviation.

(b)

Expert Solution
Check Mark

Answer to Problem 17.15AYK

  x¯=1.07s=0.379

Explanation of Solution

Given:

    ACT offACT on
    1.20.3
    1.60.2
    20.6
    10.2
    1.70.5
    1.90.4
    0.80.2
    1.70.6
    1.60.3
    0.50.2
    1.40.1

Formula used:

  x¯=dn

  s=(dx¯)2n1

Calculation:

    ACT offACT ondifference = ACT off-ACT ond-

      x¯

    (dx¯)2
    1.20.30.9-0.170.03
    1.60.21.40.330.11
    20.61.40.330.11
    10.20.8-0.270.07
    1.70.51.20.130.02
    1.90.41.50.430.18
    0.80.20.6-0.470.22
    1.70.61.10.030.00
    1.60.31.30.230.05
    0.50.20.3-0.770.60
    1.40.11.30.230.05
    Sum11.81.44

  x¯=dn=11.811=1.07s=(dx¯)2n1=1.44111=0.379

(c)

To determine

To find: the margin of error of a 95% confidence interval for the mean difference.

(c)

Expert Solution
Check Mark

Answer to Problem 17.15AYK

Margin of Error = 0.2546

Explanation of Solution

Given:

  n=11

From the part (b)

  x¯=1.07s=0.379

Formula used:

  MarginofError = tα2×sn

Calculation:

For 0.05 level of significance and

Degree of Freedom

  df=n1=111=10

  t0.052,df=10=2.228

  MarginofError = tα2×sn=2.228×0.37911=0.2546

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