Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393614046
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Stacey Lowery Bretz, Geoffrey Davies
Publisher: W. W. Norton & Company
Question
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Chapter 1, Problem 1.74QP

(a)

Interpretation Introduction

Interpretation: The mean, standard deviation for the given set of data and the given value comes under the 95% confidence interval or not is to be determined.

Concept introduction: The mean of the given set of data is calculated by using the formula,

x¯=xin

The standard deviation for the given sets of data is calculated by using the formula,

s=(xix¯)2n1

To determine: The mean and standard deviation of the given set of data.

(a)

Expert Solution
Check Mark

Answer to Problem 1.74QP

Solution

The mean of the given set of data is 105.4mg/dL and the standard deviation is 3.91mg/dL .

Explanation of Solution

Explanation

Given

The data for the diabetes from the given series of blood samples gives the results: 106,99,109,108 and 105mg/dL .

The mean of the given set of data is calculated by using the formula,

x¯=xin

Where,

  • x¯ is the mean of the given set of data.
  • xi is the ith value of the distribution.
  • n is the number of values in the data.

Substitute the values of xi and n for the given data in the above equation.

x¯=106+99+109+108+1055mg/dL=105.4mg/dL_

The standard deviation for the given sets of data is calculated by using the formula,

s=(xix¯)2n1

Where,

  • x¯ is the mean of the given set of data.
  • xi is the ith value of the distribution.
  • n is the number of values in the data.
  • s is the standard deviation of the data.

Substitute the values of xi , x¯ and n for manufacturer 1 in the above equation.

s=(106105.4)2+(99105.4)2+(109105.4)2+(108105.4)2+(105105.4)251=0.36+40.96+12.96+6.76+0.164=3.91mg/dL_

(b)

Interpretation Introduction

To determine: If the given value 120mg/dL is in the 95% confidence interval of the given data or not.

(b)

Expert Solution
Check Mark

Answer to Problem 1.74QP

Solution

The value 120mg/dL is much higher than the confidence interval of the given data.

Explanation of Solution

Explanation

Given

The mean of the given data is 105.4mg/dL .

The standard deviation of the given set of data is 3.91mg/dL .

The value of n is 5 .

The value of confidence level at 95% is 2.776 .

The confidence interval of the given set of data is calculated by using the formula,

μ=x¯±tsn

Where,

  • μ is the confidence interval.
  • x¯ is the mean of the given data.
  • t is the value of confidence level.
  • s is the standard deviation of the given data.
  • n is the number of values in the data.

Substitute the values of x¯ , t , s and n in the above equation.

μ=105.4±2.776×3.915=105.4±4.9_

Conclusion

The mean of the given set of data is 105.4mg/dL and the standard deviation is 3.91mg/dL .

The value 120mg/dL is much higher than the confidence interval of the given data

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Chapter 1 Solutions

Chemistry: The Science in Context (Fifth Edition)

Ch. 1 - Prob. 1.2VPCh. 1 - Prob. 1.3VPCh. 1 - Prob. 1.4VPCh. 1 - Prob. 1.5VPCh. 1 - Prob. 1.6VPCh. 1 - Prob. 1.7VPCh. 1 - Prob. 1.8VPCh. 1 - Prob. 1.9QPCh. 1 - Prob. 1.10QPCh. 1 - Prob. 1.11QPCh. 1 - Prob. 1.12QPCh. 1 - Prob. 1.13QPCh. 1 - Prob. 1.14QPCh. 1 - Prob. 1.15QPCh. 1 - Prob. 1.16QPCh. 1 - Prob. 1.17QPCh. 1 - Prob. 1.18QPCh. 1 - Prob. 1.19QPCh. 1 - Prob. 1.20QPCh. 1 - Prob. 1.21QPCh. 1 - Prob. 1.22QPCh. 1 - Prob. 1.23QPCh. 1 - Prob. 1.24QPCh. 1 - Prob. 1.25QPCh. 1 - Prob. 1.26QPCh. 1 - Prob. 1.27QPCh. 1 - Prob. 1.28QPCh. 1 - Prob. 1.29QPCh. 1 - Prob. 1.30QPCh. 1 - Prob. 1.31QPCh. 1 - Prob. 1.32QPCh. 1 - Prob. 1.33QPCh. 1 - Prob. 1.34QPCh. 1 - Prob. 1.35QPCh. 1 - Prob. 1.36QPCh. 1 - Prob. 1.37QPCh. 1 - Prob. 1.38QPCh. 1 - Prob. 1.39QPCh. 1 - Prob. 1.40QPCh. 1 - Prob. 1.41QPCh. 1 - Prob. 1.42QPCh. 1 - Prob. 1.43QPCh. 1 - Prob. 1.44QPCh. 1 - Prob. 1.45QPCh. 1 - Prob. 1.46QPCh. 1 - Prob. 1.47QPCh. 1 - Prob. 1.48QPCh. 1 - Prob. 1.49QPCh. 1 - Prob. 1.50QPCh. 1 - Prob. 1.51QPCh. 1 - Prob. 1.52QPCh. 1 - Prob. 1.53QPCh. 1 - Prob. 1.54QPCh. 1 - Prob. 1.55QPCh. 1 - Prob. 1.56QPCh. 1 - Prob. 1.57QPCh. 1 - Prob. 1.58QPCh. 1 - Prob. 1.59QPCh. 1 - Prob. 1.60QPCh. 1 - Prob. 1.61QPCh. 1 - Prob. 1.62QPCh. 1 - Prob. 1.63QPCh. 1 - Prob. 1.64QPCh. 1 - Prob. 1.65QPCh. 1 - Prob. 1.66QPCh. 1 - Prob. 1.67QPCh. 1 - Prob. 1.68QPCh. 1 - Prob. 1.69QPCh. 1 - Prob. 1.70QPCh. 1 - Prob. 1.71QPCh. 1 - Prob. 1.72QPCh. 1 - Prob. 1.73QPCh. 1 - Prob. 1.74QPCh. 1 - Prob. 1.75QPCh. 1 - Prob. 1.76QPCh. 1 - Prob. 1.77QPCh. 1 - Prob. 1.78QPCh. 1 - Prob. 1.79QPCh. 1 - Prob. 1.80QPCh. 1 - Prob. 1.81QPCh. 1 - Prob. 1.82QPCh. 1 - Prob. 1.83QPCh. 1 - Prob. 1.84QPCh. 1 - Prob. 1.85QPCh. 1 - Prob. 1.86QPCh. 1 - Prob. 1.87QPCh. 1 - Prob. 1.88QPCh. 1 - Prob. 1.89APCh. 1 - Prob. 1.90APCh. 1 - Prob. 1.91APCh. 1 - Prob. 1.92APCh. 1 - Prob. 1.93APCh. 1 - Prob. 1.94APCh. 1 - Prob. 1.95APCh. 1 - Prob. 1.96APCh. 1 - Prob. 1.97APCh. 1 - Prob. 1.98AP
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