4. Given the following data set.NameHomeworkMidtermFinalFrank9795929296EmilyWilliam93959088Janet8883757984BettyJoeBob86858286726372(a) Calculate the mean and standard deviation of all final scores.(b) Construct a 95% confidence interval for the mean homework score.(c) Given a=.05, determine whether the mean of Midterm is more than 85. Set up appropriatehypotheses, calculate your test statistic and report your P-value.

Name: 4. Given the following data set.NameHomeworkMidtermFinalFrank97959292
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Description: 4. Given the following data set.NameHomeworkMidtermFinalFrank9795929296EmilyWilliam93959088Janet8883757984BettyJoeBob86858286726372(a) Calculate the mean and standard deviation of all final scores.(b) Construct a 95% confidence interval for the mean

Answered: 4. Given the following data set. Final…

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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4. Given the following data set.
Name
Homework
Midterm
Final
Frank
97
95
92
92
96
Emily
William
93
95
90
88
Janet
88
83
75
79
84
Betty
Joe
Bob
86
85
82
86
72
63
72
(a) Calculate the mean and standard deviation of all final scores.
(b) Construct a 95% confidence interval for the mean homework score.
(c) Given a=.05, determine whether the mean of Midterm is more than 85. Set up appropriate
hypotheses, calculate your test statistic and report your P-value.

Expert Solution

Step 1

(a) To calculate the mean and standard deviation of all final scores:

Mean is the sum of observations divided by the total no. of observations.

$\begin{array}{rcl} \bar{x} & = & \frac{92 + 96 + 88 + 75 + 84 + 86 + 72}{7} \\ = & \frac{593}{7} \\ = & 84.71428571 \\ \approx & 84.7143 \end{array}$

Standard deviation is calculated using the formula:

$\begin{array}{rcl} s & = & \sqrt{\frac{\sum_{i} {(x_{i} - \bar{x})}^{2}}{n - 1}} \\ = & \sqrt{\frac{{(92 - 84.7143)}^{2} + {(96 - 84.7143)}^{2} + {(88 - 84.7143)}^{2} + . . . . . . . . {(72 - 84.7143)}^{2}}{7 - 1}} \\ = & 8.654754 \\ \approx & 8.6548 \end{array}$

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