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STAT 100 - C40 Elementary Statistics for Applications Assignment - 2 Due on 20 May, 2020 by 1 pm Total Marks: 25 Please upload your solutions in UR courses MARKS 1. The following data represent the...

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STAT 100 - C40 Elementary Statistics for Applications
Assignment - 2
Due on 20 May, 2020 by 1 pm
Total Marks: 25
Please upload your solutions in UR courses
MARKS
1. The following data represent the social ambivalence scores for 14 people as measured by a[9]
psychological test. (The higher the score, the stronger the ambivalence.)
XXXXXXXXXX XXXXXXXXXX
Find the interquartile range, five number summary and sketch a box and whisker plot for the
data. Find if there are any outlier(s)?
2. A social skills training program was implemented with seven mildly challenged students in[10]
a study to determine whether the program caused improvement in pre/post measures and
ehavior ratings. For one such test, the pre - and posttest scores for the seven students are
given in the table.
Subject Pretest, x Posttest, y
Earl XXXXXXXXXX
Ned 90 88
Jasper XXXXXXXXXX
Charlie 106 98
Tom 91 104
Susie 91 94
Lori 89 98
a) Calculate the linear co
elation coefficient, r. Describe the relationship.
) Calculate the line-of-best-fit(linear regression line).
c) Estimate the post test score if the pretest score was 98.
3. The linear co
elation coefficient of a set of data points is -0.85.[2]
a) Is the slope of the regression line positive or negative?
) Determine the coefficient of determination.
4. A regression analysis between sales (y in $1000) and advertising (x in $100) resulted in the[2]
following least-squares line: ŷ = 82 + 7x. Given this information, if advertising costs were
$900, what could we reasonably expect the amount of sales (in dollars) to be?
5. A regression analysis between weight (y in kg) and height (x in cm) resulted in the following[2]
least-squares line: ŷ = –20 + 0.5x. Taking this into consideration, if the height is increased
y 1 cm, what does this imply about the change in the weight, on average?
1
Answered Same Day May 20, 2021

Solution

Pooja answered on May 20 2021
162 Votes
1)
Data: 18 12 11 4 5 7 8 5 8 10 11 9 16 14
Sorted data: 4 5 5 7 8 8 9 10 11 11 12 14 16 18
IQR = Q3- Q1
IQR = QUARTILE(B2:B15,3)-QUARTILE(B2:B15,1)
IQR = 4.5
5 number summary:    
Min = 4
Q1 = (1/4*14)th observation = 3.5th observation = 25% of the observations are below this value = 7+ 0.25 = 7.25
Q2 = (2/4*14)th observation = 7th observation = 50% of the observations are below this value = 9+ 0.5 = 9.5
Q3 = (3/4*14)th observation = 10.5th observation =75% of the observations are below this value = 11+0.75 = 11.75
Max = 18
There are not evident outliers in the...
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