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1 answer:
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The question is incomplete, here is the complete question:
On May 3, 1999, 59 tornadoes hit a certain state in the largest tornado outbreak ever recorded for that state. Sixteen of these were classified as strong (F2 or F3) or violent (F4 or F5). Length of Path (miles) Intensity:
a. 7 F.
b. 9 F2.
c. 16 F3.
d. 11 F3.
e. 8 F2.
f. 9 F2.
g. 5 F2.
h. 16 F4.
i. 15 F3.
j. 38 F5.
k. 42 F4.
l. 3 F2.
m. 8 F2.
n. 1 F2.
o. 14 F3.
p. 23 F3.
Part (a) Make a box and whisker plot
Part (b) Identify outliners. Remove them and redraw the box and whisker diagram
Part (c) How does the removal of outliners effect box and whisker plot? How does is effect the median of the set?
Answer:
Given below with explanations
Step-by-step explanation:
(a)
The data is first arranged in ascending order:
1, 3, 5, 7, 8, 8, 9, 9, 11, 14, 15, 16, 16, 23, 38, 42
Total values = 16
Minimum value = 1
Maximum value = 42
Lower Quartile = 1/4 x 16 = 4th value = 7 (1, 3, 5, 7, 8, 8, 9, 9, 11, 14, 15, 16, 16, 23, 38, 42)
Upper Quartile = 3/4 x 16 = 12th value = 16 (1, 3, 5, 7, 8, 8, 9, 9, 11, 14, 15, 16, 16, 23, 38, 42)
Median = 1/2 x 16 = 8th value = 9 (1, 3, 5, 7, 8, 8, 9, 9, 11, 14, 15, 16, 16, 23, 38, 42)
The Box and whisker plot is attached in the picture below:
(b)
Outliners are 38 and 42
1, 3, 5, 7, 8, 8, 9, 9, 11, 14, 15, 16, 16, 23
Total values = 14
Minimum value = 1
Maximum value = 23
Lower Quartile = 1/4 x 14 = 3.5th value = 6 (1, 3, 5, 7, 8, 8, 9, 9, 11, 14, 15, 16, 16, 23)
(5+7)/2
Upper Quartile = 3/4 x 14 = 10.5th value = 14.5 (1, 3, 5, 7, 8, 8, 9, 9, 11, 14, 15, 16, 16, 23)
(14+15)/2
Median = 1/2 x 14 = 7th value = 9 (1, 3, 5, 7, 8, 8, 9, 9, 11, 14, 15, 16, 16, 23)
The Box and whisker plot is attached in the picture below:
(c)
The lower and upper quartile values have changed after removing the outliners but the median remains same in this case
Answer:
a. The margin of error is 2.29.
b. 19.23 to 23.81
Step-by-step explanation:
The sample size is n=64.
We start by calculating the sample mean and standard deviation with the following formulas:
The sample mean is M=21.52.
The sample standard deviation is s=6.89.
We have to calculate a 99% confidence interval for the mean.
The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.
When σ is not known, s divided by the square root of N is used as an estimate of σM:
The degrees of freedom for this sample size are:
The t-value for a 99% confidence interval and 63 degrees of freedom is t=2.656.
The margin of error (MOE) can be calculated as:
Then, the lower and upper bounds of the confidence interval are:
LL=M-t \cdot s_M = 21.52-2.29=19.23\\\\UL=M+t \cdot s_M = 21.52+2.29=23.81
The 99% confidence interval for the mean is (19.23, 23.81).