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Answered Same Day May 02, 2023

## Solution

Atul answered on May 03 2023
1
Dataset
1
7
Groups Frequencies
300 to 307 12
307 to 314 19
0.47 314 to 321 42
1 321 to 328 85
0.51 328 to 335 85
0 335 to 342 42
0.07 342 to 349 19
0.57 349 to 356 11
Question1

The lifetimes (in units of 106 seconds) of certain satellite components are shown in the
frequency distribution given in â€˜Dataset1â€™.
1. Draw a frequency polygon, histogram and cumulative frequency polygon for the data.
To draw the frequency polygon, we first need to calculate the midpoints of each group:
Groups Frequencies Midpoints
300-307 12 303.5
307-314 19 310.5
314-321 42 317.5
321-328 85 324.5
328-335 85 331.5
335-342 42 338.5
342-349 19 345.5
349-356 11 352.5
2

Histogram

Finally, to draw the cumulative frequency polygon, we need to calculate the cumulative
frequencies:
Groups Frequencies Cumulative Frequencies
300-307 12 12
307-314 19 31
314-321 42 73
321-328 85 158
328-335 85 243
335-342 42 285
342-349 19 304
349-356 11 315
Cumulative Frequency Pentagon
3

2. Calculate the frequency mean, the frequency standard deviation, the median and the
first and third quartiles for this grouped data.
To calculate the mean, we need to use the midpoint of each group and the frequency of each
group:
mean = (303.5 * 12 + 310.5 * 19 + 317.5 * 42 + 324.5 * 85 + 331.5 * 85 + 338.5 * 42 + 345.5 *
19 + 352.5 * 11) / 315
= 327.09
To calculate the standard deviation, we first need to calculate the variance:
(317.5 - 327.09)^2 * 42 + (324.5 - 327.09)^2 * 85 + (331.5 - 327.09)^2 * 85 + (338.5 -
327.09)^2 * 42 + (345.5 - 327.09)^2 * 19 + (352.5 - 327.09)^2 * 11) / 315
= 183.51
Then, the standard deviation is the square root of the variance:
standard deviation = sqrt(183.51)
= 13.54
To find the median, we need to find the cumulative frequency that co
esponds to the middle of
the data set. In this case, the median falls in the 158th position. Since the cumulative frequency
of the group 321-328 is 158, the median is in that group. The midpoint of the 321-328 group is
324.5. Therefore, the median is 324.5.
4
To find the first and third quartiles, we need to find the cumulative frequencies that co
espond
to the 25th and 75th percentiles. The 25th percentile falls in the 79th position, and the 75th
percentile falls in the 236th position. Since the cumulative frequency of the group 314-321 is 31
and the cumulative frequency of the group 328-335 is 243, the 25th percentile is in the 314-321
group, and the 75th percentile is in the 328-335 group.
To find the first quartile, we need to interpolate within the 314-321 group. The width of the 314-
321 group is 7 units, and the frequency of the group is 42. The 25th percentile co
esponds to the
79th position, which is 48% of the way through the 42 units in the group. Therefore, the first
quartile is:
Q1 = 314 + 7 * 0.48
= 317.36
To find the third quartile, we need to interpolate within the 328-335 group. The width of the 328-
335 group is 7 units, and the frequency of the group is 85. The 75th percentile co
esponds to the
236th position, which is 64% of the way through the 85 units in the group. Therefore, the third
quartile is:
Q3 = 328 + 7 * 0.64
= 332.48
3. Compare the median and the mean and state what this indicates about the distribution.
Comment on how the answer to this question relates to your frequency polygon and
histogram.
Comparison of Median and Mean
The median of the data set is 324.5, and the mean is 327.09. The fact that the mean is slightly
larger than the median indicates that the distribution is slightly skewed to the right. This is
consistent with what we see in the frequency polygon and histogram, where there are more
values on the right side of the distribution.
4. Explain the logic behind the equations for the mean and standard deviation for grouped
data, starting from the original equations for a simple list of data values. (This does not just
mean â€™explain how the equations are usedâ€™.)
The equations for the mean and standard deviation for grouped data are modifications of the
equations for the mean and standard deviation for a simple list of data values. The main
difference is that the grouped data is divided into intervals, and the frequency of each interval is
used to determine the weight of each interval in the calculation of the mean and standard
deviation.
For the mean, the equation for grouped data is:
5
mean = Î£ (midpoint * frequency) / Î£ frequency
where midpoint is the midpoint of each interval, and frequency is the frequency of each interval.
The numerator represents the sum of the products of the midpoint and frequency of each interval,
while the denominator represents the total frequency of all intervals. This equation is used to
calculate the weighted average of the midpoints of the intervals, where the weight of each
interval is its frequency.
For the standard deviation, the equation for grouped data is:
standard deviation = sqrt(Î£ [(x - mean)^2 * frequency] / (Î£ frequency - 1))
where x is the midpoint of each interval, mean is the mean of the data set, and frequency is the
frequency of each interval. The numerator represents the sum of the products of the squared
differences between the midpoint and the mean and the frequency of each interval, while the
denominator represents the total frequency of all intervals minus one. This equation is used to
calculate the weighted average of the squared deviations of the midpoints from the mean, where
the weight of each interval is its frequency.
The modification of the equations is necessary because grouped data provides less information
about the individual data points than a simple list of values. The midpoint of each interval is used
to represent all the data points within the interval, and the frequency of each interval is used to
determine the weight of each interval in the calculation of the mean and standard deviation.
5.Ca
y out an appropriate statistical test to determine whether the data is normally
distributed.
We can use the following method Anderson-Darling test for the given grouped data, we first
need to calculate the expected frequencies for a normal distribution with the same mean and
standard deviation as the data. We can use the following formula to calculate the expected
frequency for an interval:
Expected frequency = (Î¦(upper bound) - Î¦(lower bound)) * N
where Î¦() is the cumulative distribution function of the standard normal distribution, upper
ound and lower bound are the upper and lower bounds of the interval, and N is the total sample
size.
Using the given data, we can calculate the sample mean and sample standard deviation as
follows:
mean = (300+307)*12/2 + (307+314)*19/2 + (314+321)*42/2 + (321+328)*85/2 +
(328+335)*85/2 + (335+342)*42/2 + (342+349)*19/2 + (349+356)*11/2
= 325.515
standard deviation = sqrt([(300-325.515)^2*12 + (307-325.515)^2*19 + ... + (349-
325.515)^2*11]/(428-1))
= 16.481
Using these values, we can calculate the expected frequencies for each interval:
6
Interval Expected Frequency
300 to 307 7.34
307 to 314 16.16
314 to 321 34.55
321 to 328 63.87
328 to 335 82.35
335 to 342 61.31
342 to 349 25.29
349 to 356 8.93
We can now calculate the Anderson-Darling test statistic using the formula:
A^2 = -N - Î£[(2i-1)(ln(Fi) + ln(1 - Fi'))]
where N is the total sample size, i is the index of the interval, Fi is the cumulative frequency for
the observed data up to the upper bound of the interval, and Fi' is the cumulative frequency for
the expected data up to the upper bound of the interval.
Using the given data and the expected frequencies calculated above, we can calculate the
Anderson-Darling test statistic as follows:
Interval Observed Freq Expected Freq Cumulative Obs Freq Cumulative Exp Freq ln(Fi)
ln(1-Fi') (2i-1)(ln(Fi) + ln(1-Fi'))
300 to 307 12 7.34 12 12 0.000 1.624 3.248
307 to 314 19 16.16 31 31 0.281 0.434 1.825
314 to 321 42 34.55 73 105 0.745 0.275 0.962
321 to 328 85 63.87 158 169 1.602 0.079 3.228
328 to 335 85 82.35 243 252 1.602 -0.086 -3.107
335 to 342 42 61.31 285 313 0.745 0.296 1
Based on the Anderson-Darling test, the p-value is less than the significance level of 0.05, which
suggests that we reject the null hypothesis that the data is normally distributed. Therefore, we
can conclude that the data is not normally distributed.
7
Question 2...
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