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photoshop1234 [79]
3 years ago
15

Use the box plots comparing the number of males and number of females attending the latest superhero movie each day for a month

to answer the questions.
Part A: Estimate the IQR for the males' data. (2 points)

Part B: Estimate the difference between the median values of each data set. (2 points)

Part C: Describe the distribution of the data and if the mean or median would be a better measure of center for each. (4 points)

Part D: Provide a possible reason for the outlier in the data set. (2 points)

Mathematics
1 answer:
Maksim231197 [3]3 years ago
4 0

Answer:

Part A:

The interquartile range is approximately 10

Part B:

The difference between the median values for each data set  is approximately 6

Part C:

i) More widely distributed and concentrated to the beginning of the month

The better measure of the center for the male dataset is the median

ii) The skewed distribution

The mean is the better measure of center for the dataset

Part D;

A possible reason for the outlier is by chance

Step-by-step explanation:

Part A:

The interquartile range, IQR, is the width of the box in the box plot = The difference between the third quartile, Q₃ and the first quartile, Q₁ = Q₃ - Q₁

From the diagram, Q₃ ≈ 13, Q₁ ≈ 3

∴ The interquartile range, IQR = 13 - 3 = 10

Part B:

From the diagram of the box plot, the median value, of the males, M-Q₂ ≈ 12

The median value for the females, F-Q₂ ≈ 18

The difference, M-Q₂ - F-Q₂ = 18 - 12 ≈ 6

The difference between the median values for each data set, d = M-Q₂ - F-Q₂ ≈ 6

Part C:

i) The distribution of the of the male dataset for the male has more data on the left of the box plot, with the median located approximately at the center and an outlier at end of the month

The better measure of the center for the male dataset is the median

ii) The distribution of the dataset for the females is skewed (concentrated on the right of the plot). The data set has no outlier and the first quartile is 0, while the third quartile corresponds with the maximum value at 21, the first quartile is 15 and the IQR is approximately 6

The better measure of center for the dataset is the mean, as the difference between the value of the first quartile, 0, and the other data points is more accounted for by the mean

Part D;

The possible reason for the outlier in the male dataset can be attributed to chance

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Step-by-step explanation:

Hello!

The objective is to test if the two machines are filling the bottles with a net volume of 16.0 ounces or if at least one of them is different.

The parameter of interest is μ₁ - μ₂, if both machines are filling the same net volume this difference will be zero, if not it will be different.

You have one sample of ten bottles filled by each machine and the net volume of the bottles are measured, determining two variables of interest:

Sample 1 - Machine 1

X₁: Net volume of a plastic bottle filled by the machine 1.

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Sample 2 - Machine 2

X₂: Net volume of a plastic bottle filled by the machine 2

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With p-values 0.4209 (for X₁) and 0.6174 (for X₂) for the normality test and using α: 0.05, we can say that both variables of interest have a normal distribution:

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Using a statistic software I've calculated the test

F_{H_0}= 1.41

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To study the difference between the population means you can use

t= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }  ~~t_{n_1+n_2-2}

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H₁: μ₁ - μ₂ ≠ 0

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Sa= \sqrt{\frac{(n_1-1)S^2_1+(N_2-1)S^2_2}{n_1+n_2-2}} = \sqrt{\frac{9*9.2*10^{-4}+9*6.5*10^{-4}}{10+10-2} } = 0.028= 0.03

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The p-value for this test is: 0.882433

The p-value is greater than the significance level, the decision is to not reject the null hypothesis.

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