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4 years ago

Can someone please help me on 50.

Mathematics

1 answer:

Andrei [34K]4 years ago

5 0

50) MAD=1.6

Follow steps...
To find the mean absolute deviation of the data, start by finding the mean of the data set. Find the sum of the data values, and divide the sum by the number of data values. Find the absolute value of the difference between each data value and the mean: |data value – mean|. Find the sum of the absolute values of the differences. Divide the sum of the absolute values of the differences by the number of data values.

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3 years ago

The amounts (in ounces) of juice in eight randomly selected juice bottles are: 15.8, 15.6, 15.1, 15.2, 15.1, 15.5, 15.9, 15.5. C

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Answer:

The required 97.5% confidence interval is

$\text {CI} = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\\text {CI} = 15.5 \pm 2.8412\cdot \frac{0.31}{\sqrt{8} } \\\\\text {CI} = 15.5 \pm 2.8412\cdot 0.1096\\\\\text {CI} = 15.5 \pm 0.311\\\\\text {CI} = 15.5 - 0.311, \: 15.5 + 0.311\\\\\text {CI} = (15.19, \: 15.81)\\\\$

Therefore, we are 97.5% confident that the actual mean amount of juice in all such bottles is within the range of 15.19 to 15.81 ounces

Step-by-step explanation:

The amounts (in ounces) of juice in eight randomly selected juice bottles are:

15.8, 15.6, 15.1, 15.2, 15.1, 15.5, 15.9, 15.5

Let us first compute the mean and standard deviation of the given data.

Using Excel,

=AVERAGE(number1, number2,....)

The mean is found to be

$\bar{x} = 15.5$

=STDEV(number1, number2,....)

The standard deviation is found to be

$s = 0.31$

The confidence interval for the mean amount of juice in all such bottles is given by

$$ \text {CI} = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $\\\\$

Where $\bar{x}$ is the sample mean, n is the samplesize, s is the sample standard deviation and $t_{\alpha/2}$ is the t-score corresponding to a 97.5% confidence level.

The t-score corresponding to a 97.5% confidence level is

Significance level = α = 1 - 0.975 = 0.025/2 = 0.0125

Degree of freedom = n - 1 = 8 - 1 = 7

From the t-table at α = 0.0125 and DoF = 7

t-score = 2.8412

So the required 97.5% confidence interval is

Therefore, we are 97.5% confident that the actual mean amount of juice in all such bottles is within the range of 15.19 to 15.81 ounces.

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3 years ago