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KonstantinChe [14]
3 years ago
9

The owner of a manufacturing plant employs eighty people. As part of their personnel file, she asked each one to record to the n

earest one-tenth of a mile the distance they travel one way from home to work. The distances for a random sample of six employees are listed below: 26 32 29 16 45 19 Find the variance for the given data.
Mathematics
1 answer:
Mashcka [7]3 years ago
3 0

Answer:

The variance of the given data is 106.9667

Step-by-step explanation:

The variance, S² of a sample is given by the expression;

S^2 = \dfrac{\sum \left (x_{i}-\bar{x}  \right )^{2}}{n-1}

Where:

x_{i} = One of the observations

\bar{x} = The mean of the observations

n = Number in the sample = 6

The mean = Σx_{i}/n = (26  + 32 + 29  + 16  + 45  + 19 )/6 = 167/6 = 27.833

Therefore, we have;

S² = ((26 - 27.833)²+(32 - 27.833)²+(29 - 27.833)²+(16 - 27.833)²+(45 - 27.833)²+(19 - 27.833)²)/(6 - 1) = 106.9667.

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Suppose the linear regression line y=2.1x+130 predicts sales based on the money spent on advertising. If x represents the dollar
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A manufacturer of detergent claims that the contents of boxes sold weigh on average at least 20 ounces. The distribution of weig
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Answer:

z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625    

The p value for this case would be given by:

p_v =P(z  

Since the p value is higher than the significance level provided we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly less than 20 ounces.

Step-by-step explanation:

Information provided

\bar X=19.87 represent the sample mean

\sigma=0.4 represent the population deviation

n=25 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0.01 represent the significance level

z would represent the statistic

p_v represent the p value

Hypothesis to test

We want to test if the true mean is at least 20 ounces, the system of hypothesis would be:  

Null hypothesis:\mu \geq 20  

Alternative hypothesis:\mu < 20  

The statistic is given by:

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}  (1)  

Replacing the info given we got:

z=\frac{19.87-20}{\frac{0.4}{\sqrt{25}}}=-1.625    

The p value for this case would be given by:

p_v =P(z  

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