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IrinaVladis [17]
3 years ago
14

A Internet provider has implemented a new process for handling customer complaints. Based on a review of customer complaint data

for the past 2 years, the mean time for handling a customer complain was 27 minutes. Three months after implementing the plan, a random sample of the records of 50 customers who had complaints produced the following response times. Use the 50 data values to determine if the new process has reduced the mean time to handle customer complaints. 32.3 26.9 25.4 32.9 27.7 32.2 24.8 20.5 30.4 21.3 25.9 27.1 19.2 28.4 18.0 33.1 31.1 21.9 33.4 24.3 25.5 29.6 32.7 21.3 31.8 27.6 17.4 26.9 18.9 28.6 23.5 21.6 20.1 30.9 26.8 28.7 24.6 21.5 21.9 28.3 24.1 28.9 29.8 27.1 23.8 25.3 30.7 27.2 19.0 30.0
a. Estimate the mean time for handling a customer complaint under the new process using a 95% confidence interval.

b. Is there substantial evidence (a  .05) that the new process has reduced the mean time to handle a customer complaint?

c. What is the population about which inferences from these data can be made?
Mathematics
1 answer:
sergey [27]3 years ago
5 0

Answer:

<em>a)</em><em> The mean time for handling a customer complaint under the new process using 95% confidence level is between 25.0 min. and 27.5 min. </em>

<em>b) </em><em>there is no substantial evidence</em><em> </em><em>(in 95% confidence level) that the new process has reduced the mean time to handle a customer complaint.</em>

<em>c) </em><em>The population about which inferences from these data can be made is the people following the new implemented plan when handling customer complaints.</em>

Step-by-step explanation:

the random sample of the response times of 50 customers who had complaints has

size=50

mean≈26.218

standard deviation ≈4.42

a. Confidence interval can be estimated using the formula:

M±\frac{z*s}{\sqrt{N} } where

  • M is the sample mean (26.22)
  • z is the corresponding z-score for the 95% confidence interval (1.96)
  • s is the sample standard deviation (4.42)
  • N is the sample size (50)

When we put the numbers in the formula, mean time for handling a customer complain under the new process using 95% confidence interval is:

26.22±\frac{1.96*4.42}{\sqrt{50} }≈ 26.22±1.225 i.e. between 25 and 27.45

b<em>. </em>According to the customer complaint data for the past 2 years, the mean time for handling a customer complaint was 27 minutes. After the plan, estimated mean time for handling customer complaint is between <em>25.0 min. and 27.5 min. with 95% confidence. </em>Since 27 min. is within this interval, we can conclude that there is no substantial evidence (in 95% confidence level) that the new process has reduced the mean time to handle a customer complaint.

c. The population about which inferences from these data can be made is the people following the new implemented plan when handling customer complaints.

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Distance from the moon is 2.4 × 10⁵ miles

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Length of 1-inch paperclip = 1.6\times 10^{-5} miles

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Total distance from the moon = Number of paper clips used × Length of one clip

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Harold and his friend both enjoy running long distances.Yesterday Harold ran 8 2/4 miles and his friend ran 4 3/4 miles. How man
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Answer:

Harold ran 3\frac{3}{4}\ miles more than his friend.

Step-by-step explanation:

Given:

Number of miles Harold ran = 8\frac{2}{4}\ miles

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Number of miles Harold ran = \frac{34}{4}\ miles

Number of miles his friend ran = 4\frac{3}{4}\ miles

4\frac{3}{4}\ miles can be Rewritten as \frac{19}{4}\ miles

Number of miles his friend ran = \frac{19}{4}\ miles

We need to find number of miles Harold ran more than his friend.

Solution:

Now we can say that;

To find the number of miles Harold ran more than his friend we need to subtract Number of miles his friend ran from Number of miles Harold ran.

framing in equation form we get;

the number of miles Harold ran more than his friend = \frac{34}{4}-\frac{19}{4} =\frac{34-19}{4} = \frac{15}{4}\ miles

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Hence Harold ran 3\frac{3}{4}\ miles more than his friend.

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What is the equation, in point-slope form, of the line that is perpendicular to the given line and passes through the point (−4,
zheka24 [161]
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Explanation:

Given that you did not include the "given line", I can help you by explaning how to solve this kind of problems, step by step.

The procedure is based of the property of perpendicular lines: the product of the slopes of perpedicular lines is negative 1.

If you call m1, the slope of a line and m2 the slope of a perpendicular line, then:

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With that this is the procedure:

1) find the slope of the "given line". Name it m1.

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