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Julli [10]
3 years ago
6

In 2010, the average duration of long-distance telephone calls originating in one town was 9.4 minutes. A long-distance telephon

e company wants to perform a hypothesis test to determine whether the average duration of long-distance phone calls has changed from the 2010 mean of 9.4 minutes. The mean duration for a random sample of 50 calls originating in the town was 8.6 minutes. Does the data provide sufficient evidence to conclude that the mean call duration, µ, is different from the 2010 mean of 9.4 minutes? Perform the appropriate hypothesis test using a significance level of 0.01. Assume that s = 4.8 minutes.
Mathematics
1 answer:
Zepler [3.9K]3 years ago
4 0

<u>Answer</u>: No, we do not have sufficient evidence to conclude that the mean call duration, µ, is different from the 2010 mean of 9.4 minutes.

Step-by-step explanation:

As per given , we have

H_0: \mu=9.4\\\\H_a:\mu\neq9.4, since H_0 is two-tailed so , the test is a two tail test.

Since population standard deviation is unknown, so we use t-test.

Critical value (two-tailed) for significance level of 0.01= t_{n-1,\alpha/2}=t_{49, 0.005}\pm2.609228

For n =50 , \overline{x}=8.6 and s= 4.8

Test statistic : t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}

t=\dfrac{8.6-9.4}{\dfrac{4.8}{\sqrt{50}}}\approx-1.18

Since test statistic value (-1.18) lies in critical interval (-2.609228, 2.609228), it means the null hypothesis is failed to reject.

We do not have sufficient evidence to conclude that the mean call duration, µ, is different from the 2010 mean of 9.4 minutes.

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Step-by-step explanation:

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Answer:

a) The value of absolute minimum value = - 0.3536  

b) which is attained at   x = \frac{1}{\sqrt{2} }  

Step-by-step explanation:

<u>Step(i)</u>:-

Given function

                       f(x) = \frac{-x}{2x^{2} +1}     ...(i)

Differentiating equation (i) with respective to 'x'

                     f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2}  }   ...(ii)

                    f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2}  }

Equating Zero

                   f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2}  } = 0

                 \frac{2x^{2}-1}{(2x^{2}+1)^{2}  } = 0

                2 x^{2}-1 = 0

               2 x^{2} = 1

             x^{2}  = \frac{1}{2}

             x = \frac{-1}{\sqrt{2} }  , x = \frac{1}{\sqrt{2} }

<u><em>Step(ii):</em></u>-

Again Differentiating equation (ii) with respective to 'x'

f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4}  }

put

      x = \frac{1}{\sqrt{2} }

f^{ll} (x) > 0

The absolute minimum value at   x = \frac{1}{\sqrt{2} }

<u><em>Step(iii):</em></u>-

The value of absolute minimum value

                         f(x) = \frac{-x}{2x^{2} +1}

                       f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}

         on calculation we get

The value of absolute minimum value = - 0.3536      

<u><em>Final answer</em></u>:-

a) The value of absolute minimum value = - 0.3536  

b) which is attained at   x = \frac{1}{\sqrt{2} }    

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