A marathon is a foot race with a distance of 26.22 miles. It was one of the orig

Question

A marathon is a foot race with a distance of 26.22 miles. It was one of the original events of the modern Olympics, where it was a men's-only event The women's marathon did not become an Olympic event until 1984. The Olympic record for the men's marathon was set during the 2008 Olympics by Samuel Kamau Wanjiru of Kenya. With a time of 2 hours, 6 minutes, 32 seconds. The Olympic record for the women's Marathon was set during the 2000 Olympics by Naoko Takahashi of Japan, with a time of 2 hours, 23 minutes, 14 seconds. Training for a marathon typically lasts at least 6 months. The training is gradual, with increases in distance about every 2 weeks. About 1 to 3 weeks before the race, the distance run is decreased slightly. The stem-and leaf plots below show the marathon training times (in minutes) for a sample of 30 male runners and 30 female runners. Use the sample to find a point estimate for the mean training time of the male runners. female runners. Find the standard deviation of the training times for the male runners. female runners. Use the sample to construct a 95% confidence interval for the population mean training time of the male runners. Female runners. Interpret the result of Exercise 3. Use the sample to construct a 95% confidence interval for the population mean training time of all runners. How do your results differ from those in Exercise 3? Explain. A trainer wants to estimate the population mean running times for both male and female runners within 2 minutes. Determine the minimum sample size required to construct a 99% confidence interval for the population mean training time of male runners. Assume the population standard deviation is 8.9 minutes. female runners Assume the population standard deviation is 8.4 minutes.

Explanation / Answer

5.

Note that              
              
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.025          
X = sample mean =    178.45          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    13.3510934          
n = sample size =    60          
              
Thus,              
              
Lower bound =    175.0717693          
Upper bound =    181.8282307          
              
Thus, the confidence interval is              
              
(   175.0717693   ,   181.8282307   ) [ANSWER]

This result is different because we use all the 60 values as just one sample.

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6.

A)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.005  
      
Using a table/technology,      
      
z(alpha/2) =    2.575829304  
      
Also,      
      
s = sample standard deviation =    8.9  
E = margin of error =    2  
      
Thus,      
      
n =    131.3875399  
      
Rounding up,      
      
n =    132   [ANSWER]

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b)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.005  
      
Using a table/technology,      
      
z(alpha/2) =    2.575829304  
      
Also,      
      
s = sample standard deviation =    8.4  
E = margin of error =    2  
      
Thus,      
      
n =    117.039576  
      
Rounding up,      
      
n =    118   [ANSWER]

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