A company with a large fleet of cars wants to keep gasoline cost down and sets a

Question

A company with a large fleet of cars wants to keep gasoline cost down and sets a goal of attaining a fleet average of at least 27 miles per gallon. To see if the goal is being? met, they check the gasoline usage for 15 cars chosen at? random, finding a mean of 25.96 mpg and a standard deviation of 2.63 mpg with a normal shape. We want to find whether they have failed to attain their fuel economy goal.

a) Mike says that the observed sample mean of 25.96 mpg is a lot less than 27 miles per gallon, so we should conclude they failed to attain their fuel economy goal. Do you agree with his conclusion? Select One O A. No, we don't have enough sample size to apply the central limit theorem. O B. Yes, they fail because the sample size is large enough to say that the true population MPG mean is less than 27 mpg. O C. No, It is possible to get a mean of 25.96 mpg even though the true population MPG mean is 27 mpg. O D. Yes, they fail because 25.96 mpg is less than the target value of 27 miles. b) What table do we use for both the confidence interval and hypothesis test? OA. Z-table, the population standard deviation is known B. T-table, the population mean is unknown. C. T-table, the population standard deviation, ' is unknown. O D. Z-table, the sample standard deviation is known. c) Choose the correct null and alternative hypothesis A. Ho : > 27 B. Ho : = 27 O C. Ho: = 25.96 VS HA: = 27 HA: #27 VS VS HA:

Explanation / Answer

A)

We cannot be so sure. We have to do a hypothesis test first, because this can simply be due to variation.

Hence,

OPTION C: No, it is possible to get a mean of 25.96 mpg even though the true population MPG mean is 27 mpg. [ANSWER]

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b)

As n = 15 is small, and sigma is unknown, we just use

OPTION C: T table, as sigma is unknown. [ANSWER]

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c)

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   27  
Ha:    u   <   27   [ANSWER, D]

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d)

              
As we can see, this is a    left   tailed test.      
Getting the test statistic, as              
              
X = sample mean =    25.96          
uo = hypothesized mean =    27          
n = sample size =    15          
s = standard deviation =    2.63          
              
Thus, t = (X - uo) * sqrt(n) / s =    -1.531521932   [ANSWER]

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e)
              
df = n - 1 =    14          

Also, the p value is, as this is left tailed,              
              
p =    0.073960414   [ANSWER]

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f)

The P value is the probability of obtaining a sample if Ho is true. Hence,

OPTION B. [ANSWER]

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g)

As P > 0.05,

OPTION A: TRUE. [ANSWER]

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h)

Note that              
Margin of Error E = t(alpha/2) * s / sqrt(n)              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.05          
X = sample mean =    25.96          
t(alpha/2) = critical t for the confidence interval =    1.761310136          
s = sample standard deviation =    2.63          
n = sample size =    15          
df = n - 1 =    14          
Thus,              
Margin of Error E =    1.196040686          
Lower bound =    24.76395931          
Upper bound =    27.15604069          
              
Thus, the confidence interval is              
              
(   24.76395931   ,   27.15604069   ) [ANSWER]

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