Moore, 7.23 (modified) Computers in some vehicles calculate various quantities r

Question

Moore, 7.23 (modified) Computers in some vehicles calculate various quantities related to performance. One of these is miles per gallon. A person records the following mpg while a Toyota Highlander Hybrid is set to 60 mpg with cruise control. Here are the data: 37.2 21.0 17.4 24.9 27.0 36.9 38.8 35.3 32.3 23.9 19.0 26.1 25.8 41.4 34.4 32.5 25.3 26.5 28.2 22.1

(a) What is the sample mean? What is the sample standard deviation? (Hint: USE EXCEL).

(b) Use your answer from part (a) to construct both 90 percent and 95 percent confidence intervals around the sample mean

(c) Suppose your null hypothesis is that the mean number of miles per gallon is 30 mpg and that you are doing a two-sided hypothesis. Write down the null hypothesis and the alternative.

(d) What is the test-statistic?

(e) Compare your hypothesis to the relevant critical value for the 90, 95, and 99 percent confidence levels. State your conclusion

Explanation / Answer

a)

Using technology,

X = 28.8 [SAMPLE MEAN]
s = 6.893398826 [SAMPLE STANDARD DEVIATION]

******************

b)

FOR 90% CONFIDENCE:

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 =    28.8          
t(alpha/2) = critical t for the confidence interval =    1.729132812          
s = sample standard deviation =    6.893398826          
n = sample size =    20          
df = n - 1 =    19          
Thus,              
Margin of Error E =    2.665304054          
Lower bound =    26.13469595          
Upper bound =    31.46530405          
              
Thus, the confidence interval is              
              
(   26.13469595   ,   31.46530405   ) [ANSWER]

FOR 95% CONFIDENCE:

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 =    0.025          
X = sample mean =    28.8          
t(alpha/2) = critical t for the confidence interval =    2.093024054          
s = sample standard deviation =    6.893398826          
n = sample size =    20          
df = n - 1 =    19          
Thus,              
Margin of Error E =    3.22620996          
Lower bound =    25.57379004          
Upper bound =    32.02620996          
              
Thus, the confidence interval is              
              
(   25.57379004   ,   32.02620996   ) [ANSWER]

**********************

c)

Ho: u = 30
Ha: u =/= 30 [ANSWER]

*************************

d)

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   30  
Ha:    u   =/   30  
              
As we can see, this is a    two   tailed test.      
              
      
df = n - 1 =    19          
              
Getting the test statistic, as              
              
X = sample mean =    28.8          
uo = hypothesized mean =    30          
n = sample size =    20          
s = standard deviation =    6.893398826          
              
Thus, t = (X - uo) * sqrt(n) / s =    -0.778507567   [ANSWER]      
              
******************************

E)

As we can see, this t value is less than the critical value for both the 90% and 95% confidence intervals.

hence, we fail to reject Ho.

This is consistent with the confidence intrvals, as they both include 30 inside them.

Thus, there is no significant evidnece that the true mean is not 30 mpg.  

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