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Sagot :
According to the hypothesis test we find out that the t-statistic is 0.634 and thus, conclude that the manufacturer's advertising campaign is incorrect as its new small cars have an average of less than or equal to 45 miles per gallon in highway driving.
It is given to us that -
Company manufactures small automobiles that have averaged 45 miles per gallon of gasoline in highway driving
Company now advertises that its new small cars average more than 45 miles per gallon in highway driving
52 automobiles were tested
Sample average is 46.1 miles per gallon
Sample standard deviation is 5.3 miles per gallon
Significance level is 0.01
We have to conduct an appropriate hypothesis test and find the t-statistic and determine the correct conclusion.
Let us assume that -
p = population of average gasoline of new small cars in highway driving
According to the given information,
Null hypothesis represents that the advertising campaign of the manufacturer is incorrect. This implies that -
[tex]H_{0}[/tex] : [tex]p\leq 45[/tex] miles per gallon
Alternative hypothesis represents that the advertising campaign of the manufacturer is correct. This implies that -
[tex]H_{A}[/tex] : [tex]p > 45[/tex] miles per gallon
For carrying out the hypothesis test for the given situation we have to apply one sample z test statistic since the sample standard deviation information is available to us.
According to one sample z test, the value of test statistic is given as -
[tex]T = \frac{(X_{avg} -p)}{\frac{S}{\sqrt{n} } }[/tex] ------ (1)
where,
[tex]X_{avg}[/tex] = Sample mean
p = hypothesized population mean
S = Sample standard deviation
n = Sample size
According to the given information, we have
[tex]X_{avg}[/tex] = 46.1 miles per gallon
p = 45 miles per gallon
S = 5.3 miles per gallon
n = 52
Substituting the above values in equation (1), we have
[tex]T = \frac{(X_{avg} -p)}{\frac{S}{\sqrt{n} } }\\= > T = \frac{(46.1 -45)}{\frac{5.3}{\sqrt{52} } }\\= > T = \frac{1.1}{0.735}\\ = > T = 0.634[/tex]
It is given to us that the significance level is 0.01. According to the z-table, the critical value of the test statistic at 0.01 significance level is 2.58.
But, we see that the test statistic that we obtained above is less than the critical value at significance level of 0.01. i.e.,
0.634 < 2.58
This implies that there is not sufficient evidence for us to reject the null hypothesis for this particular situation.
Therefore, we can conclude from the hypothesis test that the t-statistic is 0.634 and thus, the manufacturer's advertising campaign is incorrect as its new small cars have an average of less than or equal to 45 miles per gallon in highway driving.
To learn more about hypothesis test visit https://brainly.com/question/17347077
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According to the hypothesis test we find out that the t-statistic is 0.634 and thus, conclude that the manufacturer's advertising campaign is incorrect as its new small cars have an average of less than or equal to 45 miles per gallon in highway driving.
It is given to us that -
Company manufactures small automobiles that have averaged 45 miles per gallon of gasoline in highway driving
Company now advertises that its new small cars average more than 45 miles per gallon in highway driving
52 automobiles were tested
Sample average is 46.1 miles per gallon
Sample standard deviation is 5.3 miles per gallon
Significance level is 0.01
We have to conduct an appropriate hypothesis test and find the t-statistic and determine the correct conclusion.
Let us assume that -
p = population of average gasoline of new small cars in highway driving
According to the given information,
Null hypothesis represents that the advertising campaign of the manufacturer is incorrect. This implies that -
H0 = p<= 45 : miles per gallon
Alternative hypothesis represents that the advertising campaign of the manufacturer is correct. This implies that -
HA : p > 45 : miles per gallon
For carrying out the hypothesis test for the given situation we have to apply one sample z test statistic since the sample standard deviation information is available to us.
According to one sample z test, the value of test statistic is given as -
T = (X - p)/(SD/√n)
where,
X = Sample mean
p = hypothesized population mean
S = Sample standard deviation
n = Sample size
According to the given information, we have
X = 46.1 miles per gallon
p = 45 miles per gallon
S = 5.3 miles per gallon
n = 52
Substituting the above values in equation (1), we have
T = 46.1-45/(5.2/√52)
T = 1.1/0.735
T = 0.634
It is given to us that the significance level is 0.01. According to the z-table, the critical value of the test statistic at 0.01 significance level is 2.58.
But, we see that the test statistic that we obtained above is less than the critical value at significance level of 0.01. i.e.,
0.634 < 2.58
This implies that there is not sufficient evidence for us to reject the null hypothesis for this particular situation.
Therefore, we can conclude from the hypothesis test that the t-statistic is 0.634 and thus, the manufacturer's advertising campaign is incorrect as its new small cars have an average of less than or equal to 45 miles per gallon in highway driving.
To learn more about hypothesis test visit brainly.com/question/17347077
#SPJ4
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