A local car manufacturer manufactures small automobiles that averaged 50 miles per gallon of gasoline in highway driving.The company has developed a more efficient engine for its small cars and now advertises that its new small cars average more than 50 miles per gallon in highway driving. In a sample of 36 road-tested automobiles, it showed an average of 51.5 miles per gallon and a standard deviation of 6 miles per gallon.Test to determine whether or not the manufacturer's advertising campaign is legitimate at 0.05 level of significance and using the p-value approach, Group of answer choicesWith p-value of 0.0668, therefore, do not reject Ho. There is no sufficient evidence to conclude that the new cars average more than 50 mile per gallon.With p-value of 0.0068, therefore, reject Ho. There is sufficient evidence to conclude that the new cars average more than 50 miles per gallon.With p-value of 0.0724, therefore, do not reject Ho. There is no sufficient to conclude that the new cars average more than 50 miles per gallon.With p-value of 0.0008, therefore, reject Ho. There is sufficient evidence to conclude that the new cars average more than 50 miles per gallon.
Question
A local car manufacturer manufactures small automobiles that averaged 50 miles per gallon of gasoline in highway driving.The company has developed a more efficient engine for its small cars and now advertises that its new small cars average more than 50 miles per gallon in highway driving. In a sample of 36 road-tested automobiles, it showed an average of 51.5 miles per gallon and a standard deviation of 6 miles per gallon.Test to determine whether or not the manufacturer's advertising campaign is legitimate at 0.05 level of significance and using the p-value approach, Group of answer choicesWith p-value of 0.0668, therefore, do not reject Ho. There is no sufficient evidence to conclude that the new cars average more than 50 mile per gallon.With p-value of 0.0068, therefore, reject Ho. There is sufficient evidence to conclude that the new cars average more than 50 miles per gallon.With p-value of 0.0724, therefore, do not reject Ho. There is no sufficient to conclude that the new cars average more than 50 miles per gallon.With p-value of 0.0008, therefore, reject Ho. There is sufficient evidence to conclude that the new cars average more than 50 miles per gallon.
Solution
To answer this question, we first need to set up our null and alternative hypotheses.
The null hypothesis (H0) is that the new cars do not average more than 50 miles per gallon. The alternative hypothesis (H1) is that the new cars do average more than 50 miles per gallon.
Next, we calculate the test statistic using the formula for a one-sample t-test:
t = (x̄ - μ) / (s/√n)
where x̄ is the sample mean (51.5 miles per gallon), μ is the population mean under the null hypothesis (50 miles per gallon), s is the sample standard deviation (6 miles per gallon), and n is the sample size (36 cars).
Plugging in these values, we get:
t = (51.5 - 50) / (6/√36) = 1.5
Next, we need to find the critical value for a one-tailed t-test with 35 degrees of freedom (n-1) at the 0.05 level of significance. Looking this up in a t-distribution table, we find that the critical value is approximately 1.69.
Since our test statistic (1.5) is less than the critical value (1.69), we do not reject the null hypothesis.
Finally, we need to calculate the p-value, which is the probability of observing a test statistic as extreme as ours, assuming the null hypothesis is true. Using a t-distribution calculator, we find that the p-value is approximately 0.0724.
Since the p-value (0.0724) is greater than the level of significance (0.05), we do not reject the null hypothesis.
Therefore, there is not sufficient evidence to conclude that the new cars average more than 50 miles per gallon.
So, the correct answer is: "With p-value of 0.0724, therefore, do not reject Ho. There is no sufficient evidence to conclude that the new cars average more than 50 miles per gallon."
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