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Can we conclude, using a .05 level of significance,that the miles-per-gallon (mpg) performance of Model Ais greater than the miles-per-gallon performance of Model B?  Model AModel BSample Size2428Sample Mean29.827.3Sample standard deviation2.561.81 What is the test statisticGroup of answer choices5,513.974.404.00

Question

Can we conclude, using a .05 level of significance,that the miles-per-gallon (mpg) performance of Model Ais greater than the miles-per-gallon performance of Model B?  Model AModel BSample Size2428Sample Mean29.827.3Sample standard deviation2.561.81 What is the test statisticGroup of answer choices5,513.974.404.00

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Solution

To answer this question, we need to conduct a hypothesis test for the difference between two means. Here are the steps:

  1. State the null hypothesis (H0) and the alternative hypothesis (Ha). In this case, H0: μA - μB = 0 (the mean mpg of Model A is equal to the mean mpg of Model B) and Ha: μA - μB > 0 (the mean mpg of Model A is greater than the mean mpg of Model B).

  2. Calculate the test statistic. The formula for the test statistic in a two-sample t-test is:

    t = [(x̄A - x̄B) - (μA - μB)] / sqrt[(sA^2/nA) + (sB^2/nB)]

    where x̄A and x̄B are the sample means, μA and μB are the population means (which are 0 under the null hypothesis), sA and sB are the sample standard deviations, and nA and nB are the sample sizes.

    Substituting the given values into the formula, we get:

    t = [(29.8 - 27.3) - 0] / sqrt[(2.56^2/24) + (1.81^2/28)] = 2.5 / sqrt[(0.2733) + (0.1166)] = 2.5 / 0.661 = 3.78

  3. Compare the test statistic to the critical value. The critical value for a one-tailed t-test with a significance level of 0.05 and degrees of freedom df = nA + nB - 2 = 24 + 28 - 2 = 50 can be found in a t-distribution table or calculated using a statistical software. The critical value is approximately 1.68.

  4. Make a decision. Since the test statistic (3.78) is greater than the critical value (1.68), we reject the null hypothesis.

Therefore, we can conclude, using a .05 level of significance, that the miles-per-gallon (mpg) performance of Model A is greater than the miles-per-gallon performance of Model B.

This problem has been solved

Similar Questions

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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.

A car manufacturer advertises that its new subcompact models get 47 miles per gallon (mpg). Let µ be the mean of the mileage distribution for these cars. You assume that the manufacturer will not underrate the car, but you suspect that the mileage might be overrated. State the null hypothesis and the alternate hypothesis for this case.a.H0: µ = 47mpg  and H1: µ ≠ 47mpgb.H0: µ = 47mpg  and H1: µ < 47mpgc.H0: µ = 47mpg  and H1: µ > 47mpgd.H0: µ ≥ 47mpg  and H1: µ < 47mpg

Based on the p-value, what is your conclusion (use .05 significance level)?

he p-value for the hypothesis test is 0.0139 , which means the observed difference is/ is noy statistically significant at the 5% level

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