Indicate the 4 steps. 3. Calculate the test statisticGiven:X = 26.5µ = 24.76 = 4.8N = 751-obtZobt26.5-24.7Zobt =4.8//751.8Zobt5%0.5543Zabt =3.251.6453.25(a) One-tailed test4. assess whether z,b, falls within the critical regionsince Iz.bl 1.645, it falls within the critical region for rejection of the null hypothesis.therefore, we reject the null hypothesis and conclude that the gasoline additive doesincrease miles per gallon SAMPLE PROBLEMA gasoline manufacturer believes a ne w additive will result in more m iles pergallon. A la rge number of mileage mea surements on t he gasoline without theadditive have been made by the company under rigorously controlled conditions.The results show a mean of 24.7 miles per gallon and a standard deviation of 4.8.Tests are conducted on a sa mple of 75 cars using the gasoline plus the additive.The sample mean equals 26.5 miles per gallon.a. Let's assume there is a dequate ba sis for a one -tailed test. What is the direc-tional alternative hypothesis?b. What is the null hypothesis?c. What is the conclusion? Use a = 0.05, cail-1. State the hypotheses and select an alpha levelNull Hypothesis – Ho: µ = population meanAlternative Hypothesis – H: µ # population mean2. Draw the normal curve: label the critical value andcritical region; determine the critical valuerefer to the z-table for the critical value3. Calculate the test statistic4. Make a decision regarding Ho assess whether Zee falls within the critical regionIf kel z kail reject the null hypothesis. If not, retain the nul hypothesis.ANSWER1. State the hypothesisNull hypothesis: The new additive does not increase the number of miles per gallon.The sample with the mean 26.5 is a random sample from a population with populationmean less than or equal to 24.7H us 24.7Alternative hypothesis: The new additive increases the number of miles per gallon.Therefore, the sample with the mean 26.5 is a random sample from a population with apopulation mean that is greater than 24.7H,: u 24.72. Draw the normal curve, label the critical value andregiona = .05silZer = 1.6455%1.645(a) One-tailed test

Question

Indicate the 4 steps. 3. Calculate the test statisticGiven:X = 26.5µ = 24.76 = 4.8N = 751-obtZobt26.5-24.7Zobt =4.8//751.8Zobt5%0.5543Zabt =3.251.6453.25(a) One-tailed test4. assess whether z,b, falls within the critical regionsince Iz.bl> 1.645, it falls within the critical region for rejection of the null hypothesis.therefore, we reject the null hypothesis and conclude that the gasoline additive doesincrease miles per gallon SAMPLE PROBLEMA gasoline manufacturer believes a ne w additive will result in more m iles pergallon. A la rge number of mileage mea surements on t he gasoline without theadditive have been made by the company under rigorously controlled conditions.The results show a mean of 24.7 miles per gallon and a standard deviation of 4.8.Tests are conducted on a sa mple of 75 cars using the gasoline plus the additive.The sample mean equals 26.5 miles per gallon.a. Let's assume there is a dequate ba sis for a one -tailed test. What is the direc-tional alternative hypothesis?b. What is the null hypothesis?c. What is the conclusion? Use a = 0.05, cail-1. State the hypotheses and select an alpha levelNull Hypothesis – Ho: µ = population meanAlternative Hypothesis – H: µ # population mean2. Draw the normal curve: label the critical value andcritical region; determine the critical valuerefer to the z-table for the critical value3. Calculate the test statistic4. Make a decision regarding Ho assess whether Zee falls within the critical regionIf kel z kail reject the null hypothesis. If not, retain the nul hypothesis.ANSWER1. State the hypothesisNull hypothesis: The new additive does not increase the number of miles per gallon.The sample with the mean 26.5 is a random sample from a population with populationmean less than or equal to 24.7H us 24.7Alternative hypothesis: The new additive increases the number of miles per gallon.Therefore, the sample with the mean 26.5 is a random sample from a population with apopulation mean that is greater than 24.7H,: u > 24.72. Draw the normal curve, label the critical value andregiona = .05silZer = 1.6455%1.645(a) One-tailed test

Accepted Answer

Given :

Null and Alternative Hypotheses :
H0:μ≤24.7Ha:μ>24.7 \begin{array}{ccl} H_0: \mu  &…