The EPA reports that the exhaust emissions for a certain car model has a normal distribution with a mean of 1.45 grams of nitrous oxide per mile and a standard deviation of 0.4. The car manufacturer claims their new process reduces the mean level of exhaust emitted for this car model. A SRS of 28 cars is taken and the mean level of exhaust emitted for this sample is 1.21 grams. (a) Calculate the p-value (b)What is the decision at the 0.01 significance level?
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21 - 1.45}{\frac{0.4}{\sqrt{28}}} = -3.175 \] ** Show more…
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The EPA reports that the exhaust emissions for a certain car model has a mean of 1.45 grams of nitrous oxide per mile. The car manufacturer claims their newest model of that car emits less than 1.45 grams of nitrous oxide per mile. A simple random sample of 34 cars is taken and the mean level of exhaust emitted for this sample is 1.21 grams with a standard deviation of 0.52 grams. Perform steps 2 and 3 of the hypothesis test with a level of significance of 0.05. Step 2 (as a decimal): α = Step 3 -- use Minitab Note: to change the decimal places in Minitab, right click the value you want to change. test statistic (round to two decimal places): p-value (round to four decimal places):
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The level of nitrogen oxides (NOX) in the exhaust of a particular car model varies with a mean of 0.9 grams per mile and a standard deviation of 0.17 grams per mile. (a) What sample size is needed so that the standard deviation of the sampling distribution is 0.01 grams per mile? ANSWER: (b) If a larger sample is considered, the standard deviation for x̄ would be. (NOTE: Enter ''SMALLER'', ''LARGER'' or ''THE SAME'' without the quotes.)
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The level of nitrous oxides (NOX) and nonmethane organic gas (NMOG) in the exhaust over the useful life (150,000 miles of driving) of cars of a particular model varies normally with a mean of 80 mg/mi and a standard deviation of 4 mg/mi. A company has 25 cars of this model in its fleet. What is the level L such that the probability that the average NOX + NMOG level x̄ for the fleet is greater than L is only 0.01? (Hint: This requires a backward normal calculation.)
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