Question

In: Statistics and Probability

For this problem, you will use the data in the Emission Level column. A major car...

For this problem, you will use the data in the Emission Level column. A major car manufacturer wants to test a new engine to determine whether it meets new air-pollution standards. The mean emission ? of all engines of this type must be less than 20 parts per million (ppm) of carbon. Ten engines are manufactured for testing purposes, and the emission level of each is determined. The data is given in ppm. The test will be run with ? = 0.01. Assume it is known that ? = 3.01. a. State mathematically the null and alternative hypotheses for this test. H0: _________________________ H1: _________________________ b. State the decision rule for the test. c. (1) You might need Minitab to find the descriptive statistics for you, but then find the test statistic and pvalue “by hand,” showing all work/explanations below. (2) Use Minitab to run the appropriate hypothesis test (i.e., choose between the 1-sample Z and the 1-sample t in the Stat | Basic Statistics menu). Your choice for the Hypothesized mean (the nullhypothesized value for the mean) and the sign for Alternative hypothesis (found under Options) should match the hypotheses you wrote in part (a). Paste ALL of the Minitab output into this document, from the test name all the way down to the p-value. At the end of the output, state the test statistic and the p-value. d. State your conclusion to the hypothesis test – include a statement about rejecting or retaining H0 (and why) and also a concluding statement in plain English about emission level. Module 8 – Fishbowl Assignment – Group 1 e. In order to run the above test, we needed to know that the distribution of X is (approximately) normal. How do we know it is? Justify your answer. Listed are the ten engine data 15.6 16.2 22.5 20.5 16.4 19.4 16.6 17.9 12.7 13.9

Solutions

Expert Solution

a)

Null hypothesis H?: ? = 20
Alternative hypothesis H?: ? < 20

b)

Decision Rule: The null hypothesis is rejected if the P-value is less than 0.05 at 5% significance level.

c) 1)

The descriptive statistic is obtained using the minitab by following step,

Step 1: Stat > Basic Statistic > Display Descriptive Statistic > OK. The screenshot is shown below,

Step 2: Select column with data values. The screenshot is shown below,

The result is obtained. the screenshot is shown below,

Step 3: Now the t statistic is obtained using the formula,

Step 4: The P-value can be obtained using the t distribution table for t = -3.00 and degree of freedom = 10 - 1 = 9.

c) 2)

The population varianace is unknown hence t test will be used

Appropriate test: One sample t test.

The t test is performed in minitab by following these steps,

Step 1: Write the data values in worksheet. The screesnshot is shown below,

Step 2: Stat > Basic Statistic > One Sample t test. The screenshot is shown below,

Step 3: Select column C1, tick Perform hypothesis test; Hypothesized mean = 20 and click option and select Mean < hypothesized mean and confidence level = 95.0 then OK. The screenshot is shown below,

The result is obtained. The screenshot is shown below,

d)

The P-value for the t statistic is,

Hence the null hypothesis is rejected at 5% significance level. now we can state that mean is significantly less than 20.


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