MAT 221 Practice Exam 3

Please read the questions carefully. Show all work!

Formulae:

 

X ± Za/2 (s/Ön)          X ± Ta/2 (s/Ön)           n =       s2 (Za/2) 2

                                                                              B 2

 

1.                  A laboratory technician is timed 20 times in the performance of a task, getting a sample mean time of 7.9 minutes and a standard deviation of 1.2 minutes. If the population average for a career is 7.5 minutes and a = 0.05, does this sample support the claim that his career average is different than the population career average? Be sure to state your conclusion in plain English.

 

2.                  Suppose we take a random sample from a Normal population with mean m=20 and standard deviation s=3. Find n, the sample size, if we want the probability that the sample mean falls between 19 and 21 to be 0.95.

 

3.                  Springfield is considering changing the type of tires it uses on it police cars. A random sample of 50 current type tires wore out on average after 40,000 miles with a standard deviation of 10,000 miles. The old tires wore out on average after 35,000 miles. Is there evidence at a=0.05 that the new type tires last longer? Be sure to state your conclusion in plain English.

 

4.         The Mars Company produces M&M’s. When operating properly, the production process will produce on average 200 pounds of M&M’s per hour, with a standard deviation of 50 pounds. If 100 hours are sampled:

    1. Find the probability that the sample average of M&M’s produced will be more than 210 pounds.
    2. Find the probability that the sample average of M&M’s produced will be between 193 and 210 pounds.
    3. Find the value w, such that 95% of sample means will be smaller than w.
    4. If the sample mean was found to be 189.2, draw a conclusion about the production process and justify it.

Answers:

 

1.         7.9 ± 2.093 * 1.2 / sqrt(20) = (7.33384, 8.4616)

7.5 is in the interval, so there is not sufficient evidence to say his career average is different.

 

2.         B = 1, σ = 3, Therefore  n = (1.96) 2 * 3 2 / 1 = 34.5 = 35.

 

3.         40000 ± 1.96 (10000) / sqrt(50)

            (37228, 42772)

            35000 not in interval so there is evidence that the new tires last longer.

 

4.         a.         P(Xbar > 210) = P(Z > 2) = .0228

            b.         P(193 < Xbar < 210) = P(-1.4 < Z < 2) = .9772 - .0808 = .8964

            c.         w = 1.645 (5) + 200 = 208.225

d.         P(Xbar < 189.2) = P(Z < -2.16) = .0154, < 0.05, therefore the process is out of control.