STAT 400 Discussion 12 Spring 2015 1. An economist states that 10% of Springfield's labor force is unemployed. A random sample of 400 people in the labor force is obtained, of whom 28 are unemployed. a) Test whether Springfield's unemployment rate is as the economist claims or is different at a 1% level of significance b) Find the p-value of the test in part (a). c) Using the p-value from part (b), state your decision ( Reject H 0 or Do Not Reject H 0 ) at = 0.06 and at = 0.03. 2 – 3. An examination of the records for a random sample of 16 motor vehicles in a large fleet resulted in the sample mean operating cost of 26.33 cents per mile and the sample standard deviation of 2.80 cents per mile. (Assume that operating costs are approximately normally distributed.) 2. a) The manager wants to believe that the actual mean operating cost is at most 25 cents per mile. Perform the appropriate test at a 5% level of significance. b) Using the t distribution table only, what is the p-value of the test in part (a) ? ( You may give a range. ) 3. d) Test whether the overall standard deviation of the operating costs is more than 2.30 cents per mile or not at a 5% significance level. e) Using the Chi-Square distribution table only, what is the p-value of the test in part (d) ? ( You may give a range. )1. An economist states that 10% of Springfield's labor force is unemployed. A random sample of 400 people in the labor force is obtained, of whom 28 are unemployed. a) Test whether Springfield's unemployment rate is as the economist claims or is different at a 1% level of significance H 0 : p = 0.10 vs. H 1 : p 0.10. n = 400, x = 28, 40028ˆnxp = 0.07. Test Statistic: 40090.010.0 10.007.0)1(000ˆnpppp Z = – 2.00. Rejection Region: Z < z 0.005 = 2.576 or Z > z 0.005 = 2.576. Do NOT Reject H 0 at = 0.01. OR P-value: Two – tailed. P-value = 2 × P( Z ≤ – 2.00 ) = 2 × 0.0228 = 0.0456. p-value = 0.0456 > 0.01 = . Do NOT Reject H 0 at = 0.01. b) Find the p-value of the test in part (a). p-value = 0.0456. ( See part (a) ) c) Using the p-value from part (b), state your decision ( Reject H 0 or Do Not Reject H 0 ) at = 0.06 and at = 0.03. p-value = 0.0456 < 0.06 = . p-value = 0.0456 > 0.03 = . Reject H 0 at = 0.06. Do NOT Reject H 0 at = 0.03.2 – 3. An examination of the records for a random sample of 16 motor vehicles in a large fleet resulted in the sample mean operating cost of 26.33 cents per mile and the sample standard deviation of 2.80 cents per mile. (Assume that operating costs are approximately normally distributed.) 2. a) The manager wants to believe that the actual mean operating cost is at most 25 cents per mile. Perform the appropriate test at a 5% level of significance. X = 26.33, s = 2.80, n = 16. H 0 : 25 vs. H 1 : > 25. Right – tailed. Test Statistic: 16 80.22533.26 XT sμ0n = 1.90. Rejection Region: T > t 0.05 ( 15 ) = 1.753. The value of the test statistic is not in the Rejection Region. Reject H 0 at = 0.05. b) Using the t distribution table only, what is the p-value of the test in part (a) ? ( You may give a range. ) 2.131 > 1.90 > 1.753. 0.025 < p-value < 0.05.3. d) Test whether the overall standard deviation of the operating costs is more than 2.30 cents per mile or not at a 5% significance level. s 2 = 2.80. n = 16. H 0 : 2.30 vs. H 1 : > 2.30. Right – tailed. Test Statistic: 222 022 30.280.21161 σsχ n = 22.23. Rejection Region: Right – tailed. Reject H 0 if 22 n – 1 = 15 degrees of freedom. = 0.05 2050. = 25.00. Reject H 0 if 2 > 25.00. The value of the test statistic does not fall into the Rejection Region. Do NOT Reject H 0 at = 0.05. e) Using the Chi-Square distribution table only, what is the p-value of the test in part (d) ? ( You may give a range. ) 22.31 > 22.23 > 8.547. 0.10 < p-value < 0.90. ( close to 0.10
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