8-3. Exercises

1. BASIC

Ex 1. Compute the observed significance of each test.

  1. Testing H0:μ=54.7 vs. Ha:μ<54.7,H_0:μ=54.7 \space vs. \space Ha:μ<54.7, test statistic z=−1.72.z=−1.72.

  2. Testing H0:μ=195 vs. Ha:μ≠54.7,H_0:μ=195 \space vs. \space Ha:μ \ne 54.7, test statistic z=−2.07z=−2.07.

  3. Testing H0:μ=45 vs. Ha:μ>−45,H_0:μ=45 \space vs. \space Ha:μ> -45, test statistic z=2.54.z=2.54.

Ex 2. Compute the observed significance of each test.

  1. Testing H0:μ=0 vs. Ha:μ≠0,H_0:μ=0 \space vs. \space Ha:μ \ne 0, test statistic z=2.82.z=2.82.

  2. Testing H0:μ=18.4 vs. Ha:μ<18.4,H_0:μ=18.4 \space vs. \space Ha:μ < 18.4, test statistic z=−1.74.z=-1.74.

  3. Testing H0:μ=63.85 vs. Ha:μ>63.85,H_0:μ=63.85 \space vs. \space Ha:μ > 63.85, test statistic z=1.93.z=1.93.

Ex 3. Compute the observed significance of each test. (Some of the information given might not be needed.)

  1. Testing H0:μ=27.5 vs. Ha:μ>27.5;n=49,xˉ=28.9,s=3.14,H_0:μ=27.5 \space vs. \space H_a:μ>27.5; n = 49, \bar{x}=28.9, s = 3.14, test statistic z=3.12.z=3.12.

  2. Testing H0:μ=581 vs. Ha:μ<581;n=32,xˉ=560,s=47.8,H_0:μ=581 \space vs. \space H_a:μ<581; n = 32, \bar{x}=560, s = 47.8,test statistic z=−2.49.z=-2.49.

  3. Testing H0:μ=138.5 vs. Ha:μ≠138.5;n=44,xˉ=137.6,s=2.45,H_0:μ=138.5 \space vs. \space H_a:μ \ne 138.5; n = 44, \bar{x}=137.6, s = 2.45,test statistic z=−2.44.z=−2.44.

Ex 4. Compute the observed significance of each test. (Some of the information given might not be needed.)

  1. Testing H0:μ=−17.9 vs. Ha:μ<−17.9;n=34,xˉ=−18.2,s=0.87,H_0:μ=-17.9 \space vs. \space H_a:μ<-17.9; n = 34, \bar{x}=-18.2, s = 0.87, test statistic z=−2.01.z=−2.01.

  2. Testing H0:μ=5.5 vs. Ha:μ≠5.5;n=56,xˉ=7.4,s=4.82,H_0:μ=5.5 \space vs. \space H_a:μ \ne 5.5; n = 56, \bar{x}=7.4, s = 4.82, test statistic z=2.95.z = 2.95.

  3. Testing H0:μ=1255 vs. Ha:μ>1255;n=152,xˉ=1257,s=7.5,H_0:μ=1255 \space vs. \space H_a:μ>1255; n = 152, \bar{x}=1257, s = 7.5, test statistic z=3.29.z = 3.29.

Ex 5. Make the decision in each test, based on the information provided.

  1. Testing H0:μ=82.9 vs. Ha:μ<82.9,@α=0.05,H_0:μ=82.9 \space vs. \space Ha:μ<82.9, @ α=0.05, observed significance p=0.038.p = 0.038.

  2. Testing H0:μ=213.5 vs. Ha:μ≠213.5,@α=0.01,H_0:μ=213.5 \space vs. \space Ha:μ \ne 213.5, @ α=0.01, observed significance p=0.038.p = 0.038.

Ex 6. Make the decision in each test, based on the information provided.

  1. Testing H0:μ=31.4 vs. Ha:μ>31.4,@α=0.10,H_0:μ=31.4 \space vs. \space Ha:μ>31.4, @ α=0.10, observed significance p=0.062.p = 0.062.

  2. Testing H0:μ=−75.5 vs. Ha:μ<−75.5,@α=0.05,H_0:μ=-75.5 \space vs. \space Ha:μ<-75.5, @ α=0.05, observed significance p=0.062.p = 0.062.

2. APPLICATIONS

Ex 7. A lawyer believes that a certain judge imposes prison sentences for property crimes that are longer than the state average 11.7 months. He randomly selects 36 of the judge’s sentences and obtains mean 13.8 and standard deviation 3.9 months.

  1. Perform the test at the 1% level of significance using the critical value approach.

  2. Compute the observed significance of the test.

  3. Perform the test at the 1% level of significance using the p-value approach. You need not repeat the first three steps, already done in part (a).

Ex 8. In a recent year the fuel economy of all passenger vehicles was 19.8 mpg. A trade organization sampled 50 passenger vehicles for fuel economy and obtained a sample mean of 20.1 mpg with standard deviation 2.45 mpg. The sample mean 20.1 exceeds 19.8, but perhaps the increase is only a result of sampling error.

  1. Perform the relevant test of hypotheses at the 20% level of significance using the critical value approach.

  2. Compute the observed significance of the test.

  3. Perform the test at the 20% level of significance using the p-value approach. You need not repeat the first three steps, already done in part (a).

Ex 9. The mean score on a 25-point placement exam in mathematics used for the past two years at a large state university is 14.3. The placement coordinator wishes to test whether the mean score on a revised version of the exam differs from 14.3. She gives the revised exam to 30 entering freshmen early in the summer; the mean score is 14.6 with standard deviation 2.4.

  1. Perform the test at the 10% level of significance using the critical value approach.

  2. Compute the observed significance of the test.

  3. Perform the test at the 10% level of significance using the p-value approach. You need not repeat the first three steps, already done in part (a).

Ex 10. The mean increase in word family vocabulary among students in a one-year foreign language course is 576 word families. In order to estimate the effect of a new type of class scheduling, an instructor monitors the progress of 60 students; the sample mean increase in word family vocabulary of these students is 542 word families with sample standard deviation 18 word families.

  1. Test at the 5% level of significance whether the mean increase with the new class scheduling is different from 576 word families, using the critical value approach.

  2. Compute the observed significance of the test.

  3. Perform the test at the 5% level of significance using the p-value approach. You need not repeat the first three steps, already done in part (a).

Ex 11. The mean yield for hard red winter wheat in a certain state is 44.8 bu/acre. In a pilot program a modified growing scheme was introduced on 35 independent plots. The result was a sample mean yield of 45.4 bu/acre with sample standard deviation 1.6 bu/acre, an apparent increase in yield.

  1. Test at the 5% level of significance whether the mean yield under the new scheme is greater than 44.8 bu/acre, using the critical value approach.

  2. Compute the observed significance of the test.

  3. Perform the test at the 5% level of significance using the p-value approach. You need not repeat the first three steps, already done in part (a).

Ex 12. The average amount of time that visitors spent looking at a retail company’s old home page on the world wide web was 23.6 seconds. The company commissions a new home page. On its first day in place the mean time spent at the new page by 7,628 visitors was 23.5 seconds with standard deviation 5.1 seconds.

  1. Test at the 5% level of significance whether the mean visit time for the new page is less than the former mean of 23.6 seconds, using the critical value approach.

  2. Compute the observed significance of the test.

  3. Perform the test at the 5% level of significance using the p-value approach. You need not repeat the first three steps, already done in part (a).

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