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    Address Questions 2 on page 1 (see attachment).
    Questions 4, 7, 12, and 16 on page 2
    Questions 9, 10 on page 3
    Questions 12, 16, 18, and 36 a-d on page 4
    Questions 40, 42, 44 on page 5
    Questions 2 a-c, 4 a-d, and 6 a-d on page 6
    Questions 8 a-d, 10 a-f, 12 a-c on page 7
    Questions 18, 20, and 22 a-b on page 8
    Questions 26 a-b, and 32 a-d on page 9
    Questions 34 a-c, 36 a-d, on page 10
    Questions 40, and 42 on page 11.

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    https://brainmass.com/math/consumer-mathematics/various-question-hypothesis-testing-568892

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    SOLUTION This solution is FREE courtesy of BrainMass!

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    One question needs licensed software to do. Two questions are missing data.

    Address Questions 2 on page 1
    2. Point estimate=275/785=0.35.
    At 95% confidence level, the critical value is 1.96 from standard normal table.
    A 95% confidence interval for population proportion is [0.35-1.96*sqrt(0.35*0.65/785), 0.35+1.96*sqrt(0.35*0.65/785)]=[0.317, 0.383]

    Questions 4, 7, 12, and 16 on page 2
    4. At 95% confidence level with df=17-1=16, the critical value is 2.12 from t table.
    A 95% confidence interval for population mean is [3.25-2.12*1.17/sqrt(17), 3.25+2.12*1.17/sqrt(17)]=[2.648, 3.852]
    Interpretation: we are 95% confident that population mean is between 2.648 and 3.852.

    7. At 95% confidence level with df=40-1=39, the critical value is 2.203 from t table.
    A 95% confidence interval for mean length of sentence is [54-2.203*8/sqrt(40), 54+2.203*8/sqrt(40)]=[51.213, 56.787]
    Interpretation: we are 95% confident that mean length of sentence is between 51.213 and 56.787.

    12. At 99% confidence level with df=40-1=39, the critical value is 2.708 from t table.
    A 99% confidence interval for mean amount spent diary is [93.43-2.708*15/sqrt(40), 93.43+2.708*15/sqrt(40)]=[87.007, 99.853]
    Interpretation: we are 99% confident that mean length of sentence is between 87.007 and 99.853.

    16. A confidence interval can NOT be constructed. Reason: from the box plot (right photo), there is clearly outlier. So the distribution may be not normal.

    Questions 9, 10 on page 3
    9. The hypothesis is right-tailed because of ">" sign. Parameter being tested is population mean u.

    10. The hypothesis is left-tailed because of "<" sign. Parameter being tested is population proportion p.

    Questions 12, 16, 18, and 36 a-d on page 4
    12. The hypothesis is right-tailed because of ">" sign. Parameter being tested is population proportion p.

    16. Let u be the mean charitable contribution per household in the United State in 2005. Ho: u=17072 vs. Ha: u≠17072.
    A type I error occurs when the researchers believe that the mean charitable contribution per household in the United State in 2005 is not $17072 but in fact it is.
    A type II error occurs when the researchers believe that the mean charitable contribution per household in the United State in 2005 is $17072 but in fact it is not.

    18. Let u be the mean ounce of a jar of peanut. Ho: u=32 vs. Ha: u<32.
    A type I error occurs when the researchers believe that the mean ounce of a jar of peanut is less than 32 but in fact it is 32.
    A type II error occurs when the researchers believe that the mean ounce of a jar of peanut is 32 but in fact it is less than 32.

    36. (a) Let u be the mean score on the SAT Math Reasoning exam. Ho: u=516 vs. Ha: u>516.
    (b) Conclusion: there is no evidence to say that the mean score on the SAT Math Reasoning exam is higher than 516.
    (c) Type II error would occur because we fail to reject null hypothesis but in fact the null hypothesis is wrong because 522 is larger than 516.
    And the probability of committing a type I error is just alpha which is 0.01.
    (d) We would increase the level of significance if we decrease the probability of making a type I error.

    Questions 40, 42, 44 on page 5
    40. (a) Let u be the mean increase in efficiency of engine. Ho: u=0 vs. Ha: u>0.
    (b) Conclusion: there is no evidence to say that a woman drove 4 hours without oil.

    42. We would choose the level of significance to the least acceptable value which is 0.01 because we ensure that decision leading to type I error does not occur more than once in 100 trials.

    44. My understanding is: interpreting the significance level as a probability is conditional on assuming the null hypothesis is true.

    Questions 2 a-c, 4 a-d, and 6 a-d on page 6
    2. (a) At a=0.1 with df=22-1=21, the critical value is 1.321 for right-tailed test from t table.
    (b) At a=0.01 with df=40-1=39, the critical value is -2.426 for left-tailed test from t table.
    (c) At a=0.01 with df=33-1=32, the critical value is ±2.426 for two-tailed test from t table.

    4. (a) Test statistic=(42.3-40)/[4.3/sqrt(25)]=2.674
    (b) At a=0.1 with df=25-1=24, the critical value is 1.318 from t table.
    (c)
    (d) Since 2.674>1.318, we reject null hypothesis.

    6. (a) test statistic=(76.9-80)/[8.5/sqrt(22)]=-1.711
    (b) At a=0.02 with df=22-1=21, the critical value is -2.189 from t table.
    (c)
    (d) Since -1.711>-2.189, we fail to reject null hypothesis

    Questions 8 a-d, 10 a-f, 12 a-c on page 7
    8. (a) test statistic=(4.9-4.5)/[1.3/sqrt(13)]=1.11
    (b)
    (c) From t table, p-value=P(T>1.11)=0.144
    Interpret: p-value is the area under the t-distribution with df=12 to the right of 1.11.
    (d) Since p-value=0.144>0.1=a, we fail to reject null hypothesis

    10. (a) No, since n=40 is large enough, population does not have to be normally distributed
    (b) test statistic=(48.3-45)/[8.5/sqrt(40)]=2.455
    (c)
    (d) From t table, p-value=2*P(T>2.455)=0.0187
    Interpret: p-value is the area under the t-distribution with df=39 between -2.455 and 2.455.
    (e) Since p-value=0.0187>0.1=a, we fail to reject null hypothesis
    (f) At 99% confidence level with df=40-1=39, the critical value is 2.708 from t table.
    A 99% confidence interval is [48.3-2.708*8.5/sqrt(40), 48.3+2.708*8.5/sqrt(40)]=[44.66, 51.94]. Since 45 is contained in the interval, we fail to reject null hypothesis

    9. 12. (a) Let u be the mean height of women 20 years of age or older. Ho: u=63.7 vs. Ha: u>63.7
    (b) P-value is 0.35: it means that the probability of obtaining a sample mean of 63.9 inches or taller from a population whose mean is 63.7 inches is 0.35.
    (c) Since p-value=0.35>0.10=a, we fail to reject null hypothesis

    Questions 18, 20, and 22 a-b on page 8
    18. Test statistic=(128.7-154.8)/[46.5/sqrt(50)]=-3.969
    At a=0.05 with df=50-1=49, the critical value is -1.677 from t table.
    Since -1.677>-3.969, we reject null hypothesis
    Conclusion: there is evidence to say that internet users spend less time watching TV.

    20. At 95% confidence level with df=35-1=34, the critical value is 2.032 from t table.
    A 95% confidence interval for mean energy expenditure is [1618-2.032*321/sqrt(35), 1618+2.032*321/sqrt(35)]=[1507.746, 1728.254]
    Interpretation: we are 95% confident that mean energy expenditure is between 1507.746 and 1728.254.

    22. (a) Yes, the requirements are satisfied because from probability plot, the distribution is normal and from box plot, there is no outlier.
    (b) From excel, xbar=208.4 and sd=9.3832. So test statistic=(208.4-198)/[9.3832/sqrt(10)]=3.505.
    But at a=0.10 with df=10-1=9, the critical value=1.383<3.505. We reject Ho.
    Conclusion: there is evidence to say that class is effective.

    Questions 26 a-b, and 32 a-d on page 9
    26. (a) Because from the histogram, the distribution is skewed and from the box plot, there is outlier.
    (b) Test statistic=(3.28-5.44)/[1.68/sqrt(35)]=-7.606
    But at a=0.05 with df=35-1=34, the critical value=-2.032>-7.606. We reject Ho.
    Conclusion: there is evidence to say that the volume of stock has changed.

    32. (a) Let u be the mean weight loss after 8 weeks. Ho: u=0 vs. Ha: u>0
    (b) Test statistic=(0.9-0)/[7.2/sqrt(950)]=3.853.
    At a=0.1 and df=950-1=949, the critical value is 1.282<3.853. We reject Ho.
    There is evidence to say that mean weight loss of 0.9 pounds is significant.
    (c) No, weight loss does not have any practical significance because 0.9 is too small.
    (d) Test statistic=(0.9-0)/[7.2/sqrt(40)]=0.791.
    At a=0.1 and df=40-1=39, the critical value is 1.304>0.791. We fail to reject Ho.
    There is no evidence to say that mean weight loss of 0.9 pounds is significant.
    Conclusion: as the sample size decrease, p-value increase. Hence, the probability of rejecting Ho decreases.

    Questions 34 a-c, 36 a-d, on page 10
    34. (a) The data is missing in this question.
    (b)
    (c)

    36. (a) Simulation can only be done with licensed software.
    (b)
    (c)
    (d)

    Questions 40, and 42 on page 11
    40. This means that minor departures from normality will not adversely affect the results of the test.

    42. The data is missing in this question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 11:26 pm ad1c9bdddf>
    https://brainmass.com/math/consumer-mathematics/various-question-hypothesis-testing-568892

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