Sample Company Credentials Part A 1 When trying by wzt64827

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									   Part A:

1. When trying to hire managers and executives, companies sometimes verify
the academic credentials described by the applicants. One company that
performs these checks summarized its findings for a six-month period. Of the 84
applicants whose credentials were checked, 15 lied about having a degree.
   a.) Find the large-sample estimate of the proportion of applicants who lied
       about having a degree and the standard error of the estimate.
   b.) Consider these data to be a random sample of credentials from a large
       collection of similar applicants. Give a 90% confidence interval for the true
       proportion of applicants who lie about having a degree.
   c.) Suppose we want margin of error to be .03. What would our sample size
       have to be?



2. A member of Congress wants to determine her popularity in a certain part of
the state. She indicates that the proportion of voters who will vote for her must be
estimated within plus or minus 2 percent of the population proportion. Further, a
95% confidence interval will be used. In past elections, the representative
received 40 percent of the popular vote in that area of the state. She doubts
whether it has changed much. How many registered voters should be sampled?
   (A) 1651      (B) 2000       (C) 2301      (D) 3842      (E) None of the above.


  3. Suppose we want to test Ho: p = 0.5 vs Ha: p < 0.5. Which of the following would
     be a Type II error?
          a. We fail to conclude that p = 0.5 when it really equals 0.5.
          b. We fail to conclude that p = 0.5 when it is really less than 0.5.
          c. We fail to conclude that p < 0.5 when it is really less than 0.5.
          d. We conclude that p < 0.5 when it really equals 0.5.
          e. We conclude that p = 0.5 when it really equals 0.5.


   4. Of 500 respondents in a Christmas tree market telephone survey, 44% had no
      children at home and 56% had at least 1 child at home. The corresponding figures
      for the most recent census are 48% with no children and 52% with at least 1 child.
      Test the alternative hypothesis that the telephone survey technique had a
      probability of selecting a household with no children that is less the value
      obtained by the census. Give the z-statistic and the p-value. What do you
      conclude?
5. In the following situation, a significance test is called for. State the null and
   alternative hypothesis.

       a. The census bureau reports that households spend an average of 31% of
          their total spending on housing. A homebuilders association in Cleveland
          wonders of the national finding applies in their area. They interview a
          sample of 40 households in the Cleveland metropolitan area to learn what
          percent of their spending goes toward housing.


6. Suppose we have found that 11 customers from a random sample of 40 would be
   willing to buy a software upgrade that costs $100. If the upgrade is to be
   profitable, you will need to sell it to more than 20% of your customers. Do the
   sample data give good evidence that more than 20% are willing to buy?
       a. Formulate this problem as a hypothesis test. Give the null and alternative
           hypothesis. Will you use a one or two-sided hypothesis? Why?



       b. Carry out the significance test at level 0.10. Report the test statistic and
          the p-value.



       c. Should you proceed with plans to produce and market the upgrade?



Part B:
1. The t distribution is used because:
        a. We don’t know the distribution of our data and the sample size is small.
        b. We don’t know the true mean of our data and the sample size is small, but
           the data is normal.
        c. We don’t know the true standard deviation of our data and our sample size
           is small, but the data is normal.
        d. We don’t know the sample mean of our data and our sample size is small,
           but the data is normal.
        e. We don’t know the sample standard deviation of our data and our sample
           size is small, but the data is normal.


2. We want to test whether eating a good breakfast improves the mental performance
   of students on exams. If we decide to give a sample of 50 students an exam
   without letting them eat and then the next day giving them a similar exam (but
   having taken the first exam wouldn’t help them on the next one) after they all ate,
   what type of test procedure would we use?
        a. A one sample t test since we don’t know the true standard deviation.
        b. A one sample test of proportions using the success rate as the sample
           proportion.
        c. A two sample test of proportions comparing the success rates.
        d. A two sample t test comparing the exam averages.
        e. A paired t test, pairing the exam scores by student.


3. A sportswriter wished to see if a football filled with helium travels farther, on
   average, than a football filled with air. He used 18 adult males, randomly divided
   into 2 groups: one kicked the helium filled footballs, the other kicked the air
   filled footballs. What are the correct hypotheses to test?
        a. H 0 :   X vs. H A :   X , where X is the true average distance an air
            filled football travels when kicked.
        b. H 0 :  air   helium vs. H A :  air   helium
        c. H 0 :  air   helium vs. H A :  air   helium
        d. H 0 :  air   helium vs. H A :  air   helium
        e. H 0 : p air  p helium vs. H A : p air  p helium where p is the proportion of field
            goals kicked.




4. You will have the complete sales information for last month in a week but right
   now you only have data from a random sample of 50 stores. The mean change of
   sales in the sample is +4.8% and the standard deviation of the changes is 15%.
   Are the average sales for all stores different from last month?
       a. State the appropriate null and alternative hypothesis. Explain how you
           would decide between the one- and two-sided alternatives.



        b. Find the t statistic and its p-value. State your conclusion.
             c. If the test gives strong evidence against the null hypothesis, would you
                conclude that sales are up in each of your stores?



             d. Would the appropriate confidence interval provide the same conclusion?




5. Pat wants to compare the cost of one and two bedroom apartments in the area of your
campus. She collects data from a random sample of 10 advertisements of each type. The
following table gives the data for rent in dollars for the two samples.
                                    Des criptive Statis tics

                           N           Minimum     Max imum     Mean      Std. Deviation
   TWOBED                      10        495.00      750.00    609.0000       89.31219
   ONEBED                      10        395.00      650.00    531.0000       82.79157
   V alid N (lis tw ise)       10


Suppose data are normal. Find a 95% confidence interval for the additional cost of a
second bedroom.




   6. Pat wonders if two bedroom apartments rent for significantly more than one
      bedroom apartments.
          a. State the appropriate null and alternative hypotheses.



             b. Report the test statistic, its degrees of freedom, and the p-value. What do
                you conclude?



             c. Can you conclude every one bedroom apartment costs less than every two
                bedroom apartment?

             d. In the previous exercise you found a confidence interval. In this exercise
                you performed a significance test. Which do you think is more useful to
                someone planning to rent an apartment? Why?
7. According to a poll conducted for Men’s Health magazine/CNN by Opinion Research,
67% of men and 77% of women said that a good diet is very important to good health.
Suppose that this study is based on samples of 900 men and 1200 women.

   a. Test whether the percentages of all men and all women who hold this view are
      different. Use the 5% significance level. Based on this answer will the interval
      contain 0?


   b. Construct a 95% Confidence Interval.



8. A simple random sample of 25 male faculty members at a large university found that
10 felt that the university was supportive of female and minority faculty. An independent
SRS of 20 female faculty found that only 5 felt that the university was supportive of
female and minority faculty. A 95% Confidence Interval for the difference in the true
proportions of males and female faculty at the university who felt that the university was
supportive of female and minority faculty at the time of the survey is 0.15±0.270. If we
want to test Ho: pM=pF vs. Ha: pM  pF, what conclusion should we make at the 5%
significance level?

           A. We cannot decide without running a test of hypothesis.
           B. Since both sample proportions (.4 for males and .25 for females) are not
              less than 0.05, we fail to reject Ho and conclude we could not prove that
              the true proportions are different.
           C. Since the sample proportions .4 for males and .25 for females) are not
              equal, we reject Ho and conclude that the true proportions are not equal.
           D. Since the confidence interval includes 0, we fail to reject Ho and conclude
              that we could not prove the true proportions are different.
           E. Since the confidence interval does not include 0, we reject Ho and
              conclude that the true proportions are different.



9. Which of the following is/are necessary assumptions for the data and test procedure
above?
   A. We must have at least 30 in each sample
   B. Since the sample sizes are less than 30, the data must be normal
   C. The standard deviations (variances) must be equal
   D. All of the above are necessary
   E. None of the above are necessary
Part A keys:
     1. a) 15/84=0.1786, sqrt(15/84*(1-15/84)/84)=0.0418
        b) p_hat +/- 1.645*SE = (0.11, 0.25)
        c) (1.645/0.03)^2*.1786*(1-.1786)=441.088=442
     2. C. sample size (1.96/0.02)^2*.4*.6=2304.96=2305
3. c
4. z=-1.79, p-value=0.0367, reject H0.
5. Ho: p = 31% vs. Ha: p  31%
6. a. Ho: p = 20% vs. Ha: p > 20%
   b. z=1.19, p-value=0.117
   c. no


Part B Keys:
1.c
2.e
3.d
4.a. two-sided hypothesis; b. t=2.26, p-value = between 0.02 and 0.04, reject null
hypothesis; c. no; d. yes
5. 78  87.11
6. t = -2.025, df = 9, p-value = between 0.025 and 0.05, reject null hypothesis;
  c. no; d. problem 6 is more useful
7. a. z = -5.09, p-value = 0, reject null
  b. (-0.14,-0.06)
8.d
9.e

								
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