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Review Problems for Quiz 4 with Solutions
  1. Why do we generally expect some error when estimating a parameter (such as a population mean) by a statistic (such as a sample mean)?


    2. Sampling variability (error)
     

  2. Explain why increasing the sample size results in a tendency for smaller sampling error when using a sample mean to estimate a population mean.

    3.  

     

    The standard deviation of the mean will decrease as the sample size increases.
     

  3. Define sampling error (also called sampling variability).
Sampling error is the variability that occurs because the value of a sample statistic will vary from sample to sample.
  1. What exactly is a confidence interval and why is it better to report a confidence interval instead of a single number for estimating the population mean, ??

    2.  

     

    A confidence interval is simply an estimate +/- a margin of error. It is better to report a confidence interval because it allows us to include some estimate of the sampling variability present when estimating a population parameter by a sample statistic.
     

  2. Suppose that we have obtained data by taking a simple random sample from a population. What should we do (e.g., what assumptions should we verify) before we construct a confidence interval from the sample?


    3. Verify that the sample mean has a normal distribution.
     

  3. Suppose that we have obtained data by taking a simple random sample from a population and we intend to find a confidence interval for the population mean. We will either use a 95% confidence interval or a 99% confidence interval. Which confidence level will give us a narrower interval?

    4.  

     

    The 95% confidence interval will give a narrower interval for estimating the mean.
     

  4. The Gallup Organization conducts annual national surveys on home gardening. Results are published by the national Association for Gardening. A random sample is taken of 25 households with vegetable gardens. The size of vegetable gardens is normally distributed. The mean size of the vegetable gardens from the sample was 643 sq ft.
  1. Find a 90% confidence interval for the mean size of all household vegetable gardens in the United States. Assume that sigma=247 sq ft.


    2. x-bar +/- z*(sigma / sqrt(n))

    643 +/- (1.645)(247 / sqrt(25))

    643 +/- 81.263
     

  2. Explain in words what the confidence interval from part (a) means.
We are 90% confident that the true average size of vegetable gardens in the U.S. is between 561.74 and 724.26 square feet.
  1. A quality-control engineer in a bakery goods plant needs to estimate the mean weight of bags of potato chips that are packed by a machine. He knows from experience that sigma=0.1 oz for this machine. Weights of bags are normally distributed. A random sample of 12 bags has a mean weight of 16.01 oz.
  1. Find a 99% confidence interval for the mean weight bags of potato chips.


    2. x-bar +/- z*(sigma / sqrt(n))

    16.01 +/- (2.576)(0.1 / sqrt(12))

    16.01 +/- 0.074
     
     
     

  2. Explain in words what the confidence interval from part (a) means.
We are 99% confident that the true mean weight of the bags of potato chips is between 15.94 and 16.08 ounces.
 
  1. Suppose that we are considering a hypothesis test for a population mean. In each part below, express the alternative hypothesis symbolically and give the formula for the p-value that would be used for the test:
  1. We want to decide whether the population mean is different from a specified value mo .

    2.  

     

    Ha: m ¹mo
    p-value = 2*P(Z ³ |z|)
     

  2. We want to decide whether the population mean is less than a specified value mo

    3.  

     

    Ha: m < mo
    p-value = P(Z £ z)
     

  3. We want to decide whether the population mean is greater than a specified value mo .
Ha: m > mo
p-value = P(Z ³ z)
  1. The Radio Advertising Bureau of New York reports in Radio Facts that in 1994 the mean number of radios per U.S. household was 5.6. A random sample of 45 U.S. households taken this year showed that the average number of radios owned is 5.9. Do the data provide sufficient evidence to conclude that this year’s mean number of radios per U.S. household has changed from the 1994 mean of 5.6? Assume that the standard deviation of this year’s number of radios per U.S. household is 1.9. Use the following steps to answer the question.
  1. State the null and alternative hypotheses.

    2.  

     

    Ho: m = 5.6

    Ha: m ¹ 5.6
     

  2. Discuss the logic of conducting the hypothesis test (e.g., how will you determine whether you have enough evidence to reject the null hypothesis).

    3.  

     

    We will calculate the test statistic and then calculate the p-value. The p-value will tell us whether or not the test statistic is extreme enough for us to reject the null hypothesis.
     

  3. Identify the distribution of the variable x-bar; that is, the sampling distribution of the mean for samples of size 45.

    4.  

     

    Since n>30, the Central Limit Theorem (CLT) tells us that the sampling distribution of the mean is NORMAL.
     

  4. Obtain a precise criterion for deciding whether to reject the null hypothesis in favor of the alternative hypothesis (e.g., pick out a significance level that you will use for conducting the test. This can be any value that you would like to use, but most of the time, a 5% significance level is used).

    5.  

     

    We will use 0.05 to conduct the test.
     

  5. Calculate your p-value.

    6.  

     

    To do this, we must first calculate the test statistic:

    z = (x-bar -  mo ) / {s / sqrt(n)} = (5.9 - 5.6)/{1.9/sqrt(45)} = 1.07
     

    p-value = 2*P(Z ³ 1.07) = 2{1 – P(Z £ 1.07)} = 2*(1 – 0.8577) = 0.2846
     

  6. Apply the criterion in part (d) to the problem and state your conclusion.
Since the p-value (0.2846) is greater than a (0.05), fail to reject the null hypothesis and conclude that the average number of radios per household has not changed since 1994.