Solutions
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 The population standard deviation is unknown, and the sample size is small (13), so we need to use a t procedure. The problem tells us that the sample is random. A histogram of the data shows no significant departure from normality:
 This is a twoproportion situation. We are told that the groups were randomly selected, but we need to check that the samples are sufficiently large:
 In this problem, H_{0}: μ = 5 and H_{A}: μ < 5, so we are only interested in the area to the left of our finding of = 4.55. since the hotel believes that the average stay is less than 5 days. We are interested in the area shaded in the graph:
 The 99% confidence interval will be more likely to contain the population value being estimated, but will be wider than a 95% confidence interval.


 We are 95% confident that the true difference between the mean age of male statistics teachers and female statistics teachers is between –4.5 years and 3.2 years.
 Since 0 is contained in this interval, we do not have evidence of a statistically significant difference between the mean ages of male and female statistics teachers.

 We are 95% confident that the true difference between the mean age of male statistics teachers and female statistics teachers is between 2.1 years and 3.9 years.
 Since 0 is not in this interval, we do have evidence of a real difference between the mean ages of male and female statistics teachers. In fact, since the interval contains only positive values, we have evidence that the mean age of male statistics teachers is greater than the mean age of female statistics teachers.

 We are 95% confident that the true difference between the mean age of male statistics teachers and female statistics teachers is between –5.2 years and –1.7 years.
 Since 0 is not in this interval, we have evidence of a real difference between the mean ages of male and female statistics teachers. In fact, since the interval contains only negative values, we have evidence that the mean age of male statistics teachers is less than the mean age of female statistics teachers.

 t procedures are appropriate because the population is approximately normal. n = 20 df = 20  1 = 19 t* = 2.861 for C = 0.99.

 A TypeI error is made when we mistakenly reject a true null hypothesis. In this situation, that means that we would mistakenly reject the true hypothesis that the available housing is sufficient. The risk would be that a lot of money would be spent on building additional housing when it wasn't necessary.
 A TypeII error is made when we mistakenly fail to reject a false hypothesis. In this situation that means we would fail to reject the false hypothesis that the available housing is sufficient. The risk is that the university would have insufficient housing for its students.

 We are 95% confident that the true proportion of subjects helped by a new antiinflammatory drug is (0.56, 0.65).
 The process used to construct this interval will capture the true population proportion, on average, 95 times out of 100.
 We have . Because the hypothesis is two sided, we are concerned about the probability of being in either tail of the curve even though the finding was larger than expected.
 The problem states that the samples were randomly selected and that the scores were approximately normally distributed, so we can construct a twosample t interval using the conservative approach. For C = 0.99, we have df = min{231, 262} = 22 t* = 2.819. Using a TI84 with the invT function, t* = invT(0.995,22).
 The power of the test is our ability to reject the null hypothesis. In this case, we reject the null if > 36.5 when μ = 38. We are given = 0.99. Thus
 For a given margin of error using P* = 0.5:
 The finding is statistically significant because the Pvalue is less than the significance level. In this situation, it is unlikely that we would have obtained a value of as different from 0 as we got by chance alone if, in fact, = 0.
 Trick question! The sample size needed for a 95% confidence interval (or any Clevel interval for that matter) is not a function of population size. The sample size needed is given by
 Because n = 50, we could use a largesample confidence interval. For n = 50, z* = 2.33 (that's the upper critical zvalue if we put 1% in each tail).
 The problem states that we have an SRS and n = 49. Since we do not know s, but have a large sample size, we are justified in using t procedures.
 C = 0.95 t* = 2.021 (from Table B with df = 40; from a TI84 with the invT function, the exact value of t* = 2.0106 with df = 48). Thus, 188 ± 2.021 = (186.27, 189.73). Using the TI83/84, we get (186.28, 189.72) based on 48 degrees of freedom. Since 190 is not in this interval, it is a likely population value from which we would have gotten this interval. There is some doubt about the company's claim.
 Let μ = the true mean operating time of the company's pens.
 H_{0}: μ = 190.
 H_{A}: μ ≠ 190.
 We will use a onesample ttest. Justify the conditions needed to use this procedure
 Determine the test statistic (t) and use it to identify the Pvalue
 Compare the Pvalue with α. Give a conclusion in the context of the problem.
(The wording of the questions tells us H_{A} is two sided.)
 It is not appropriate because confidence intervals use sample data to make estimates about unknown population values. In this case, the actual difference in the ages of actors and actresses is known, and the true difference can be calculated.
You would need to survey at least 271 students.
Because 100 is contained in this interval, we do not have strong evidence that the mean number of bushels per acre differs from 100, even though the sample mean is only 98.1.
Since all values are greater than or equal to 5, we are justified in constructing a twoproportion z interval. For a 90% z confidence interval, z* = 1.645.
Thus, .
We are 90% confident that the true difference between the dropout rates is between 0.02 and 0.38. Since 0 is not in this interval, we have evidence that the counseling program was effective at reducing the number of dropouts.
Since we do not know s, but do have a large sample size, we will use a t procedure. , df = 100  1. Using df = 80 from Table B (rounding down from 99), we have 0.05 < Pvalue < 0.10. Using a TI83/84 with df = 99, the Pvalue = tcdf(100,1.45,99)= 0.075
The conditions are present to construct a oneproportion z interval.
We are 95% confident that the true proportion of people who will get the flu despite getting the vaccine is between 11.9% and 19.5%. To say that we are 95% confident means that the process used to construct this interval will, in fact, capture the true population proportion 95% of the time (that is, our "confidence" is in the procedure, not in the interval we found!).
Then, the Pvalue = 2(0.1788) = 0.3576. Using the TI83/84, the Pvalue = 2normalcdf(0.92,100).
0 is a likely value for the true difference between the means because it is in the interval. Hence, we do not have evidence that there is a difference between the true means for males and females.
Using the STAT TESTS 2SampTInt function on the TI83/84, we get an interval of (5.04, 7.24), df = 44.968. Remember that the larger number of degrees of freedom used by the calculator results in a somewhat narrower interval than the conservative method (df = min{n_{1}  1, n_{2}  1}) for computing degrees of freedom.
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If the true mean is really 38, we are almost certain to reject the false null hypothesis.
.
To reduce the margin of error by a factor of 0.5, we have
We need to quadruple the sample size to reduce the margin of error by a factor of 1/2.
Hence,
The department should ask at least 35 students about their intentions.
n is a function of z* (which is determined by the confidence level), M (the desired margin of error), and P* (the estimated value of ). The only requirement is that the population size be at least 20 times as large as the sample size.
Anna tells her father that he can be 98% confident that the true average test score for his 41 years of teaching is between 71.16 and 75.84.
Since we do not know s, however, a t interval is more appropriate. The TI83/84 calculator returns a tinterval of (71.085, 75.915). Here, t* = 2.4049, so that the resulting interval is slightly wider, as expected, than the interval obtained by assuming an approximately normal distribution.
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