What is a percentile?

When we look up a z score on the z table, what information can we report?

How do we calculate the percentage of scores below a particular positive z score?

How is calculating a percentile for a mean from a distribution of means different from doing so for a score from a distribution of scores?

In statistics, what do we mean by assumptions?

What sample size is recommended in order to meet the assumption of a normal distribution of means, even when the underlying population of scores is not normal?

What is the difference between parametric tests and nonparametric tests?

What are the six steps of hypothesis testing?

What are critical values and the critical region?

What is the standard size of the critical region used by most statisticians?

What does statistically significant mean to statisticians?

What do these symbolic expressions mean: H₀: μ₁ = μ₂ and H₁: μ₁ ≠ μ₂?

Using everyday language rather than statistical language, explain why the words critical region might have been chosen to define the area in which a z statistic must fall in order for a researcher to reject the null hypothesis.

Using everyday language rather than statistical language, explain why the word cutoff might have been chosen to define the point beyond which we reject the null hypothesis.

What is the difference between a one-tailed hypothesis test and a two-tailed hypothesis test in terms of critical regions?

Why do researchers typically use a two-tailed test rather than a one-tailed test?

Write the symbols for the null hypothesis and research hypothesis for a one-tailed test.

Calculate the following percentages for a z score of –1.61, with a tail of 5.37%:

Calculate the following percentages for a z score of 0.74, with a tail of 22.96%:

Using the z table in Appendix B, calculate the following percentages for a z score of –0.08:

Using the z table in Appendix B, calculate the following percentages for a z score of 1.71:

Rewrite each of the following percentages as probabilities, or p levels:

Rewrite each of the following probabilities, or p levels, as percentages:

If the critical values for a hypothesis test occur where 2.5% of the distribution is in each tail, what are the cutoffs for z?

For each of the following p levels, what percentage of the data will be in each critical region for a two-tailed test?

State the percentage of scores in a one-tailed critical region for each of the following p levels:

You are conducting a z test on a sample of 50 people with an average SAT verbal score of 542 (assume we know the population mean to be 500 and the standard deviation to be 100). Calculate the mean and the spread of the comparison distribution ( μ_M and σ_M).

You are conducting a z test on a sample of 132 people for whom you observed a mean SAT verbal score of 490. The population mean is 500, and the standard deviation is 100. Calculate the mean and the spread of the comparison distribution ( μ_M and σ_M).

If the cutoffs for a z test are –1.96 and 1.96, determine whether you would reject or fail to reject the null hypothesis in each of the following cases:

If the cutoffs for a z test are –2.58 and 2.58, determine whether you would reject or fail to reject the null hypothesis in each of the following cases:

Use the cutoffs of –1.65 and 1.65 and a p level of approximately 0.10, or 10%. For each of the following values, determine whether you would reject or fail to reject the null hypothesis:

You are conducting a z test on a sample for which you observe a mean weight of 150 pounds. The population mean is 160, and the standard deviation is 100.

Percentiles and unemployment rates: The U.S. Bureau of Labor Statistics’ annual report published in 2011 provided adjusted unemployment rates for 10 countries. The mean was 7, and the standard deviation was 1.85. For the following calculations, treat 7 as the population mean and 1.85 as the population standard deviation.

Height and the z distribution, question 1: Elena, a 15-year-old girl, is 58 inches tall. Based on what we know, the average height for girls at this age is 63.80 inches, with a standard deviation of 2.66 inches.

Height and the z distribution, question 2: Kona, a 15-year-old boy, is 72 inches tall. According to the CDC, the average height for boys at this age is 67.00 inches, with a standard deviation of 3.19 inches.

Height and the z statistic, question 1: Imagine a class of thirty-three 15-year-old girls with an average height of 62.6 inches. Remember, μ = 63.8 inches and σ = 2.66 inches.

Height and the z statistic, question 2: Imagine a basketball team comprised of thirteen 15-year-old boys. The average height of the team is 69.5 inches. Remember, μ = 67 inches and σ = 3.19 inches.

The z distribution and statistics test scores: Imagine that your statistics professor lost all records of students’ raw scores on a recent test. However, she did record students’ z scores for the test, as well as the class average of 41 out of 50 points and the standard deviation of 3 points (treat these as population parameters). She informs you that your z score was 1.10.

The z statistic, distributions of means, and height, question 1: Using what we know about the height of 15-year-old girls (again, μ = 63.8 inches and σ = 2.66 inches), imagine that a teacher finds the average height of 14 female students in one of her classes to be 62.4 inches.

The z statistic, distributions of means, and height, question 2: Another teacher decides to average the heights of all 15-year-old male students in his classes throughout the day. By the end of the day, he has measured the heights of 57 boys and calculated an average of 68.1 inches (remember, for this population μ = 67 inches and σ = 3.19 inches).

Directional versus nondirectional hypotheses: For each of the following examples, identify whether the research has expressed a directional or a nondirectional hypothesis:

Null hypotheses and research hypotheses: For each of the following examples (the same as those in Exercise 7.41), state the null hypothesis and the research hypothesis, in both words and symbolic notation:

The z distribution and Hurricane Katrina: Hurricane Katrina hit New Orleans on August 29, 2005. The National Weather Service Forecast Office maintains online archives of climate data for all U.S. cities and areas. These archives allow us to find out, for example, how the rainfall in New Orleans that August compared to that in the other months of 2005. The table below shows the National Weather Service data (rainfall in inches) for New Orleans in 2005.

Steps 1 and 2 of hypothesis testing for a study of the Wechsler Adult Intelligence Scale-Revised: Boone (1992) examined scores on the Wechsler Adult Intelligence Scale-Revised (WAIS-R) for 150 adult psychiatric inpatients. He determined the “intrasubtest scatter” score for each inpatient. Intrasubtest scatter refers to patterns of responses in which respondents are almost as likely to get easy questions wrong as hard ones. In the WAIS-R, we expect more wrong answers near the end, as the questions become more difficult, so high levels of intrasubtest scatter would be an unusual pattern of responses. Boone wondered if psychiatric patients have different response patterns than nonpatients have. He compared the intrasubtest scatter for 150 patients to population data from the WAIS-R standardization group. Assume that he had access both to means and standard deviations for this population. Boone reported that “the standardization group’s intrasubtest scatter was significantly greater than those reported for the psychiatric inpatients” and concluded that such scatter is normal.

Step 1 of hypothesis testing for a study of college football: Let’s consider whether U.S. college football teams are more likely or less likely to be mismatched in the upper National Collegiate Athletic Association (NCAA) divisions. Overall, the 53 Football Bowl Subdivision (FBS) games (the highest division) had a mean spread (winning score minus losing score) of 16.189 in a particular week, with a standard deviation of 12.128. We took a sample of 4 games that were played that week in the next-highest league, the Football Championship Subdivision (FCS), to see if the mean spread was different; one of the many leagues within the FCS, the Patriot League, played 4 games that weekend.

Steps 2 through 6 of hypothesis testing for a study of college football: Refer to the scenario described in Exercise 7.45.

Passwords, red heads, and a z test: The BBC reported that red-haired women were more likely than others to choose strong passwords (Ward, 2013). How might a researcher study this? Computer scientist Cynthia Kuo and her colleagues conducted a study in which they gave passwords a score, with higher scores going to passwords that are not in “password crack dictionaries” that are used by hackers, that are longer, and that use a mix of letters, numbers, and symbols (2006). They found a mean score of 15.7 with a standard deviation of 7.3. For the purposes of this exercise, treat these numbers as the population parameters. Based on your knowledge of the z test, explain how you might design a study to test the hypothesis that red-haired women create stronger passwords than others.

The Graded Naming Test and sociocultural differences: Researchers often use z tests to compare their samples to known population norms. The Graded Naming Test (GNT) asks respondents to name objects in a set of 30 black-and-white drawings. The test, often used to detect brain damage, starts with easy words like kangaroo and gets progressively more difficult, ending with words like sextant. The GNT population norm for adults in England is 20.4. Roberts (2003) wondered whether a sample of Canadian adults had different scores than adults in England. If they were different, the English norms would not be valid for use in Canada. The mean for 30 Canadian adults was 17.5. For the purposes of this exercise, assume that the standard deviation of the adults in England is 3.2.

Patient adherence and orthodontics: A research report (Behenam & Pooya, 2007) begins, “There is probably no other area of health care that requires. . .cooperation to the extent that orthodontics does,” and explores factors that affected the number of hours per day that Iranian patients wore their orthodontic appliances. The patients in the study reported that they used their appliances, on average, 14.78 hours per day, with a standard deviation of 5.31. We’ll treat this group as the population for the purposes of this example. Let’s say a researcher wanted to study whether a video with information about orthodontics led to an increase in the amount of time patients wore their appliances, but decided to use a two-tailed test to be conservative. Let’s say he studied the next 15 patients at his clinic, asked them to watch the video, and then found that they wore their appliances, on average, 17 hours per day.

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Radiation levels on Japanese farms: Fackler (2012) reported in the New York Times that Japanese farmers have become skeptical of the Japanese government’s assurances that radiation levels were within legal limits in the wake of the 2011 tsunami and radiation disaster at Fukushima. After reports of safe levels in Onami, more than 12 concerned farmers tested their crops and found dangerously high levels of cesium.