Ch. 6 - Normal Probability Distributions

Triola - Elementary Statistics 14th Edition

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Triola 14th Edition

Ch. 6 - Normal Probability Distributions

Problem 6.6.18c

Chapter 6, Problem 6.6.18c

Sleepwalking Assume that 29.2% of people have sleepwalked (based on “Prevalence and Comorbidity of Nocturnal Wandering in the U.S. Adult General Population, by Ohayon et al., Neurology, Vol. 78, No. 20). Assume that in a random sample of 1480 adults, 455 have sleepwalked.

c. What does the result suggest about the rate of 29.2%?

Verified step by step guidance

Step 1: Define the null hypothesis (H₀) and the alternative hypothesis (H₁). The null hypothesis states that the proportion of people who have sleepwalked is 29.2% (p = 0.292). The alternative hypothesis states that the proportion is different from 29.2% (p ≠ 0.292).

Step 2: Calculate the sample proportion (p̂). The sample proportion is given by the formula:

\frac{x}{n}

, where x is the number of people who have sleepwalked (455) and n is the sample size (1480).

Step 3: Compute the standard error (SE) of the sample proportion. The formula for the standard error is:

\sqrt{\frac{p (1 - p)}{n}}

, where p is the hypothesized proportion (0.292) and n is the sample size (1480).

Step 4: Calculate the test statistic (z-score). The formula for the z-score is:

\frac{p̂ - p}{SE}

, where p̂ is the sample proportion, p is the hypothesized proportion, and SE is the standard error.

Step 5: Compare the calculated z-score to the critical z-value for the chosen significance level (e.g., α = 0.05 for a two-tailed test). If the z-score falls outside the range of the critical z-values, reject the null hypothesis. Otherwise, fail to reject the null hypothesis. Interpret the result in the context of whether the sample proportion suggests a significant difference from the hypothesized rate of 29.2%.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Sampling Distribution

The sampling distribution refers to the probability distribution of a statistic (like a sample proportion) obtained from a large number of samples drawn from a specific population. It helps in understanding how sample statistics vary and is crucial for making inferences about the population based on sample data.

Hypothesis Testing

Hypothesis testing is a statistical method used to make decisions about a population based on sample data. It involves formulating a null hypothesis (e.g., the population proportion is 29.2%) and an alternative hypothesis, then using sample data to determine whether to reject the null hypothesis based on a significance level.

Confidence Intervals

A confidence interval is a range of values, derived from sample statistics, that is likely to contain the population parameter with a specified level of confidence (e.g., 95%). It provides an estimate of the uncertainty around the sample proportion and helps assess whether the observed sample proportion significantly deviates from the hypothesized population proportion.

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Related Practice

Textbook Question

In Exercises 1 and 2, use the following wait times (minutes) at 10:00 AM for the Tower of Terror ride at Disney World (from Data Set 33 “Disney World Wait Times” in Appendix B).

35 35 20 50 95 75 45 50 30 35 30 30

Tower of Terror Wait Times

a. Find Q1, Q2 and Q3.

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Textbook Question

Continuity Correction In testing the assumption that the probability of a baby boy is 0.512, a geneticist obtains a random sample of 1000 births and finds that 502 of them are boys. Using the continuity correction, describe the area under the graph of a normal distribution corresponding to the following. (For example, the area corresponding to “the probability of at least 502 boys” is this: the area to the right of 501.5.)

c. The probability of more than 502 boys

Textbook Question

In Exercises 1 and 2, use the following wait times (minutes) at 10:00 AM for the Tower of Terror ride at Disney World (from Data Set 33 “Disney World Wait Times” in Appendix B).

35 35 20 50 95 75 45 50 30 35 30 30

d. Find the variance.

Textbook Question

Hershey Kisses Based on Data Set 38 “Candies” in Appendix B, weights of the chocolate in Hershey Kisses are normally distributed with a mean of 4.5338 g and a standard deviation of 0.1039 g

d. What is the value of the variance?

Textbook Question

Significance For bone density scores that are normally distributed with a mean of 0 and a standard deviation of 1, find the percentage of scores that are

c. not significant (or less than 2 standard deviations away from the mean).

Textbook Question

Ergonomics. Exercises 9–16 involve applications to ergonomics, as described in the Chapter Problem.

Safe Loading of Elevators The elevator in the car rental building at San Francisco International Airport has a placard stating that the maximum capacity is “4000 lb—27 passengers.” Because 4000/27=148, this converts to a mean passenger weight of 148 lb when the elevator is full. We will assume a worst-case scenario in which the elevator is filled with 27 adult males. Based on Data Set 1 “Body Data” in Appendix B, assume that adult males have weights that are normally distributed with a mean of 189 lb and a standard deviation of 39 lb.

c. What do you conclude about the safety of this elevator?