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Chapter 8: Problem 23

A 2003 study of dreaming published in the journal Perceptual and Motor Skills found that out of a random sample of 113 people, 92 reported dreaming in color. However, the proportion of people who reported dreaming in color that was established in the 1940 s was \(0.29\) (Schwitzgebel 2003). Check to see whether the conditions for using a one-proportion z-test are met assuming the researcher wanted to see whether the proportion dreaming in color had changed since the \(1940 \mathrm{~s}\).

Short Answer

Expert verified
All the conditions to run a one-proportion z-test are met. The sample was random, the size was large enough to meet the Normality conditions, and the Independence condition is met.

Step by step solution

01

Identify the sample and population

From the problem, it is understood that a random sample of 113 people have been taken, making this the sample size. Of these, 92 reported dreaming in colour. This leads to a sample proportion of \(\frac{92}{113} = 0.814\). The established proportion in the 1940s is given as 0.29.
02

Check for Randomness

The problem states that the sample was random. Therefore, the condition of randomness is met
03

Check for Normality

The normality condition can be checked by ensuring that np and n(1-p) are both greater than 10. Here, n is 113, p is 0.29 (the old proportion from the 1940s). Checking these values, np = 113*0.29 = 32.77 and n(1-p) = 113*(1-0.29) = 80.23. Both these values are greater than 10. Therefore, the sample size is sufficiently large, and the Normality condition is met.
04

Check for Independence

The question states that the data was gathered from a random sample. Presuming that the population who dreams is more than 10 times the size of the sample (1130), the independence condition is met.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Random Sampling
In the context of hypothesis testing, random sampling is crucial to ensure unbiased and representative data. A random sample allows each member of the population an equal chance of being selected. This means the data collected will more accurately reflect the population as a whole. In our exercise, a random sample of 113 people was selected. This ensures that the findings related to dreaming in color are not skewed by selection bias.
  • Using random sampling helps avoid bias.
  • It increases the validity of the conclusions.
Without a random sample, our results might not be applicable to the broader population, affecting the reliability of the z-test.
Normality Condition
The normality condition is vital for performing a one-proportion z-test. This condition ensures that the sampling distribution of the sample proportion is approximately normal. To check for normality, both np and n(1-p) must be greater than 10, where n is the sample size and p is the population proportion. In our scenario:
  • \( np = 113 \times 0.29 = 32.77 \)
  • \( n(1-p) = 113 \times (1-0.29) = 80.23 \)
Both values exceed 10, satisfying the normality condition. This implies that the sample size is large enough for the sampling distribution to resemble a normal distribution, allowing for effective use of a z-test.
Independence Condition
The independence condition ensures that each individual's response is not influenced by others. For a one-proportion z-test, we assume that the sample observations are independent. Since the sample was random and presumably less than 10% of the population, this condition is likely met.
  • Random selection supports independence.
  • The sample should be less than 10% of the total population.
Assuming our sample is considerably smaller than the total dreaming population, we can proceed confidently with our hypothesis test. It reinforces the reliability and accuracy of the z-test results.

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Most popular questions from this chapter

Choose one of the answers given. The null hypothesis is always a statement about a ______ (sample statistic or population parameter).

If we do not reject the null hypothesis, is it valid to say that we accept the null hypothesis? Why or why not?

The researchers in a Pew study interviewed two random samples, one in 2015 and one in 2018 . Both samples were asked, "Have you read a print book in the last year?" The results are shown in the table below. $$\begin{array}{|lrrl|}\hline \text { Read a print book } & \mathbf{2 0 1 5} & \mathbf{2 0 1 8} & \text { Total } \\ \hline \text { Yes } & 1201 & 1341 & 2542 \\\\\hline \text { No } & 705 & 661 & 1366 \\ \hline \text { Total } & 1906 & 2002 & \\\\\hline\end{array}$$ a. Find and compare the sample proportions that had read a print book for these two groups. b. Find a pooled estimate of the sample proportion. c. Has the proportion who read print books increased? Find the observed value of the test statistic to test the hypotheses \(\mathrm{H}_{0}: p_{2015}=p_{2018}\) and \(\mathrm{H}_{\mathrm{a}}: p_{2015}

By establishing a small value for the significance level, are we guarding against the first type of error (rejecting the null hypothesis when it is true) or guarding against the second type of error?

According to one source, \(50 \%\) of plane crashes are due at least in part to pilot error (http://www.planecrashinfe .com). Suppose that in a random sample of 100 separate airplane accidents, 62 of them were due to pilot error (at least in part.) a. Test the null hypothesis that the proportion of airplane accidents due to pilot error is not \(0.50\). Use a significance level of \(0.05\). b. Choose the correct interpretation: i. The percentage of plane crashes due to pilot error is not significantly different from \(50 \%\). ii. The percentage of plane crashes due to pilot error is significantly different from \(50 \%\)
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