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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

In a Northeastern University/Gallup poll of 461 young Americans aged 18 to 35,152 reported they would be comfortable riding in a self-driving car. Suppose we are testing the hypothesis that more than \(30 \%\) of Americans in this age group would be comfortable riding in a self-driving car, using a significance level of \(0.05 .\) Which of the following figures correctly matches the alternative hypothesis \(p>0.30 .\) Report and interpret the correct p-value.

Votes for Independents Judging on the basis of experience, a politician claims that \(50 \%\) of voters in Pennsylvania have voted for an independent candidate in past elections. Suppose you surveyed 20 randomly selected people in Pennsylvania, and 12 of them reported having voted for an independent candidate. The null hypothesis is that the overall proportion of voters in Pennsylvania that have voted for an independent candidate is \(50 \%\). What value of the test statistic should you report?

A hospital readmission is an episode when a patient who has been discharged from a hospital is readmitted again within a certain period. Nationally the readmission rate for patients with pneumonia is \(17 \% .\) A hospital was interested in knowing whether their readmission rate for pneumonia was less than the national percentage. They found 11 patients out of 70 treated for pneumonia in a two-month period were readmitted. a. What is \(\hat{p}\), the sample proportion of readmission? b. Write the null and alternative hypotheses. c. Find the value of the test statistic and explain it in context. d. The p-value associated with this test statistic is \(0.39 .\) Explain the meaning of the \(\mathrm{p}\) -value in this context. Based on this result, does the \(\mathrm{p}\) -value indicate the null hypothesis should be doubted?

St. Louis County is \(24 \%\) African American. Suppose you are looking at jury pools, each with 200 members, in St. Louis County. The null hypothesis is that the probability of an African American being selected into the jury pool is \(24 \%\). a. How many African Americans would you expect on a jury pool of 200 people if the null hypothesis is true? b. Suppose pool A contains 40 African American people out of 200 , and pool B contains 26 African American people out of 200 . Which will have a smaller p-value and why?

If we reject the null hypothesis, can we claim to have proved that the null hypothesis is false? Why or why not?
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