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How Often Do You Use Cash? In a survey \(^{13}\) of 1000 American adults conducted in April 2012 . \(43 \%\) reported having gone through an entire week without paying for anything in cash. Test to see if this sample provides evidence that the proportion of all American adults going a week without paying cash is greater than \(40 \%\). Use the fact that a randomization distribution is approximately normally distributed with a standard error of \(S E=0.016 .\) Show all details of the test and use a \(5 \%\) significance level.

Short Answer

Expert verified
Yes, the sample provides sufficient evidence to conclude that the proportion of all American adults going a week without paying cash is greater than \(40\% \) at a \(5 \%\)% significance level.

Step by step solution

01

Define the hypotheses

We want to test if the proportion of American adults not using cash for a week is greater than \(40 \%\). Hence, we can define the null hypothesis \(H_0: p = 0.40\) and the alternative hypothesis \(H_A: p > 0.40\). 'p' denotes the proportion of American adults not using cash for a week.
02

Calculate the Test Statistic

The test statistic in this case is a Z-score, which compares the observed sample proportion to the assumed population proportion under the null hypothesis. This can be calculated as \(Z = (p - p_0) / SE\), where \(p = 0.43\) is the sample proportion, \(p_0 = 0.40\) is the assumed population proportion under the null hypothesis and \(SE = 0.016\) is the Standard Error. Substituting the values, we get \(Z = (0.43 - 0.40) / 0.016 = 1.875\).
03

Determine the Rejection Region

At \(5 \%\)% significance level, the rejection region is \(Z > Z_{0.05}\) where \(Z_{0.05}\) is the Z-value with cumulative probability \(0.95\), which equals \(1.645\). If the computed test statistic \(Z = 1.875\) falls in the rejection region, we reject the null hypothesis.
04

Make the Decision

Since the computed test statistic \(Z = 1.875\) is greater than \(Z_{0.05} = 1.645\), it falls in the rejection region. Hence, we reject the null hypothesis. This means that the sample provides sufficient evidence to conclude that the proportion of all American adults going a week without paying cash is greater than \(40\% \) at a \(5 \%\)% significance level.

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Key Concepts

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

Statistical Significance
Statistical significance is a concept used in hypothesis testing to determine if the results from a study are likely to be true or occurred by chance. When we conduct a hypothesis test, we're generally trying to assess whether a sample's results reflect a real effect in the larger population.

In the context of the exercise, the researchers set a significance level of 5%. This level is also called the alpha level and represents the probability of rejecting the null hypothesis when it is actually true. By setting the significance level at 5%, the researchers accepted a 5% risk of concluding that a higher proportion of adults are not using cash when, in fact, the proportion might be at 40%.
  • Significance Level: 5% (伪 = 0.05)
  • Null Hypothesis (H鈧): The true population proportion is 40%.
  • Alternate Hypothesis (H鈧): More than 40% of adults do not use cash.
When the result falls in the rejection area, it is deemed statistically significant, meaning our sample gives enough evidence to move away from the null hypothesis. Thus, in our example, rejecting the null hypothesis at this level suggests the proportion is statistically greater than the assumed 40%.
Proportion Testing
Proportion testing is a statistical method to determine whether the proportion of a certain characteristic within a population is equal to a certain value. It often uses a null hypothesis to test against an alternative hypothesis.

In our problem, the researchers wanted to test the hypothesis that more than 40% of adults go a week without using cash. The sample provided a proportion of 43%, above the claimed population proportion of 40%. By comparing the sample's proportion to the hypothesized population proportion, we determine if the observed data supports the alternative hypothesis.
  • Sample Proportion: 43% (p = 0.43)
  • Hypothesized Proportion: 40% (p鈧 = 0.40)
The key part of testing for a proportion is calculating the standard error (SE). This represents how much we expect the sample proportion to vary from the population proportion due to chance. In this case, the SE was given as 0.016. This standard error helped in calculating the Z-score, for determining how far the observed proportion is from the hypothesized proportion in terms of standard errors.
Z-score
A Z-score is a measure that describes a value's position in relation to the mean of a group of values. It is expressed in terms of standard deviations from the mean. A Z-score summarizes how far away a particular point is from the mean in standard error units.

In our scenario, the Z-score indicates how many standard errors the sample proportion of 43% is from the hypothesized value of 40%. The formula used is:\[ Z = \frac{(p - p_0)}{SE} \]Here, \( p = 0.43 \) is the sample proportion, \( p_0 = 0.40 \) is the hypothesized population proportion, and \( SE = 0.016 \) is the standard error.

The calculated Z-score was 1.875, which is greater than the critical Z-value of 1.645 for a 5% significance level. This tells us that our sample proportion is significantly greater than the hypothesized proportion. A higher Z-score in the rejection region suggests that the difference we observe is not just due to random variation鈥攊t's likely due to a true difference in proportions. Thus, statistical methods affirm evidence against the null hypothesis in favor of the alternative.

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Most popular questions from this chapter

Exercises 5.7 to 5.12 include a set of hypotheses, some information from one or more samples, and a standard error from a randomization distribution. Find the value of the standardized \(z\) -test statistic in each situation. Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1}>\mu_{2}\) when the samples have \(n_{1}=n_{2}=50, \bar{x}_{1}=35,4, \bar{x}_{2}=33.1, s_{1}=\) 1.28 , and \(s_{2}=1.17\). The standard error of \(\bar{x}_{1}-\bar{x}_{2}\) from the randomization distribution is \(0.25 .\)

Exercise and Gender The dataset ExerciseHours contains information on the amount of exercise (hours per week) for a sample of statistics students. The mean amount of exercise was 9.4 hours for the 30 female students in the sample and 12.4 hours for the 20 male students, A randomization distribution of differences in means based on these data, under a null hypothesis of no difference in mean exercise time between females and males, is centered near zero and reasonably normally distributed. The standard error for the difference in means, as estimated from the randomization distribution, is \(S E=2.38\). Use this information to test, at a \(5 \%\) level, whether the data show that the mean exercise time for female statistics students is less than the mean exercise time of male statistics students.

Find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately normally distributed. A \(95 \%\) confidence interval for a proportion \(p\) if the sample has \(n=100\) with \(\hat{p}=0.43,\) and the standard error is \(S E=0.05\).

Incentives to Exercise: How Much More Effective Is It to Lose Money? In Exercise \(5.30,\) we see that overweight participants who lose money when they don't meet a specific exercise goal meet the goal more often, on average, than those who win money when they meet the goal, even if the final result is the same financially. In particular, participants who lost money met the goal for an average of 45.0 days (out of 100 ) while those winning money or receiving other incentives met the goal for an average of 33.7 days. In Exercise 5.30 we see that the incentive does make a difference. In this exercise, we ask how big the effect is between the two types of incentives. Find and interpret a \(95 \%\) confidence interval for the difference in mean number of days meeting the goal, between people who lose money when they don't meet the goal and those who win money or receive other similar incentives when they do meet the goal. The standard error for the difference in means from a bootstrap distribution is 4.14 .

Incentives for Quitting Smoking: Group or Individual? In a smoking cessation program, over 2000 smokers who were trying to quit were randomly assigned to either a group program or an individual program. After six months in the program, 148 of the 1080 in the group program were successfully abstaining from smoking. while 120 of the 990 in the individual program were successful. \({ }^{15}\) We wish to test to see if this data provide evidence of a difference in the proportion able to quit smoking between smokers in a group program and smokers in an individual program. (a) State the null and alternative hypotheses, and give the notation and value of the sample statistic. (b) Use a randomization distribution and the observed sample statistic to find the p-value. (c) Give the mean and standard error of the normal distribution that most closely matches the randomization distribution, and then use this normal distribution with the observed sample statistic to find the p-value. (d) Use the standard error found from the randomization distribution in part (b) to find the standardized test statistic, and then use that test statistic to find the p-value using a standard normal distribution. (e) Compare the p-values from parts (b), (c), and (d). Use any of these p-values to give the conclusion of the test.

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