/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Smoking among Seniors. Since the... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Smoking among Seniors. Since the 1990 s, daily cigarette smoking among high school seniors has dropped from around \(12 \%\) to slightly over \(3 \%\). In 2017, random samples of 1725 female and 1564 male high school seniors found that \(3.7 \%\) of the females and \(3.1 \%\) of the males had smoked cigarettes daily in the 30 days before the survey. \(\underline{2}\) Give a \(95 \%\) confidence interval for the difference between the proportions of the populations of male and female high school seniors who had smoked cigarettes daily in the 30 days before the survey. Follow the four-step process as illustrated in Examples \(23.2\) and \(23.3\) (pages 518 and 520 ).

Short Answer

Expert verified
The 95% confidence interval for the difference is \((-0.0126, 0.0246)\).

Step by step solution

01

Define Variables

Let \( p_f \) be the proportion of female high school seniors who smoked daily, and \( p_m \) be the proportion of male high school seniors who smoked daily.
02

Calculate Proportions

The proportion of female smokers is \( \hat{p}_f = \frac{3.7}{100} = 0.037 \), and for males \( \hat{p}_m = \frac{3.1}{100} = 0.031 \).
03

State the Formula for Confidence Interval

We use the formula for the confidence interval of the difference of proportions: \[ (\hat{p}_f - \hat{p}_m) \pm z^* \sqrt{\frac{\hat{p}_f(1-\hat{p}_f)}{n_f} + \frac{\hat{p}_m(1-\hat{p}_m)}{n_m}} \] where \( z^* \) is the z-score for a 95% confidence level.
04

Determine Critical Value

For a 95% confidence level, \( z^* \approx 1.96 \).
05

Plug in the Values

Substitute the values into the formula:- \( \hat{p}_f = 0.037 \), \( n_f = 1725 \)- \( \hat{p}_m = 0.031 \), \( n_m = 1564 \)- Calculate the standard error: \[ \sqrt{\frac{0.037(1-0.037)}{1725} + \frac{0.031(1-0.031)}{1564}} \approx 0.0095 \]
06

Calculate the Confidence Interval

The difference in sample proportions is \( 0.037 - 0.031 = 0.006 \).Calculate the margin of error: \[ 1.96 \times 0.0095 \approx 0.0186 \]Thus, the 95% confidence interval is: \[ 0.006 \pm 0.0186 = (-0.0126, 0.0246) \]
07

Conclusion

We are 95% confident that the true difference in smoking rates between female and male high school seniors lies within the interval \((-0.0126, 0.0246)\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Difference of Proportions
In statistics, the **difference of proportions** is a measure used to compare two population proportions. It is calculated by subtracting one sample proportion from another. For instance, consider the proportions of female and male high school seniors who smoke daily. Here, the sample proportions are
  • For females, the proportion is 0.037
  • For males, it's 0.031.
To find the difference of proportions, subtract the male proportion from the female proportion. This results in the difference of 0.037 - 0.031 = 0.006.

This difference helps statisticians understand the extent of the variation in behavior (in this case, smoking habits) between two groups, not just within one. However, this difference alone does not account for variability in the data, which is why we use further statistical tools like confidence intervals and z-scores.
Z-Score
The **z-score** is an important statistic that measures how far a data point is from the mean in terms of standard deviations. In the context of confidence intervals, the z-score indicates how many standard errors a value is away from the population mean.

A z-score of 0 implies the value is exactly average, while positive or negative scores denote deviations above or below the mean, respectively.

For a 95% confidence interval, a common critical z-score is approximately 1.96. This implies that if a distribution is normal, 95% of the data will fall within 1.96 standard deviations from the mean.
  • The critical z-score is used to calculate the margin of error in confidence intervals.
  • In our example, using this z-score helps determine the range in which the true proportion difference lies, accounting for variability.
Understanding z-scores allows one to interpret how extreme a result is in relation to a normal distribution.
Random Samples
**Random samples** are crucial in statistics to ensure that the findings from sample data are generalizable to the entire population. In a random sample, each member of the population has an equal chance of being selected.

This randomness removes systematic biases and improves the reliability of the results. In the smoking survey, both male and female high school seniors were chosen randomly, meaning the sample represents the entire student population fairly.
  • This method ensures that variation in smoking habits isn’t due to preformed groups but rather actual differences in the population behavior.
  • Random sampling is a foundational principle in statistical research. It leads to more accurate and reliable statistical inferences.
Confidence Level
The **confidence level** reflects the degree of certainty or reliability in an estimation. A 95% confidence level means that if we were to take 100 different samples and make confidence intervals for each, we would expect 95 of them to contain the true population parameter.

It is a widely accepted standard in statistics for balance between reliability and precision.
  • Higher confidence levels (e.g., 99%) offer more certainty but result in wider confidence intervals.
  • Conversely, lower levels (e.g., 90%) tighten the interval but with less certainty.
In the context of our survey, the 95% confidence level gives us a range in which we believe the true difference in smoking rates between female and male high school seniors will fall. This helps in making informed guesses about population behaviors with a known level of assurance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Take \(p_{9}\) and \(p_{12}\) to be the proportions of all ninth- and 12 thgraders who ate breakfast daily. The hypotheses to be tested are a. \(H_{0}: p_{9}=p_{12}\) versus \(H_{a}: p_{9} \neq p_{12}\). b. \(H_{0}: p_{9}=p_{12}\) versus \(H_{a}: p_{9}>p_{12}\). c. \(H_{0}: p_{9}=p_{12}\) versus \(H_{a}: p_{9}

Significant Does Not Mean Important. Never forget that even small effects can be statistically significant if the samples are large. To illustrate this fact, consider a sample of 148 small businesses. During a three-year period, 15 of the 106 businesses headed by men and seven of the 42 businesses headed by women failed. 22 a. Find the proportions of failures for businesses headed by women and businesses headed by men. These sample proportions are quite close to each other. Give the \(P\). value for the \(z\) test of the hypothesis that the same proportion of women's and men's businesses fail. (Use the two-sided alternative.) The test is very far from being significant. b. Now suppose that the same sample proportions came from a sample 30 times as large. That is, 210 out of 1260 businesses headed by women and 450 out of 3180 businesses headed by men fail. Verify that the proportions of failures are exactly the same as in part (a). Repeat the \(z\) test for the new data and show that it is now significant at the \(\alpha=0.05\) level. c. It is wise to use a confidence interval to estimate the size of an effect rather than just giving a P-value. Give the large sample \(95 \%\) confidence intervals for the difference between the proportions of women's and men's businesses that fail for the settings of both parts (a) and (b). What is the effect of larger samples on the confidence interval? Do you think the size of the difference between the proportions is an important difference? In responding to Exercises \(23.26\) through \(23.36\) follow the Plan, Solve, and Conclude steps of the four-step process.

(Optional Topic) The plus four \(90 \%\) confidence interval for the difference between the proportion of those receiving the spray and those receiving the cream who had avobenzone levels exceeding \(0.5 \mathrm{ng} / \mathrm{mL}\) is a. \(0.375 \pm 0.451\) b. \(0.375 \pm 0.378\). c. \(0.375 \pm 0.230\).

{ Protecting Skiers and Snowboarders. Most alpine }\( skiers and snowboarders do not use helmets. Do helmets reduce the risk of head injuries? A study in Norway compared skiers and snowboarders who suffered head injuries with a control group who were not injured. Of 578 injured subjects, 96 had worn a helmet. Of the 2992 in the control group, 656 wore helmets. 9 Is helmet use less common among skiers and snowboarders who have head injuries? Follow the four-step process as illustrated in Examples 23.4 and \)23.5$ (pages 524 and 525 ). (Note that this is an observational study that compares injured and uninjured subjects. An experiment that assigned subjects to helmet and no- helmet groups would be more convincing.)

Abecedarian Early Childhood Education Program: Adult Outcomes. The Abecedarian Project is a randomized controlled study to assess the effects of intensive early childhood education on children who were at high risk based on several sociodemographic indicators. 2 The project randomly assigned some children to a treatment group that was provided with early educational activities before kindergarten and the remainder to a control group. A recent follow-up study interviewed subjects at age 30 and evaluated educational, economic, and socioemotional outcomes to learn if the positive effects of the program continued into adulthood. The follow-up study included 52 individuals from the treatment group and 49 from the control group. Out of these, 39 from the treatment group and 26 from the control group were considered "consistently" employed (working 30 + hours per week in at least 18 of the 24 months prior to the interview). Does the study provide significant evidence that subjects who had early childhood education have a higher proportion of consistent employment than those who did not? How large is the difference between the proportions in the two populations that are consistently employed? Do inference to answer both questions. Be sure to explain exactly what inference you choose to do.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.