Module 17 - Distribution of Sample Proportions (6 of 6 discussion 3)20 20 unread replies. 20 20 replies.Learn by DoingSome features of this activity may not work well on a cell phone or tablet. We highly recommend that you complete this activity on a computer.ContextRecall the use of data from the National Health Survey to estimate behaviors such as alcohol consumption, cigarette smoking, and hours of sleep for all U.S. adults. In the 2005-2007 report, they estimated that 30% of all current smokers started smoking before the age of 16.PromptSuppose that we randomly select 100 U.S. adults who are smokers and find that 25% of this sample started smoking before the age of 16. In this random sample, the sample proportion (25%) differs from the estimated population proportion (30%) by 5%. In other words, there is a 5% error in the sample proportion; this sample under-estimates the population proportion by 5%.Is this much error surprising? To answer this question, find the probability that a sample proportion will over- or under-estimate the parameter by more than 5%.Show your work by checking normality conditions, calculating a z-score, explaining how the area under the normal curve is used to answer the question, and stating a conclusion in the context of this problem. Module 17 - Distribution of Sample Proportions (6 of 6 discussion 3)20 20 unread replies. 20 20 replies.Learn by DoingSome features of this activity may not work well on a cell phone or tablet. We highly recommend that you complete this activity on a computer.ContextRecall the use of data from the National Health Survey to estimate behaviors such as alcohol consumption, cigarette smoking, and hours of sleep for all U.S. adults. In the 2005-2007 report, they estimated that 30% of all current smokers started smoking before the age of 16.PromptSuppose that we randomly select 100 U.S. adults who are smokers and find that 25% of this sample started smoking before the age of 16. In this random sample, the sample proportion (25%) differs from the estimated population proportion (30%) by 5%. In other words, there is a 5% error in the sample proportion; this sample under-estimates the population proportion by 5%.Is this much error surprising? To answer this question, find the probability that a sample proportion will over- or under-estimate the parameter by more than 5%.Show your work by checking normality conditions, calculating a z-score, explaining how the area under the normal curve is used to answer the question, and stating a conclusion in the context of this problem. m the National Health Survey to estimate behaviors such as alcohol consumption, cigarette smoking, and hours of sleep for all U.S. adults. In the 2005-2007 report, they estimated that 30% of all current smokers started smoking before the age of 16.PromptSuppose that we randomly select 100 U.S. adults who are smokers and find that 25% of this sample started smoking before the age of 16. In this random sample, the sample proportion (25%) differs from the estimated population proportion (30%) by 5%. In other words, there is a 5% error in the sample proportion; this sample under-estimates the population proportion by 5%.Is this much error surprising? To answer this question, find the probability that a sample proportion will over- or under-estimate the parameter by more than 5%.Show your work by checking normality conditions, calculating a z-score, explaining how the area under the normal curve is used to answer the question, and stating a conclusion in the context of this problem. Content by the Open Learning Initiative (Links to an external site.) and licensed under CC BY (Links to an external site.).Search entries or author

[tex]\begin{gathered} \sigma p^{\prime}=\sqrt{\frac{p(1-p)}{n}} \\ \\ \sigma p^{\prime}=\sqrt{\frac{0.3(1-0.3)}{100}} \\ \\ \sigma p^{\prime}=\sqrt{\frac{0.3(0.7)}{100}}=\sqrt{\frac{0.21}{100}}=\sqrt{0.0021}=0.0458 \end{gathered}[/tex]

Now, to find the probability that a sample proportion will be over or underestimate more than 5% will be:

[tex]\begin{gathered} p(p^{\prime}<0.3-0.05)+p(p^{\prime}>0.3+0.05) \\ \\ p(p^{\prime}<0.25)+p(p^{\prime}>0.35) \\ \\ z=\frac{p^{\prime}-\mu p^{\prime}}{\sigma p} \\ \\ Where:\mu p^{\prime}=0.3 \\ \end{gathered}[/tex]

Hence, we have:

[tex]\begin{gathered} p(z<\frac{0.25-0.3}{0.0458})+p(z>\frac{0.35-0.3}{0.0458}) \\ \\ p(z<\frac{-0.05}{0.0458})+p(z>\frac{0.05}{0.0458}) \\ \\ p(z<-1.09)+p(z>1.09) \end{gathered}[/tex]

Using the standard normal distribution table, we have:

NORMSDIST(-1.09) =0.1379

NORMSDIST(1.09) = 0.8621

Hence, we have:

p(z<-1.09) = 0.1379

p(z>1.09) = 1 - 0.8621 = 0.1379

p(z<-1.09) + p(z>1.09) = 0.1379 + 0.1379 = 0.2758

Therefore, the probability is 0.2758.

ANSWER:

0.2758