Use the research described below to answer the questions that follow. Shargorodsky, Curhan, Curhan, and Eavey...

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Use the research described below to answer the questions thatfollow. Shargorodsky, Curhan, Curhan, and Eavey (2010) examinedhearing loss data for participants from the National Health andNutrition Examination Survey (NHANES), aged 12-19 years. (NHANESprovides nationally representative cross-sectional data on thehealth status of the civilian, non-institutionalized U.S.population.) The researchers compared the more recent hearing lossrate among teens (12-19 years) in 2005 to previous levels in 1988.Their goal was to see whether teen hearing loss is increasing,possibly due to heavier use of ear buds. They examined data from1771 participants in the NHANES 2005 study (333 with some level ofhearing loss), as well as data on 2928 teens from the NHANES 1988study, with 480 in this group showing some level of hearingloss.
QUESTION 1: Calculate the p-value
QUESTION 2: Calculate an appropriate confidence interval for thechange (if it exists) in hearing loss.

Answer & Explanation Solved by verified expert

4.4 Ratings (901 Votes)

Q1) P-value =0.04252

Q2) 95% confidence interval for the change =(0.0018 ,0.0464)

Solution:

For 2005 x1 = 333 n1=1771 p1=333/1771=0.188029

For 1988 x2=480 n2=2928, p2= 480/2928=0.163934

p-hat = ( x1 + x2) / ( n1 + n2). = (333+480)/(1771+2928) =0.173016

Here Null and alternative hypothesis are below:

H0 : p1=p2

H1: p1 > p2

To calculate we first need to calculate test statistics z and then calculate p(Z>z)

Test statistics z=(p1-p2)/sqrt(p-hat*(1-p-hat)*(1/n1+1/n2

z=(0.188029- 0.163934)/sqrt(0.173016*(1-0.173016)*((1/1771)+(1/2928)))

=2.1160

P-Value

For Ha: p1 - p2 > 0, we calculate the proportion of the normal distribution that is greater than Z.

i.e p(Z> 2.1160)=0.04252

P-value =0.04252

appropriate confidence interval

95% Confidence interval

((p1-p2) - Z_alpha*sqrt(p-hat*(1-p-hat)*(1/n1+1/n2,(p1-p2) + Z_alpha*sqrt(p-hat*(1-p-hat)*(1/n1+1/n2)

at alpha=0.05Â Â Z_alpha=1.96

CI= ((0.188029- 0.163934)-1.96*sqrt(0.173016*(1-0.173016)*((1/1771)+(1/2928))),(0.188029- 0.163934)+1.96*sqrt(0.173016*(1-0.173016)*((1/1771)+(1/2928))))

=(0.0018 ,0.0464)

95% confidence interval for the change =(0.0018 ,0.0464)

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