CAN SOMEONE HELP ME WITH THIS?
A recent study found that, of the 1771 participants aged 12 to 19 in the National Health and Nutrition Examination Survey, 19.5% had some hearing loss (defined as a loss of 15 decibels in at least one ear). This is a dramatic increase from a decade ago. The sample size is large enough to use the normal distribution, and a bootstrap distribution shows that the standard error for the proportion SE = 0.009. Find and interpret a 90% confidence interval for the proportion of teenagers with some hearing loss.
2 answers
0.18 to 0.21
Normal percentiles for common confidence levels
Confidence level
80% 90% 95% 98% 99%
z∗ 1.282 1.645 1.960 2.326 2.576
.195 ± 1.645(.009)
.195 ± .014805
0.180195 to 0.209805
Confidence level
80% 90% 95% 98% 99%
z∗ 1.282 1.645 1.960 2.326 2.576
.195 ± 1.645(.009)
.195 ± .014805
0.180195 to 0.209805
1 answer