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.
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
To find the 90% confidence interval for the proportion of teenagers with some hearing loss, we can use the formula:
Confidence interval = p̂ ± z * SE
Where:
- p̂ is the sample proportion (19.5% = 0.195)
- z is the z-value associated with the desired confidence level (90% confidence corresponds to a z-value of 1.645)
- SE is the standard error for the proportion (0.009)
Let's calculate the confidence interval step-by-step:
Step 1: Calculate the margin of error:
Margin of error = z * SE = 1.645 * 0.009 = 0.0148
Step 2: Calculate the lower bound:
Lower bound = p̂ - margin of error = 0.195 - 0.0148
Step 3: Calculate the upper bound:
Upper bound = p̂ + margin of error = 0.195 + 0.0148
Now, let's calculate the confidence interval:
Lower bound = 0.195 - 0.0148 = 0.1802
Upper bound = 0.195 + 0.0148 = 0.2098
Therefore, the 90% confidence interval for the proportion of teenagers with some hearing loss is (0.1802, 0.2098).
Interpretation: We are 90% confident that the true proportion of teenagers with some hearing loss lies within the range of 0.1802 to 0.2098.
To find a 90% confidence interval for the proportion of teenagers with some hearing loss, we can use the sample proportion and the standard error.
Step 1: Find the sample proportion
In this case, the sample proportion is 19.5%, which we can write as 0.195.
Step 2: Find the critical value
To find the critical value, we use the Z-score associated with a 90% confidence level. For a two-tailed Z-test, which is appropriate for calculating a confidence interval, the critical value is 1.645. This value can be obtained from a Z-table or using statistical software.
Step 3: Calculate the margin of error
The margin of error is calculated by multiplying the critical value by the standard error. In this case, the standard error is given as SE = 0.009. Therefore, the margin of error (ME) is 1.645 * 0.009.
Step 4: Calculate the confidence interval
To calculate the confidence interval, we need to subtract and add the margin of error to the sample proportion. The confidence interval formula is:
Confidence Interval = Sample Proportion ± Margin of Error
Plugging in the values, we get:
Confidence Interval = 0.195 ± (1.645 * 0.009)
Step 5: Interpret the confidence interval
The confidence interval represents the range of values within which we can be 90% confident that the true population proportion lies. In this case, the interval would be interpreted as follows:
We are 90% confident that the true proportion of teenagers with some hearing loss is between (lower bound) and (upper bound).
Calculating the confidence interval using these steps will give you the specific values for the lower and upper bounds of the interval.