we developed a 95% z confidence interval for the mean BMI of women aged 20 to 29 years, based on a national random sample of 654 such women. we assumed there that the population standard deviation was known to be 7.5. in fact the sample data had mean BMI of 26.8 and standard deviation s=7.42. what is the 95%t confidence intervalfor the mean BMI of all young women?
CI95 = mean + or - 1.96(sd divided by √n)
...where + or - 1.96 represents the 95% confidence interval using a z-table, sd = standard deviation, √ = square root, and n = sample size.
Substitute the values into the formula and calculate the interval.
To calculate the 95% t confidence interval for the mean BMI of all young women, we need to use the following formula:
Confidence Interval = sample mean ± (t-score * standard error)
The t-score depends on the sample size and the desired level of confidence. Since we are given the sample size and the desired level of confidence is 95%, we can use the t-distribution table to find the appropriate t-score.
However, we first need to calculate the standard error, which is the estimated standard deviation of the sampling distribution. The formula to calculate the standard error is:
Standard Error = population standard deviation / square root of the sample size
In this case, the population standard deviation is given as 7.5, and the sample size is 654. Therefore:
Standard Error = 7.5 / √(654) ≈ 0.2934
Now, we need to find the t-score for a 95% confidence level. With 653 degrees of freedom (since n-1=654-1=653), the t-score can be found by referring to the t-distribution table.
Using the table, we find that the t-score for a 95% confidence level with 653 degrees of freedom is approximately 1.962.
Now plug the values into the confidence interval formula:
Confidence Interval = 26.8 ± (1.962 * 0.2934)
Simplifying the equation:
Confidence Interval = 26.8 ± 0.5751
Therefore, the 95% t confidence interval for the mean BMI of all young women is approximately 26.225 to 27.375.