A medical researcher is interested in the prenatal care received by pregnant women in inner cities. She interviews 35 randomly selected women with children on the streets of Baltimore and finds that the average number of gynecological checkups per pregnancy was 3, with a standard deviation of 1. Using a 95% level of confidence, estimate the population mean number of gynecological visits per pregnancy.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.025) and its Z score.
95% = mean ± Z SEm
SEm = SD/√n
To estimate the population mean number of gynecological visits per pregnancy with a 95% confidence level, we can use the formula for calculating the confidence interval for a population mean.
The formula is:
Confidence Interval = Sample Mean ± (Critical Value) * (Standard Deviation / √n)
Where:
- Sample Mean: The average number of gynecological checkups per pregnancy in the sample (3 in this case).
- Critical Value: The value obtained from the t-distribution table for the desired confidence level.
- Standard Deviation: The standard deviation of the population (1 in this case).
- n: The sample size (35 in this case).
First, we need to find the critical value for a 95% confidence level. Since the sample size is small (< 30) and the population standard deviation is unknown, we use the t-distribution.
For a 95% confidence level with 35 degrees of freedom (n - 1), the critical value can be found from the t-distribution table to be approximately 2.03.
Now we have all the values we need:
Sample Mean = 3
Standard Deviation = 1
Sample Size = 35
Critical Value = 2.03
Plugging these values into the formula, we get:
Confidence Interval = 3 ± (2.03) * (1 / √35)
Calculating this gives us:
Confidence Interval ≈ 3 ± 0.343
Therefore, the 95% confidence interval estimate for the population mean number of gynecological visits per pregnancy is approximately 2.657 to 3.343.
To estimate the population mean number of gynecological visits per pregnancy with a 95% level of confidence, we can use a confidence interval.
The formula for the confidence interval for a population mean, based on a sample, is:
Confidence interval = sample mean ± t * (sample standard deviation / √sample size)
Where:
- Sample mean is the average number of gynecological visits per pregnancy in the sample (3 in this case)
- Sample standard deviation is the standard deviation of the sample (1 in this case)
- Sample size is the number of observations in the sample (35 in this case)
- t is the critical value from the t-distribution corresponding to the desired confidence level (95% in this case)
First, we need to find the critical value (t) for a 95% confidence level. The degrees of freedom for this sample would be (n-1), where n is the sample size. In this case, the degrees of freedom would be (35-1) = 34.
Using a t-table or a statistical software, we find that the critical value for a 95% confidence level with 34 degrees of freedom is approximately 2.032.
Now we can calculate the confidence interval:
Confidence interval = 3 ± 2.032 * (1 / √35)
To calculate the square root of 35, we find that it is approximately 5.92.
Confidence interval = 3 ± 2.032 * (1 / 5.92)
Calculating the expression inside the parentheses, we get approximately 0.169.
Confidence interval = 3 ± 2.032 * 0.169
Multiplying 2.032 by 0.169, we get approximately 0.344.
Confidence interval = 3 ± 0.344
Therefore, the confidence interval for the population mean number of gynecological visits per pregnancy is (2.656, 3.344).
In other words, we are 95% confident that the true population mean number of gynecological visits per pregnancy lies between 2.656 and 3.344 visits.