Help me solve this please. One of the few on my hw list that I had no idea how to solve

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 9% level of confidence, estimate the population mean number of gynecological visits per pregnancy.

I meant to put 95% confidence level. Not 9%

To estimate the population mean number of gynecological visits per pregnancy, we can use the confidence interval formula.

The formula for the confidence interval is:

Confidence Interval = sample mean ± (critical value * standard deviation / √n)

Where:
- sample mean is the average number of gynecological checkups per pregnancy (given as 3)
- critical value corresponds to the desired level of confidence (given as 9% or 0.09)
- standard deviation is the standard deviation of the sample (given as 1)
- n is the sample size (given as 35)

Step 1: Find the critical value
Since the level of confidence is 9%, we need to find the critical value corresponding to this level in the standard normal distribution table. Subtract 1 minus the level of confidence to find the area in the tails: 1 - 0.09 = 0.91. Divide this value by 2 since we are interested in a two-tailed test: 0.91 / 2 = 0.455. This gives us a cumulative probability of 0.455. Referencing the standard normal distribution table, we can find the critical value as the z-score closest to 0.455, which is 1.70 (rounded to two decimal places).

Step 2: Calculate the confidence interval
Using the formula, plug in the given values:
Confidence Interval = 3 ± (1.70 * 1 / √35)

Step 3: Simplify the equation
Calculate the square root of the sample size: √35 ≈ 5.92
Confidence Interval = 3 ± (1.70 * 1 / 5.92)

Step 4: Evaluate the expression
Confidence Interval = 3 ± 0.29

Step 5: Finalize the result
The lower bound of the confidence interval is obtained by subtracting the calculated value (0.29) from the sample mean (3): 3 - 0.29 = 2.71
The upper bound of the confidence interval is obtained by adding the calculated value (0.29) to the sample mean (3): 3 + 0.29 = 3.29

Therefore, with a 9% level of confidence, the estimated population mean number of gynecological visits per pregnancy is between 2.71 and 3.29.

To estimate the population mean number of gynecological visits per pregnancy with a 9% level of confidence, we can use the t-distribution since we have a small sample size (n=35) and the population standard deviation is unknown.

Here's how you can calculate the confidence interval:

Step 1: Determine the critical value.
To find the critical value, we need to determine the degrees of freedom (df) which is the sample size minus 1. In this case, df = 35 - 1 = 34.

Using a 9% level of confidence, we need to find the critical value corresponding to the area in the tails (α/2) of 0.09/2 = 0.045. With a t-distribution, we can look up the critical value in a t-table or use software. For a two-tailed test with 34 degrees of freedom and a probability of 0.45 in each tail, the critical value is approximately 1.691.

Step 2: Calculate the margin of error.
The margin of error (E) measures the maximum likely difference between the sample mean and the population mean. To calculate it, we need the standard deviation of the sample (s) and the square root of the sample size (n).

In this case, the standard deviation (s) is given as 1, and the square root of the sample size (n) is √35 ≈ 5.92.

E = (Critical value) * (Standard deviation / √Sample size)
E = 1.691 * (1 / 5.92)
E ≈ 0.286

Step 3: Calculate the confidence interval.
The confidence interval is calculated by subtracting and adding the margin of error from the sample mean (x̄).

Confidence interval = Sample Mean ± Margin of Error
Confidence interval = 3 ± 0.286

Therefore, the 9% confidence interval estimate for the population mean number of gynecological visits per pregnancy is approximately (2.714, 3.286).

This means we can be 90% confident that the true population mean lies within this range.