Help me solve this please. One of the few on my 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 95% level of confidence, estimate the population mean number of gynecological visits per pregnancy.
In the end I got the confidence level to be around 1.96 - 4.04. I also got 1.96 to 4.03 at one point. But when I check the answer on the back of the book it says 2.65-3.35. Can someone go step by step on how to solve this.
95% = mean ± 1.96 SEm
SEm = SD/√n = 1/√35 = 1/5.92 = .17
.17 * 1.96 = .33
95% = mean ± 1.96 SEm = 3 ± .33 = 2.67-3.33
Rounding error?
Oh, I see you're looking for some statistical help. Well, I'll do my best to guide you through this!
To estimate the population mean number of gynecological visits per pregnancy, we can use the confidence interval formula:
Confidence interval = Sample mean ± (Critical value * Standard deviation / Square root of sample size)
First, let's find the critical value. Since we want a 95% level of confidence, we need to find the z-value that corresponds to that level. This is typically found using a z-table or calculator, and for a 95% confidence level, the z-value is approximately 1.96.
Next, we know the sample mean is 3 and the standard deviation is 1. We also have a sample size of 35. Plugging these values into the formula, we get:
Confidence interval = 3 ± (1.96 * 1 / √35)
Calculating this will give us the range of the confidence interval. However, let's not forget that 1.96 * 1 / √35 is approximately 0.331 as rounded to three decimal places.
So, the confidence interval becomes:
Confidence interval = 3 ± 0.331
Now, let's calculate the upper and lower bounds of the interval:
Upper bound = 3 + 0.331 = 3.331
Lower bound = 3 - 0.331 = 2.669
Therefore, the 95% confidence interval for the population mean number of gynecological visits per pregnancy is approximately 2.669 to 3.331.
Based on the answer you found in the back of the book (2.65-3.35), it seems they rounded the values slightly differently. The range is practically the same, so both answers are quite close!
Remember, statistics can be a bit tricky, but with practice, you'll get the hang of it. Keep those numbers in check and happy estimating!
To estimate the population mean number of gynecological visits per pregnancy with a 95% level of confidence, you can follow these steps:
Step 1: Identify the given information:
- Sample size (n) = 35
- Sample mean (x̄) = 3
- Sample standard deviation (s) = 1
- Confidence level = 95%
Step 2: Determine the critical value:
Since the sample size is small (< 30), we will use a t-distribution instead of the Z-distribution. To find the critical value, we need to look up the value from the t-table using the confidence level and (n-1) degrees of freedom. In this case, (n-1) = (35-1) = 34.
For a 95% confidence level, we need to find the t-critical value with the area of 0.025 in the upper tail (0.975 in the lower tail because it is a two-tailed test). The table value for 34 degrees of freedom and 0.025 in the upper tail is approximately 2.032.
Step 3: Calculate the margin of error:
The margin of error can be calculated using the formula:
Margin of Error = Critical value * (Standard deviation / √sample size)
In this case:
Margin of Error = 2.032 * (1 / √35) ≈ 0.345
Step 4: Calculate the confidence interval:
The confidence interval can be calculated using the formula:
Confidence Interval = Sample mean ± Margin of Error
In this case:
Confidence Interval = 3 ± 0.345
Step 5: Calculate the lower and upper bounds of the confidence interval:
Lower Bound = Sample mean - Margin of Error
Upper Bound = Sample mean + Margin of Error
In this case:
Lower Bound = 3 - 0.345 ≈ 2.655
Upper Bound = 3 + 0.345 ≈ 3.345
Therefore, the 95% confidence interval for the population mean number of gynecological visits per pregnancy is approximately 2.655 to 3.345.
It seems there might be an error in the answer provided in the back of the book. The interval you calculated (1.96 to 4.03) based on a Z-distribution is not appropriate for this problem. The correct solution uses a t-distribution and yields a confidence interval of 2.655 to 3.345.
To estimate the population mean number of gynecological visits per pregnancy with a 95% level of confidence, you can use the formula for a confidence interval:
Confidence Interval = sample mean ± (critical value) * (standard deviation / square root of sample size)
Let's go step by step to solve this problem:
1. First, you mentioned that the average number of gynecological checkups per pregnancy from the sample is 3, with a standard deviation of 1.
2. Since your sample size is 35 and the population standard deviation is unknown, we need to use the t-distribution instead of the standard normal distribution. This is because we are estimating the population mean using a sample.
3. The critical value for a 95% confidence level with a sample size of 35 is approximately 2.03.
4. Now, calculate the margin of error by multiplying the critical value by the standard deviation divided by the square root of the sample size:
Margin of Error = 2.03 * (1 / sqrt(35))
5. Calculate the confidence interval by adding and subtracting the margin of error from the sample mean:
Confidence Interval = 3 ± Margin of Error
6. Finally, calculate the upper and lower limits of the confidence interval:
Upper limit = sample mean + Margin of Error
Lower limit = sample mean - Margin of Error
Now let's perform the calculation:
Margin of Error = 2.03 * (1 / sqrt(35)) ≈ 0.343
Upper limit = 3 + 0.343 ≈ 3.343
Lower limit = 3 - 0.343 ≈ 2.657
Therefore, the 95% confidence interval estimate for the population mean number of gynecological visits per pregnancy is approximately 2.66 to 3.34, which is similar to the answer provided in the book (2.65-3.35).
Keep in mind that the exact critical value and the resulting confidence interval may vary slightly depending on the statistical software or table used for calculations.