student conduct a study and reported that 95% confidence intervak for themean ranged from 46-54. He was surethat the mean of the sample was 50,that the standard deviation of the sample was 16and that the sample was atleast 30 but could not remember exact number. can you help me
Of course, I can help you! Based on the information you provided, it seems like the student conducted a study and reported a 95% confidence interval for the mean. The reported confidence interval is from 46 to 54.
To explain, a confidence interval is a range of values within which we estimate the population parameter (in this case, the population mean) to lie. The confidence level, in this case 95%, indicates the level of confidence we have in our estimate.
Now, let's go step-by-step to find the sample mean and the sample size based on the given information.
1. The student mentioned that they were sure the mean of the sample was 50. So, the sample mean (x̄) is 50.
2. The standard deviation (σ) of the sample is given as 16. This represents the variability or spread of the data within the sample.
3. The student mentioned that the sample size is at least 30 but they couldn't remember the exact number. Since the sample size should be at least 30, we will use this as the minimum requirement.
Now, since we have the sample mean (50) and the standard deviation (16), we need to find the sample size (n). To do this, we can rearrange the formula for the margin of error in a confidence interval:
Margin of Error = (Z * σ) / √n
In a 95% confidence interval, the Z-value is approximately 1.96.
Using the given information:
1.96 * 16 / √n = (54 - 50) / 2
Now, solving for n:
1.96 * 16 / 2 = 2 / √n
(1.96 * 16 * √n) / 2 = 2
(1.96 * 16 * √n) = 4
Simplifying further, we get:
31.36 * √n = 4
√n = 4 / 31.36
√n = 0.127658
n = (0.127658)^2
n ≈ 0.016276
As the sample size (n) represents the number of individuals in the sample, it must be a whole number. Therefore, we need to round up the calculated value of n to the nearest whole number.
n ≈ 1
Therefore, the minimum required sample size should be 30, as mentioned by the student.
In conclusion, for a sample with a mean of 50, a standard deviation of 16, and a sample size of at least 30, the 95% confidence interval for the population mean ranges from 46 to 54.