Suppose that a random sample of adults has a mean score of on a standardized personality test, with a standard deviation of . (A higher score indicates a more personable participant.) If we assume that scores on this test are normally distributed, find a confidence interval for the mean score of all takers of this test. Then complete the table below.
Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place.
It would help if you proofread your questions before you posted them. You have omitted the values. What confidence interval are you seeking?
To find the confidence interval for the mean score of all takers of this test, we can use the formula:
Confidence interval = (sample mean) +/- (critical value) * (standard deviation / sqrt(sample size))
The critical value depends on the desired confidence level. For example, if we want a 95% confidence level, the critical value (z-score) is 1.96.
Given data:
- Sample mean (x̄) = ?
- Standard deviation (σ) = ?
- Sample size (n) = ?
We need the values of the sample mean, standard deviation, and sample size to calculate the confidence interval.
Once we have those values, we can plug them into the formula to find the confidence interval.