Scores on a certain test are normally distributed with a variance of 91.

A researcher wishes to estimate the mean score achieved by all adults on the test.
Find the sample size needed to assure with 95.44 percent confidence that the
sample mean will not differ from the population mean by more than 2 units.

To find the sample size needed to assure a certain level of confidence, we can use the formula for the confidence interval:

Sample Size = [ (Z * σ) / E ]^2

Where:
Z is the Z-score corresponding to the desired level of confidence (95.44% confidence corresponds to a Z-score of 1.81).
σ is the standard deviation of the population (in this case, the square root of the variance, so √91).
E is the maximum error, which is the difference between the sample mean and the population mean (in this case, 2 units).

Plugging in the values, we have:

Sample Size = [ (1.81 * √91) / 2 ]^2

Calculating this formula gives us the sample size needed to assure a 95.44% confidence level that the sample mean will not differ from the population mean by more than 2 units.