From the information, observe that the level of confidence is 0.95

The sample size is, n=64

The sample standard deviation is, s=2.2

The degrees of freedom is,

\(df=n-1\)

\(=64-1\)

\(=63\)

The level of significance is,

Level of significance=1- Level of confidence

\(=1-0.95\)

\(=0.95\)

From the standard t table values, observe that the critical value of t for the two tail test at the 5% level of significance and 63 degrees of freedom is 1.998

Using these value, observe that the t critical value as observing 0.05 in two tail at column wise and 63 degrees of freedom in row wise and identify the critical value (1.998) corresponding to the level of significance (0.05) and 63 degrees of freedom.

The calculation of the margin of error is,

\(ME=\frac{t_{a/2}\times s}{\sqrt{64}}\)

\(=\frac{(1.998)(2.2)}{8}\)

\(=\frac{4.396}{8}\)

\(=0.549\)

The calculated value of the margin of error is 0.549

The sample size is, n=64

The sample standard deviation is, s=2.2

The degrees of freedom is,

\(df=n-1\)

\(=64-1\)

\(=63\)

The level of significance is,

Level of significance=1- Level of confidence

\(=1-0.95\)

\(=0.95\)

From the standard t table values, observe that the critical value of t for the two tail test at the 5% level of significance and 63 degrees of freedom is 1.998

Using these value, observe that the t critical value as observing 0.05 in two tail at column wise and 63 degrees of freedom in row wise and identify the critical value (1.998) corresponding to the level of significance (0.05) and 63 degrees of freedom.

The calculation of the margin of error is,

\(ME=\frac{t_{a/2}\times s}{\sqrt{64}}\)

\(=\frac{(1.998)(2.2)}{8}\)

\(=\frac{4.396}{8}\)

\(=0.549\)

The calculated value of the margin of error is 0.549