Question

A random sample of 16 ATM transactions at the Last National Bank of Flat Rock revealed a mean transaction time of 2.8 minutes with a standard deviation of 1.2 minutes. The width (in minutes) of the 95 percent confidence interval for the true mean transaction time is

Answer 1

The width (in minutes) of the 95 percent confidence interval for the true mean transaction time is 1.2786 minutes.

Sample size (n) = 16.

Sample mean = 2.8 minutes.

Sample standard deviation (σ) = 1.2 minutes.

Confidence interval given = 95%.

The degree of freedom (df) can be calculated as n - 1.

df = n - 1 = 16 - 1 = 15.

Since, the level of confidence is 95%, the level of significance α = 1 - 0.95 = 0.05.

From the T-Distribution table, we take the value at critical value α/2 with the degree of freedom 15.

$t_{\frac{\alpha }{2} ,15}$ = 2.131.

Now, the margin of error can be shown as follows:

$E = t_{\frac{\alpha }{2} ,15} * \frac{\sigma}{\sqrt{n} } \\\Rightarrow E = 2.131 * \frac{1.2}{\sqrt{16} }\\ \Rightarrow E = 0.6393$

The width at the confidence interval of 95%, is twice the margin of error, that is:

Width = 2*E = 2*0.6393 = 1.2786.

Thus, the width (in minutes) of the 95 percent confidence interval for the true mean transaction time is 1.2786 minutes.

Learn more about the width of distribution at

brainly.com/question/19338886

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