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.
[tex]t_{\frac{\alpha }{2} ,15}[/tex] = 2.131.
Now, the margin of error can be shown as follows:
[tex]E = t_{\frac{\alpha }{2} ,15} * \frac{\sigma}{\sqrt{n} } \\\Rightarrow E = 2.131 * \frac{1.2}{\sqrt{16} }\\ \Rightarrow E = 0.6393[/tex]
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.
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