Answer:
The 85% confidence interval for the mean is between 5.5 and 5.7 fast food meals consumed per week.
Step-by-step explanation:
We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:
[tex]\alpha = \frac{1-0.85}{2} = 0.075[/tex]
Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].
So it is z with a pvalue of [tex]1-0.075 = 0.925[/tex], so [tex]z = 1.44[/tex]
Now, find M as such
[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]
In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.
[tex]M = 1.44*\frac{1.3}{\sqrt{548}} = 0.08[/tex]
The lower end of the interval is the sample mean subtracted by M. So it is 5.6 - 0.08 = 5.52 = 5.5.
The upper end of the interval is the sample mean added to M. So it is 5.6 + 0.08 = 5.7.
The 85% confidence interval for the mean is between 5.5 and 5.7 fast food meals consumed per week.
