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3 years ago

By comparing medians which person went to the movies the most per month

Mathematics

1 answer:

Nimfa-mama [501]3 years ago

3 0

I think they all went to the movies equally per month if you simply compare medians? Since in the second chart, they each have a median of 2 times per month, that makes it equal. If you aren’t sure if this is correct, you have to order how many times each person went to the movies, from least to greatest. For example, John’s would be 1,1,1,2,2,2,2,3,3,3,4,5. Then you would start crossing numbers off from each end. First you would cross off 1, then 5, and then 1, and then 4, and keep going until you reach the number in the very middle, which is 2. Do this with all of the other people, and check if everybody has 2 for a median. I believe they do, however it’s always best to check and be safe. Good luck

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Rasek [7]

Answer: a. 1.981 < μ < 2.18

b. Yes.

Step-by-step explanation:

A. For this sample, we will use t-distribution because we're estimating the standard deviation, i.e., we are calculating the standard deviation, and the sample is small, n = 12.

First, we calculate mean of the sample:

$\overline{x}=\frac{\Sigma x}{n}$

$\overline{x}=\frac{2.31=2.09+...+1.97+2.02}{12}$

$\overline{x}=$ 2.08

Now, we estimate standard deviation:

$s=\sqrt{\frac{\Sigma (x-\overline{x})^{2}}{n-1} }$

$s=\sqrt{\frac{(2.31-2.08)^{2}+...+(2.02-2.08)^{2}}{11} }$

s = 0.1564

For t-score, we need to determine degree of freedom and $\frac{\alpha}{2}$ :

df = 12 - 1

df = 11

$\alpha$ = 1 - 0.95

α = 0.05

$\frac{\alpha}{2}=$ 0.025

Then, t-score is

$t_{11,0.025}$ = 2.201

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$\overline{x}$ ± $t.\frac{s}{\sqrt{n} }$

2.08 ± $2.201\frac{0.1564}{\sqrt{12} }$

2.08 ± 0.099

The 95% two-sided CI on the mean is 1.981 < μ < 2.18.

B. We are 95% confident that the true population mean for this clinic is between 1.981 and 2.18. Since the mean number performed by all clinics has been 1.95, and this mean is less than the interval, there is evidence that this particular clinic performs more scans than the overall system average.

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