Question

In: Math

Jarrid Medical, Inc., is developing a compact machine for kidney dialysis, but the company's chief engineer,...

Jarrid Medical, Inc., is developing a compact machine for kidney dialysis, but the company's chief engineer, Mike Crowe, has trouble controlling the variability of how quickly the fluid moves through the device Medical employers require that the flow per hour be 4.25 liters, 98% of the time. Mr. Crowe, tests the prototype with the following results for each hourly flow test:

4.17

4.32

4.21

4.22

4.29

4.19

4.29

4.34

4.33

4.22

4.28

4.33

4.52

4.29

4.43

4.39

4.44

4.34

4.3

4.41

With the information obtained from the sample, it is necessary to find the confidence interval that reflects the expected hourly flow in 98% of the time of the developed compact machine and determine if it is satisfying the medical standards. Satisfy the prototype of medical patterns? Present all the steps required to build your confidence interval and an analysis that justifies your answer to the established question.

Solutions

Expert Solution

For the given data the mean is calculated as:

Mean = (4.17 + 4.19 + 4.21 + 4.22 + 4.22 + 4.28 + 4.29 + 4.29 + 4.29 + 4.3 + 4.32 + 4.33 + 4.33 + 4.34 + 4.34 + 4.39 + 4.41 + 4.43 + 4.44 + 4.52)/20
= 86.31/20
Mean = 4.3155 and Sample standard deviation as:

Sample Standard Deviation σ or S= √(1/20 - 1) x ((4.17 - 4.3155)2 + ...................+ ( 4.52 - 4.3155)2)
= √(1/19) x ((-0.1455)2 +..........................+ (0.2045)2)
= √(0.0526) x ((0.02117025) + ................ + (0.04182025))
= √(0.0526) x (0.156295)
= √(0.008221117)
= 0.0907

Now using the confidence interval
Since the sample size is <<30 and also the population interval is unknown so we will use t-distribution.

The formula for estimation is:

μ = M ± t(sM)

where:

= sample mean
t = t statistic determined by the confidence level
sM = standard error = √(s2/n)

based on the given data, the degree of freedom is n-1=20-1=19 and at 0.02 significance level, the t value is 2.539 which is computed using t-table shown below.

So, the confidence interval is

Conclusion:

Since the 98% confidence interval does not include the assumed value that is 4.25 hence we can say that we do not have sufficient evidence to support the assumption.


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