Question

In: Statistics and Probability

a. all Hypothesis Tests must include all four steps, clearly labeled; b. all Confidence Intervals must...

a. all Hypothesis Tests must include all four steps, clearly labeled;

b. all Confidence Intervals must include all output as well as the CI itself

c. include which calculator function you used for each problem.

3. The authors of the paper “Statistical Methods for Assessing Agreement Between Two Methods of Clinical Measurement” (International Journal of Nursing Studies [2010]: 931-936) compared two different instruments for measuring a person’s ability to breathe out air. (This measurement is helpful in diagnosing various lung disorders.) The two instruments considered were a Wright peak flow meter and a mini-Wright peak flow meter. Seventeen people were randomly selected to participate in the study, and for each person air flow was measured once using the Wright meter and once using the mini-Wright meter.

Mini-Wright Meter (x) Wright Meter (y)
512 494
430 395
520 516
364 413
380 442
658 650
445 433
432 417
626 656
428 434
500 476
600 557
260 267
477 478
259 178
350 423
451 427

(this table is color-coded for a reason; please pay close attention to it)

Suppose that the Wright meter is considered to provide a better measure of air flow, but the mini-Wright meter is easier to transport and to use. If the two types of meters produce different readings but there is a strong relationship between the readings, it would be possible to use a reading from the mini-Wright meter to predict the reading that the larger Wright meter would have given.

a. Calculate and interpret the value of the correlation coefficient. In your interpretation, you should answer the question: “Based

on the data, would you be comfortable using the mini-Wright meter reading to predict the larger Wright meter reading?”

b. Run the hypothesis test (at α = 5%) to determine if there is a correlation between the Wright meter and the mini-Wright meter.

c. Use the given data to find an equation to predict the Wright meter reading (y) using a reading from the mini-Wright meter (x).

d. What would you predict for the Wright meter reading for a person whose mini-Wright meter reading was:

600? 500? 200? Do any of these predictions pose any problems? Explain.

Solutions

Expert Solution

a). Correlation coefficient:

where

Let us form a table as below:

x y x^2 y^2 x*y
512 494 262144 244036 252928
430 395 184900 156025 169850
520 516 270400 266256 268320
364 413 132496 170569 150332
380 442 144400 195364 167960
658 650 432964 422500 427700
445 433 198025 187489 192685
432 417 186624 173889 180144
626 656 391876 430336 410656
428 434 183184 188356 185752
500 476 250000 226576 238000
600 557 360000 310249 334200
260 267 67600 71289 69420
477 478 227529 228484 228006
259 178 67081 31684 46102
350 423 122500 178929 148050
451 427 203401 182329 192577
Total 7692 7656 3685124 3664360 3662682

The correlation coefficent is 0.9433 which is very high. Therefore,

Based on the data, I will be comfortable using the mini-Wright meter reading to predict the larger Wright meter reading?”

b. Null Hypothesis: There is no correlation or the correlation is absent.

Alternate Hypothesis: The correlation is present

Level of significance:

Test statistic: will have t distribution with n-2 df.

The critical value of t(15,5%) is 2.1314. Since, the t calculated value is >tcritical, we reject the Null hypotheis and conclude that the correlation coefficient is significant.

c. Use the given data to find an equation to predict the Wright meter reading (y) using a reading from the mini-Wright meter (x).

we just need two more quantities:

and

The estimate of slope

The estimate of intercept:

  

The estimated regression equation is

d. What would you predict for the Wright meter reading for a person whose mini-Wright meter reading was:

When x=600, y=11.4817+0.9699*600=593.4217

x=500, y=11.4817+0.9699*500=496.4317

x=200,y=11.4817+0.9699*200=205.4617

It can be observed that in our data when x=600, the observed value of y=557 and the estimated value is 593

  x=500, the observed value of y=476 and the estimated value is 496

There are diffferences between actual and estimated values.


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