In: Statistics and Probability
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. |
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.