How would you report the results? Group of answer choices 1. The derived t = 0.74 was not significant at p = .05 with df = 19. Therefore, H0 was not rejected, and it was concluded that the mean number of errors for subjects after 8 hours of sleep deprivation (2.02) was not different than the mean number of errors after 24 hours of sleep deprivation (13.90), t(19) = 0.74, p > .05. In terms of the research question, it appears that sleep deprivation did not increase the number of errors in this sample. 2. The derived t = 3.30 was significant at p = .05 with df = 18. Therefore, H0 was rejected, and it was concluded that the mean number of errors for subjects after 24 hours of sleep deprivation (2.02) was different than the mean number of errors after 8 hours of sleep deprivation (2.12), t(18) = 3.30, p < .05. In terms of the research question, it appears that sleep was effective in reducing number of errors in this sample. 3. The derived t = –9.70 was significant at p = .05 with df = 19. Therefore, H0 was rejected, and it was concluded that the mean number of errors for subjects after 24 hours of sleep deprivation (13.90) was significantly higher than the mean number of errors after 8 hours of sleep deprivation (6.75), t(19) = –9.70, p < .001. In terms of the research question, it appears that sleep deprivation increased the number of errors in this sample. 4. The derived t = –7.15 was significant at p = .05 with df = 1. Therefore, H0 was rejected, and it was concluded that the mean number of errors for subjects after 8 hours of sleep deprivation (6.75) was significantly lower than the mean number of errors after 24 hours of sleep deprivation (13.90), t(1) = –7.15, p < .000. In terms of the research question, it appears that sleep was effective in reducing number of errors in this sample. 5. The derived t = 0.74 was significant at p = .05 with df = 20. Therefore, H0 was rejected, and it was concluded that the mean number of errors for subjects after 8 hours of sleep deprivation (2.02) was significantly lower than the mean number of errors after 24 hours of sleep deprivation (2.12), t(20) = 0.74, p < .01. In terms of the research question, it appears that sleep was effective in reducing number of errors in this sample.

How would you report the results? Group of answer choices 1. The derived t = 0.74 was not significant at p = .05 with df = 19. Therefore, H0 was not rejected, and it was concluded that the mean number of errors for subjects after 8 hours of sleep deprivation (2.02) was not different than the mean number of errors after 24 hours of sleep deprivation (13.90), t(19) = 0.74, p > .05. In terms of the research question, it appears that sleep deprivation did not increase the number of errors in this sample. 2. The derived t = 3.30 was significant at p = .05 with df = 18. Therefore, H0 was rejected, and it was concluded that the mean number of errors for subjects after 24 hours of sleep deprivation (2.02) was different than the mean number of errors after 8 hours of sleep deprivation (2.12), t(18) = 3.30, p < .05. In terms of the research question, it appears that sleep was effective in reducing number of errors in this sample. 3. The derived t = –9.70 was significant at p = .05 with df = 19. Therefore, H0 was rejected, and it was concluded that the mean number of errors for subjects after 24 hours of sleep deprivation (13.90) was significantly higher than the mean number of errors after 8 hours of sleep deprivation (6.75), t(19) = –9.70, p < .001. In terms of the research question, it appears that sleep deprivation increased the number of errors in this sample. 4. The derived t = –7.15 was significant at p = .05 with df = 1. Therefore, H0 was rejected, and it was concluded that the mean number of errors for subjects after 8 hours of sleep deprivation (6.75) was significantly lower than the mean number of errors after 24 hours of sleep deprivation (13.90), t(1) = –7.15, p < .000. In terms of the research question, it appears that sleep was effective in reducing number of errors in this sample. 5. The derived t = 0.74 was significant at p = .05 with df = 20. Therefore, H0 was rejected, and it was concluded that the mean number of errors for subjects after 8 hours of sleep deprivation (2.02) was significantly lower than the mean number of errors after 24 hours of sleep deprivation (2.12), t(20) = 0.74, p < .01. In terms of the research question, it appears that sleep was effective in reducing number of errors in this sample.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Question

How would you report the results?

Group of answer choices

1. The derived t = 0.74 was not significant at p = .05 with df = 19. Therefore, H₀ was not rejected, and it was concluded that the mean number of errors for subjects after 8 hours of sleep deprivation (2.02) was not different than the mean number of errors after 24 hours of sleep deprivation (13.90), t(19) = 0.74, p > .05. In terms of the research question, it appears that sleep deprivation did not increase the number of errors in this sample.

2. The derived t = 3.30 was significant at p = .05 with df = 18. Therefore, H₀ was rejected, and it was concluded that the mean number of errors for subjects after 24 hours of sleep deprivation (2.02) was different than the mean number of errors after 8 hours of sleep deprivation (2.12), t(18) = 3.30, p < .05. In terms of the research question, it appears that sleep was effective in reducing number of errors in this sample.

3. The derived t = –9.70 was significant at p = .05 with df = 19. Therefore, H₀ was rejected, and it was concluded that the mean number of errors for subjects after 24 hours of sleep deprivation (13.90) was significantly higher than the mean number of errors after 8 hours of sleep deprivation (6.75), t(19) = –9.70, p < .001. In terms of the research question, it appears that sleep deprivation increased the number of errors in this sample.

4. The derived t = –7.15 was significant at p = .05 with df = 1. Therefore, H₀ was rejected, and it was concluded that the mean number of errors for subjects after 8 hours of sleep deprivation (6.75) was significantly lower than the mean number of errors after 24 hours of sleep deprivation (13.90), t(1) = –7.15, p < .000. In terms of the research question, it appears that sleep was effective in reducing number of errors in this sample.

5. The derived t = 0.74 was significant at p = .05 with df = 20. Therefore, H₀ was rejected, and it was concluded that the mean number of errors for subjects after 8 hours of sleep deprivation (2.02) was significantly lower than the mean number of errors after 24 hours of sleep deprivation (2.12), t(20) = 0.74, p < .01. In terms of the research question, it appears that sleep was effective in reducing number of errors in this sample.

Paired Samples Statistics
Std. Error
Mean
Std. Deviation
Mean
Errors after 8 hours sleep
deprivation
Pair 1
6.75
20
2.124
.475
Errors after 24 hours
13.90
20
2.024
.452
sleep deprivation
Paired Samples Correlations
Significance
N
Correlation
One-Sided p
Two-Sided p
Pair 1
Errors after 8 hours sleep
20
-.263
.131
.262
deprivation & Errors after
24 hours sleep
deprivation
Paired Samples Test
Paired Differences
Significance
95% Confidence Interval of the
Difference
Std. Error
Mean
Std. Deviation
Mean
Lower
Upper
t
df
One-Sided p Two-Sided p
Errors after 8 hours sleep
deprivation - Errors after
24 hours sleep
deprivation
Pair 1
-7.150
3.297
.737
-8.693
-5.607
-9.698
19
<.001
<.001
Paired Samples Effect Sizes
95% Confidence Interval
Point
Standardizer
Estimate
Lower
Upper
Errors after 8 hours sleep
deprivation - Errors after
24 hours sleep
Pair 1
Cohen's d
3.297
-2.169
-2.972
-1.348
Hedges' correction
3.364
-2.125
-2.913
-1.321
deprivation
a. The denominator used in estimating the effect sizes.
Cohen's d uses the sample standard deviation of the mean difference.
Hedges' correction uses the sample standard deviation of the mean difference, plus a correction factor.

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Expert Solution

Step by step

Solved in 3 steps with 2 images

SEE SOLUTION Check out a sample Q&A here