The student conduct data for seconds and speed. She found the following, seconds
144,149.2,148.2,149,149.6,146.8,148.6,149.8,149,148.2 and for speed 37.5,36.193029,36.437247,36.241611,36.096257,36.784741,36.339166,36.048064,36.241611,36.437247.Calculate paired sample t test for . Using alpha 0.05.
To calculate the paired sample t-test, we first need to calculate the differences between the two variables (seconds and speed). Then, we can proceed with the t-test calculation.
Let's calculate the differences:
Seconds difference = (144 - 37.5) = 106.5
Seconds difference = (149.2 - 36.193029) = 113.006971
Seconds difference = (148.2 - 36.437247) = 111.762753
Seconds difference = (149 - 36.241611) = 112.758389
Seconds difference = (149.6 - 36.096257) = 113.503743
Seconds difference = (146.8 - 36.784741) = 110.015259
Seconds difference = (148.6 - 36.339166) = 112.260834
Seconds difference = (149.8 - 36.048064) = 113.751936
Seconds difference = (149 - 36.241611) = 112.758389
Seconds difference = (148.2 - 36.437247) = 111.762753
Now, let's calculate the mean of the differences and the standard deviation:
Mean of differences (Mdiff) = (106.5 + 113.006971 + 111.762753 + 112.758389 + 113.503743 + 110.015259 + 112.260834 + 113.751936 + 112.758389 + 111.762753)/10 = 111.3930243
Standard deviation (s) of the differences = sqrt(((106.5 - 111.3930243)^2 + (113.006971 - 111.3930243)^2 + ... + (111.762753 - 111.3930243)^2)/9) = 1.989624995
Next, let's calculate the t-test statistic:
t = (Mdiff - µdiff) / (s / sqrt(n))
Where,
µdiff = 0 (null hypothesis: no difference between the mean seconds and speed)
n = 10 (number of paired measurements)
t = (111.3930243 - 0) / (1.989624995 / sqrt(10)) = 21.90220996
Finally, we need to compare the calculated t-value with the critical t-value at alpha level 0.05. Since we don't have the degrees of freedom or the exact table value, we can use statistical software or online calculators to find the critical t-value. If the calculated t-value is greater than the critical t-value, we can reject the null hypothesis.
Please note that this is a general calculation process and actual t-value or software result may vary.
To calculate the paired sample t-test for the given data, follow these steps:
Step 1: Calculate the difference between the two sets of data (seconds and speed):
- Difference = seconds - speed
The calculated differences are:
-0.5, 13.006971, 11.762753, 12.758389, 13.503743, 10.015259,
12.260834, 13.751936, 12.758389, 11.762753
Step 2: Calculate the mean of the differences:
- Mean = sum of differences / number of differences
Mean = ( -0.5 + 13.006971 + 11.762753 + 12.758389 + 13.503743 + 10.015259
+ 12.260834 + 13.751936 + 12.758389 + 11.762753 ) / 10
= 121.150635 / 10
= 12.1150635
Step 3: Calculate the standard deviation of the differences:
- Standard Deviation = square root of( (sum of (difference - mean)^2) / (n-1) )
Standard Deviation = sqrt( ( (-0.5 - 12.1150635)^2 + (13.006971 - 12.1150635)^2 +
(11.762753 - 12.1150635)^2 + (12.758389 - 12.1150635)^2 +
(13.503743 - 12.1150635)^2 + (10.015259 - 12.1150635)^2 +
(12.260834 - 12.1150635)^2 + (13.751936 - 12.1150635)^2 +
(12.758389 - 12.1150635)^2 + (11.762753 - 12.1150635)^2 ) / 9 )
= sqrt( ( 182.2130375 + 0.6454975 + 0.45993725 + 0.40732975 +
0.538871 + 4.07323425 + 0.024168 + 1.48677125 +
0.41402025 + 0.47024975 ) / 9 )
= sqrt( 18.732391 )
= 4.326529091
Step 4: Calculate the t-statistic using the formula:
- t = (Mean of differences - 0) / ( Standard Deviation / sqrt(n) )
t = ( 12.1150635 - 0 ) / ( 4.326529091 / sqrt(10) )
t = 12.1150635 / 1.369306393
t = 8.860467329
Step 5: Determine the degrees of freedom:
- Degrees of Freedom = n - 1
- Degrees of Freedom = 10 - 1
- Degrees of Freedom = 9
Step 6: Look up the critical t-value using the t-distribution table or software.
- For a significance level of 0.05 and 9 degrees of freedom, the critical value is approximately 2.262.
Step 7: Compare the t-statistic with the critical t-value.
- If the t-statistic > critical t-value, reject the null hypothesis (H0). Otherwise, fail to reject H0.
In this case, the t-statistic (8.860467329) is greater than the critical t-value (2.262). Hence, we reject the null hypothesis.
Please note that this calculation assumes the data is normally distributed and the variances of the differences are equal.