In: Statistics and Probability
The table below lists the weights of some colleage students in September and later in February of their freshman year.
September weight | 62 | 58 | 71 | 67 | 51 | 60 | 72 | 50 | 66 | 57 | 70 | 82 | 62 | 56 | 73 | 51 | 71 | 55 | 70 | 82 | 56 | 60 | 69 | 58 |
February weight | 68 | 56 | 67 | 72 | 56 | 58 | 77 | 55 | 69 | 56 | 65 | 76 | 57 | 59 | 77 | 56 | 68 | 56 | 64 | 76 | 56 | 58 | 75 | 54 |
At the 0.05 significance level, test the claim of no difference
between September weights and February weights.
The test statistic is________
The critical value is ________
Solution:
Here, we have to use paired t test.
The null and alternative hypotheses for this test are given as below:
Null hypothesis: H0: There is no difference between September weights and February weights.
Alternative hypothesis: Ha: There is a difference between September weights and February weights.
H0: µd = 0 versus Ha: µd ≠ 0
This is a two tailed test.
Test statistic for paired t test is given as below:
t = (Dbar - µd)/[Sd/sqrt(n)]
From given data, we have
Dbar = -0.0833
Sd = 4.3928
n = 24
df = n – 1 = 23
α = 0.05
t = (Dbar - µd)/[Sd/sqrt(n)]
t = (-0.0833 – 0)/[ 4.3928/sqrt(24)]
t = -0.0929
Test statistic = t = -0.0929
The test statistic is -0.0929.
The critical value is ±2.0687
(by using t-table)
The p-value by using t-table is given as below:
P-value = 0.9268
P-value > α = 0.05
So, we do not reject the null hypothesis
There is sufficient evidence to conclude that there is no difference between September weights and February weights.