In: Statistics and Probability
Independent random samples from two normal distributions
returned the values:
Sample 1: 45, 78, 121, 45, 67, 89, 95, 99, 79, 74
Sample 2: 67, 32, 67, 45, 21, 67, 89, 78, 77, 96, 94, 99
Find a 95% confidence interval for the difference in population
means.
Answer:
The 95% confidence interval for the difference in population means is = ( -12.0663 , 31.79972 )
= ( -12.1 , 31.8 ) ... rounded to 1 decimal place
Explanation:
Here by considering that both the samples having equal variance, we use the independent sample t test to find 95% confidence interval for the difference in population means. ( We can check equality of variances of two samples by usinf F- test)
The 95% confidence interval for the difference in population means can be given by:
where and are sample means, n1 and n2 are sample sizes, Sp is pooled standard deviation and t is table value with (n1 +n2 -2) df
Sp is given as :
where s1^2 and s2^2 are sample variances
Now the summary statistics of two samples from excel ( we need sample means and sample variance that can also be calculared manually)
sample1 | sample2 | ||
Mean | 79.2 | Mean | 69.33333 |
Standard Error | 7.454752 | Standard Error | 7.312643 |
Median | 78.5 | Median | 72 |
Mode | 45 | Mode | 67 |
Standard Deviation | 23.574 | Standard Deviation | 25.33174 |
Sample Variance | 555.7333 | Sample Variance | 641.697 |
Kurtosis | -0.07293 | Kurtosis | -0.37501 |
Skewness | 0.039448 | Skewness | -0.74227 |
Range | 76 | Range | 78 |
Minimum | 45 | Minimum | 21 |
Maximum | 121 | Maximum | 99 |
Sum | 792 | Sum | 832 |
Count | 10 | Count | 12 |
Here S1^2 = 555.7333 and S2^2 = 641.679
hence
Sp = 24.556329
Now the 95% confidence interval is
Here = 79.2 and = 69.3333 and t = 2.086 ( from table with alpha=0.05 and df = 20)
= ( -12.0663 , 31.79972 ) = ( -12.1 , 31.8 ) ... rounded to 1 decimal place
MINITAB output for reference:
Two-Sample T-Test and CI: sample1, sample2
Two-sample T for sample1 vs sample2
N Mean StDev SE Mean
sample1 10 79.2 23.6 7.5
sample2 12 69.3 25.3 7.3
Difference = μ (sample1) - μ (sample2)
Estimate for difference: 9.9
95% CI for difference: (-12.1, 31.8) # Our
Answer
T-Test of difference = 0 (vs ≠): T-Value = 0.94 P-Value = 0.359 DF
= 20
Both use Pooled StDev = 24.5563