Respuesta :
The Test value for difference between the mean numbers of calls for the two shifts at the 0.05 level of significance is 31.94
According to the question,
Karen's data:
Sample size : n₁ = 37
Sample mean = 4.5
Sample standard deviation : s₁ = 1.1
Jodi's data ,
Sample size : n₂ = 32
Sample mean = 5.3
Sample standard deviation : s₂ = 1.5
To calculate significance difference between two means we use t-test
t = difference of mean / √(pooled variance / n₁ + n₂)
Pooled Standard deviation = [tex]\sqrt(\frac{(n_1 - 1)s_1^{2} + (n_2 - 1)s_2^{2})}{n_1 + n_2 -2}[/tex]
=> [tex]\sqrt(\frac{(37 - 1)1.21+ (32 - 1)2.25)}{37 + 32 -2}[/tex]
=> [tex]\sqrt(\frac{(43.56+ 69.75)}{67}[/tex]
=>[tex]\sqrt(\frac{(113.31)}{67}[/tex]
=> √1.6911
t = 37 - 32 / (√1.6911/37+32)
=> 5 / √0.0245
=> 31.94 is test value
To know more about Test statistic here
https://brainly.com/question/14128303
#SPJ4
