Answer:
z= 0.278
Step-by-step explanation:
Given data
n1= 60 ; n2 = 100
mean 1= x1`= 10.4; mean 2= x2`= 9.7
standard deviation 1= s1= 2.7 pounds ; standard deviation 2= s2 = 1.9 lb
We formulate our null and alternate hypothesis as
H0 = x`1- x`2 = 0 and H1 = x`1- x`2 ≠ 0 ( two sided)
We set level of significance α= 0.05
the test statistic to be used under H0 is
z = x1`- x2`/ √ s₁²/n₁ + s₂²/n₂
the critical region is z > ± 1.96
Computations
z= 10.4- 9.7/ √(2.7)²/60+( 1.9)²/ 100
z= 10.4- 9.7/ √ 7.29/60 + 3.61/100
z= 0.7/√ 0.1215+ 0.0361
z=0.7 /√0.1576
z= 0.7 (0.396988)
z= 0.2778= 0.278
Since the calculated value of z does not fall in the critical region so we accept the null hypothesis H0 = x`1- x`2 = 0 at 5 % significance level. In other words we conclude that the difference between mean scores is insignificant or merely due to chance.
