Answer:
Explanation:
Given that
sample size =n= 50
x bar = sample mean = 10.8
s = sample std deviation = 3.3
Mean difference = 50-10 = 40
Std error of sample = std dev / sqrt n=
![(3.3)/(√(40) ) =0.5218](https://img.qammunity.org/2020/formulas/mathematics/high-school/cpa4ux0kwenf296vx2coawsu8oxhlfp6yo.png)
Since population std dev is not known, we have to use t test
Test statistic t = Mean diff/std error =
![(40)/(0.5218) =76.66](https://img.qammunity.org/2020/formulas/mathematics/high-school/4zkeygbi257i0q6k5fpv2jgibznqls8t4y.png)
df = 39
p value = 0.0000
Since p < 0.09 our significance level, we reject our null hypothesis.
A. There is sufficient evidence to reject Upper H 0 for alpha greater than 0.09