Answer:
Step-by-step explanation:
Given that for a sample size of 25, sample mean = 102 lbs. and sample std deviation = 10 lbs.
[tex]H_0: \bar x = 100\\H_a: \bar x >100[/tex]
(Right tailed test)
Given that we assume population is normally distributed
Test statistic:
Mean difference = +2
Std error of sample = [tex]\frac{10}{\sqrt{25} } =2[/tex]
t = test statistic = Mean diff/Std error [tex]= 1[/tex]
degree of freedom[tex]= n-1 =24[/tex]
Critical value of t = 1.711
Our test statistic >1.71
Accept alternate hypothesis
Because there is significant difference between the test statistic and test statistic lies above critical value.