Let [tex]\mu[/tex] be the mean response time of a species of pigs to a stimulus.
As per given , we have
[tex]H_0 : \mu=0.8\\\\ H_a: \mu\neq0.8[/tex]
Since , the alternative hypothesis is two-tailed and population standard deviation is unknown , so we perform two-tailed t-test.
Also, we have n= 28
Sample mean =[tex]\overline{x}=1.0[/tex]
sample standard deviation= s= 0.3
Test statistic : [tex]t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}[/tex]
[tex]t=\dfrac{1.0-0.8}{\dfrac{0.3}{\sqrt{28}}}=3.53[/tex]
P-value : [tex]2P(t>|3.53|)=2(1-P(t\leq3.53))[/tex]
[tex]=2(1-0.9992)= 0.0016[/tex] [using p-value table for t]
Decision : Since p-value < 0.05 ( significance level ), so we reject the null hypothesis .
Conclusion : We have enough evidence to accept that alcohol affects the mean response time.
