Answer:
The test is a two -tailed test
Step-by-step explanation:
From the question we are told that
The sample size is n = 31
The sample mean is [tex]\= x =11[/tex]
The sample standard deviation is [tex]\sigma = 3[/tex]
The null hypothesis is [tex]H_o: \mu \le 10[/tex]
The alternative hypothesis is [tex]H_1 : \mu > 10[/tex]
The level of significance is [tex]\alpha = 0.05[/tex]
The test statistics is mathematically represented as
[tex]t = \frac{ \= x - \mu }{ \frac{\sigma }{\sqrt{n} } }[/tex]
substituting values
[tex]t = \frac{ 11 - 10 }{ \frac{3}{\sqrt{ 31} } }[/tex]
[tex]t = 1.85[/tex]
The p- value is mathematically represented as
[tex]p-value = p( t > 1.856) = 0.0317[/tex]
Looking at the value of [tex]p-value \ and \ \alpha[/tex] we see that [tex]p-value < \alpha[/tex] hence we reject the null hypothesis
Given the that the p value is less than 0.05 it mean the this is a two-tailed test
