Answer:
0.0239
Step-by-step explanation:
[tex]H_0;\mu=\sigma[/tex]
[tex]H_a;\mu>6[/tex]
[tex]\bar{x}=6.08[/tex] = Mean
[tex]s=0.44[/tex] = Standard deviation
[tex]n=121[/tex] = Sample size
Test statistic is given by
[tex]\dfrac{\bar{x}-\mu}{\dfrac{s}{\sqrt{n}}}=\dfrac{6.08-6}{\dfrac{0.44}{\sqrt{121}}}\\ =2[/tex]
From T table we know it is a right tailed test.
Degree of freedom = [tex]n-1=121-1=120[/tex]
[tex]P(t>2)=1-P(t<2)\\\Rightarrow P(t>2)=1-0.9761\\\Rightarrow P(t>2)=0.0239[/tex]
Hence, p-value is 0.0239.