Answer:
a) NULL Hypothesis
[tex]H_0:\mu \leq 77[/tex]
ALTERNATIVE hypotheses
[tex]H_0: \mu>77[/tex]
b) Therefore t Critical value from table X is
[tex]X=\pm 1.717[/tex]
c) Hence,we reject the Null hypothesis
Step-by-step explanation:
From the question we are told that:
Claim:
[tex]\mu=77[/tex]
Level of significance [tex]\alpha=0.05[/tex]
Sample mean [tex]\=x=78.6[/tex]
Sample Standard deviation [tex]\sigma=3.6[/tex]
Sample size [tex]n=23[/tex]
a)
Generally the equation for The null and alternative hypotheses is mathematically given by
NULL Hypothesis
[tex]H_0:\mu \leq 77[/tex]
ALTERNATIVE hypotheses
[tex]H_0: \mu>77[/tex]
b)
Generally the equation for The standardized test statistics t is mathematically given by
[tex]t=\frac{(\=x-\mu)}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]t=\frac{(78.6-77)}{\frac{3.7}{\sqrt{23}}}[/tex]
[tex]t=2.03[/tex]
Therefore Critical Value X is
[tex]X=(\alpha,df)[/tex]
Where
[tex]df=(23-1)\\\\df=22[/tex]
Therefore t Critical value from table X is
[tex]X=\pm 1.717[/tex]
c)
The Test statistics is outside the Critical Value
Hence,we reject the Null hypothesis
