Answer:
a) [tex]t_{0.035},11=\pm2.201[/tex]
b) [tex]t=-2.963[/tex]
c) [tex]Reject\ H_0\ when\ \alpha=0.05[/tex]
d) [tex]t<t_{\alpha/2},d_t[/tex]
[tex]-2.963<2.201[/tex]
Step-by-step explanation:
From the question we are told that:
Sample size [tex]n=12[/tex]
Mean [tex]\=x=42.5[/tex]
Standard deviation [tex]\sigma=3.8[/tex]
Population mean [tex]\mu=45.75[/tex]
Significance [tex]\alpha=0.05[/tex]
Generally the hypothesis given by
[tex]H_0;\mu=45.75\\H_1:\neq =45.75[/tex]
Generally the equation for test statistics is mathematically given by
[tex]t=\frac{\=x-\mu}{\sigma/\sqrt{n} }[/tex]
[tex]t=\frac{42.5-45.75}{3.8/\sqrt{12} }[/tex]
[tex]t=-2.963[/tex]
Generally the Critical value is mathematically given by
[tex]t_{\alpha/2},d_t[/tex]
[tex]\alpha=0.05 \\\alpha/2=0.025\\d_t=n-1=11[/tex]
[tex]t_{0.035},11[/tex]
From table
[tex]t_{0.035},11=\pm2.201[/tex]
Therefore
[tex]t<t_{\alpha/2},d_t[/tex]
[tex]-2.963<2.201[/tex]
[tex]Reject\ H_0\ when\ \alpha=0.05[/tex]
