Respuesta :
Answer:
It is not enough evidence to reject null hypothesis. It is not enough evidence to say that mean score is not equal to 35.
Step-by-step explanation:
[tex]\mu=35\\n=23\\\=x=33\\s=5\\\alpha=0.1[/tex]
1. Null and alternative hypothesis
[tex]H_{0}:\mu=35\\H_{1}:\mu<35[/tex]
2. Significance level
[tex]\alpha=0.1\\1-\alpha=0.99[/tex]
Freedom degrees is given by:
[tex]v=n-1\\v=23-1=22[/tex]
For a sgnificance level of 0,01 and 22 freedom degrees, t-student distribution value is:
[tex]t_{0.01;23}=-2. 5083[/tex]
3. Test statistic
[tex]t=\frac{\=x-\mu}{\frac{s}{\sqrt{n} } }[/tex]
[tex]t=\frac{33-35}{\frac{5}{\sqrt{23} } }[/tex]
[tex]t=-1,918[/tex]
In this case, we have an one left tailed analysis, it means that null hypothesis is rejected if [tex]t<t_{\alpha;n-1}[/tex]
[tex]-1.918>-2.5083[/tex]
Conclusion:
It is not enough evidence to reject null hypothesis. It is not enough evidence to say that mean score is not equal to 35.
