Question

A professor thinks that the students in her statistics class this term are less creative than most students at this university. A previous study found that students at this university had a mean score of 35 on a standard creativity test and scores were normally distributed. The professor finds that her class of 23 students scores a mean of 33 on this scale, with a standard deviation of 5. Using a .01 significance level, what is the correct decision and conclusion

Answer 1

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:

$\mu=35\\n=23\\\=x=33\\s=5\\\alpha=0.1$

1. Null and alternative hypothesis

$H_{0}:\mu=35\\H_{1}:\mu$

2. Significance level

$\alpha=0.1\\1-\alpha=0.99$

Freedom degrees is given by:

$v=n-1\\v=23-1=22$

For a sgnificance level of 0,01 and 22 freedom degrees,  t-student distribution value is:

$t_{0.01;23}=-2. 5083$

3. Test statistic

$t=\frac{\=x-\mu}{\frac{s}{\sqrt{n} } }$

$t=\frac{33-35}{\frac{5}{\sqrt{23} } }$

$t=-1,918$

In this case, we have an one left tailed analysis, it means that null hypothesis is rejected if $t$

$-1.918>-2.5083$

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.