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

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:

1. Null and alternative hypothesis

2. Significance level

Freedom degrees is given by:

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

3. Test statistic

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

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.

6mm^3 or .6cm^3

      216

      /   \

  36     6

  /   \     /

6     6   6

factor trees are nice.