A math teacher claims that she has developed a review course that increases the scores of students on the math portion of a coll
ege entrance exam. Based on data from the administrator of the exam, scores are normally distributed with mu equalsμ=522522. The teacher obtains a random sample of 20002000 students, puts them through the review class, and finds that the mean math score of the 20002000 students is 527527 with a standard deviation of 110110. Complete parts (a) through (d) below. (a) State the null and alternative hypotheses. Let muμ be the mean score. Choose the correct answer below. A. Upper H 0 : mu equals 522H0: μ=522, Upper H 1 : mu not equals 522H1: μ≠522 B. Upper H 0 : mu less than 522H0: μ<522, Upper H 1 : mu greater than 522H1: μ>522 C. Upper H 0 : mu greater than 522H0: μ>522, Upper H 1 : mu not equals 522H1: μ≠522 D. Upper H 0 : mu equals 522H0: μ=522, Upper H 1 : mu greater than 522
The null hypothesis (H0) tries to show that no significant variation exists between variables or that a single variable is no different than its mean. While an alternative Hypothesis (Ha) attempt to prove that a new theory is true rather than the old one. That a variable is significantly different from the mean.
So, for this case;
The null hypothesis is that the mean score equals to 522
H0: μ=522
The alternative hypothesis is that the mean score is greater than 522.
Since 1*7=7, 1 4/7= (7+4)/7=11/7. When dividing fractions, we switch the numerator and denominator for the bottom number, so in this case we'd have (11/7)*(6/5)=66/35=around 1.89
