Answer the following questions showing all work. Full credit will not be given t

Question

Answer the following questions showing all work. Full credit will not be given to answers without work shown. If you use StatKey or Minitab Express include the appropriate output (copy + paste) along with an explanation. Output without explanation will not receive full credit. Round all answers to at least 3 decimal places. If you have any questions, post them to the course discussion board.

4. In the population ACT scores are normally distributed with a mean of 18 and a standard deviation of 6. Suppose that we are taking a simple random sample of 60 students from one high school. [30 points]

              A. Calculate the standard error of the mean.

B. If we were to repeatedly pull samples of 60 individuals from the population of all ACT test takers, the distribution of sample means would have a mean of ____ and a standard deviation of ____.

C. Given the values from part B, 95% of random samples of n=60 will have sample means between ___ and ___.

D. What is the probability that you would pull a random sample of 60 individuals from the population of all test takers and they would have a sample mean of 19 or higher?

E. Suppose that the high school in question boasts that their students (i.e., the population of all of their students) have an average ACT score above the national average of 18. Given your results from part D, do you believe that there is evidence that the mean ACT score at this high school is greater than 18? Explain your reasoning.

Explanation / Answer

Number of students = 60

Mean ACT scores = 18

Standard deviation of ACT scores = 6

(a) Standard error of the mean se= /n = 6/ 60 = 0.7746

(b) The mean will be equal to population mean which is equls to 18 and standard deviation will be equal to standard error 0.7746.

(c) 95% of sample mean will lie = 95% confidence interval for sample mean

= xbar +- z95% se = 18 +- 1.96 * 0.7746 = (16.4818, 19.5182)

(d) so Here we have to calculate Pr(xbar > 19; 18; 0.7746) by normal distribution table.

Z = ( 19 - 18)/ 0.7746 = 1.291

so Pr(xbar > 19; 18; 0.7746) = 1 - Pr(xbar < 19; 18; 0.7746) = 1 - (1.291)

where is the cumulative standard normal probability distribtution.

Pr(xbar > 19; 18; 0.7746) = 1 - Pr(xbar < 19; 18; 0.7746) = 1 - (1.291) = 1- 0.9015 = 0.0985

(E) Lets assume null hypothesis that mean of high school in question have an avergae ACT score less than or equal to 18 marks. The alternative is that mean of high school average ACT score is above 18 marks. Now, in question D we can see that even if the mean score of high school is 19 or above, then also we cannot reject the null hypothesis at standard alpha = 0.05 significance level so we can reject the high school claim that their mean score of ACT is higher than 18.

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