Question

For Parts #1 through #9, consider the following information: The superintendent’s office for the Jersey Dunes Regional School District report that its middle school students perform well on the mathematics portion of a certain nationally recognized standardized test, earning average scores of about 765 out of a maximum of 850 points. The county superintendent’s office is double checking the reasonability of this claim, and has collected a random sample of 25 student test scores. The standardized test results on the mathematics portion were as follows:

750	770	765	760	740
775	780	750	765	820
785	755	785	820	825
775	795	745	795	740
800	825	755	785	790

Test the claim by the superintendent’s office with the equivalence of 98% confidence. [COMMENTS & HINTS: When calculating sample statistics, round any answers to the nearest thousandth, if necessary. You should round any distribution values or standard scores and any probabilities to the nearest thousandth, if necessary.]

#1: What are the null and alternative hypotheses, stated concisely with proper symbols?

#2: What type of hypothesis test is involved: two-tail, one-tail left, or one-tail right? Explain.

#3: Which type of test statistic is generated (i.e., z or t) and why? Explain.

#4: Determine the critical value(s) that designate the rejection region(s). Indicate the critical value(s) on an appropriate probability curve (i.e., z or t) so to represent the test in graph form.

#5: Determine the value of the test statistic (i.e., z or t).

#6: Determine the p-value that corresponds to the test statistic.

#7: Concisely state the technical conclusion using appropriate and complete statistical language.

#8: Interpret the findings in complete thoughts/phrases and within the context of the problem.

#9: Construct the corresponding confidence interval which affirms the conclusion.

Answer 1

Sample mean = 778

Sample s.d= 26.34

n= 25

Null Hypothesis H0:

Alternative Hypothesis Ha:

alpha= 0.02

Since population standard deviation is unknown we will use t test.

Rejection rule if t cal > t critical= 2.17 then reject null hypothesis H0.

t= xbar-mean/s/sqrt (n)

t= 778-765/26.34/sqrt (25)

t= 13/26.34/5

t= 13/5.268

t= 2.47

Since t cal greater than t critical we therefore reject null hypothesis H0

The P-Value is .010504 since p value smaller than level of significance 0.02 therefore significant.

Conclusion: The students are doing well in mathematics.

98% confidence interval is 778± 2.17*26.34/sqrt (25)

778+/-2.17*5.268

778+/-11.43

(766.57,789.43)

We are 98% confident that true population mean will lie between 766.57 to 789.43