Question

The following data is the math test scores of students graduating from a particular high school. The government uses these scores to determine if there will be accreditation awarded. In order for this to occur the mean score must be above 780. A sample of students' scores is drawn at random and they take the test. The scores are in the following table and the population is considered a normal distribution. Test at the .01 level.

980 764 798 760 796 760 798 980 796 796 798 790 960 960 900

1. Perform a hypotheses test to determine if this school should get accreditation. State the appropriate hypothesis. Determine critical values. State clearly what the results of the hypothesis test mean.

2. One of the two people that got a 980 cheated and should be removed from the analysis (you will need a new t critical) Did the cheater affect the accreditation?

Answer 1

1.

Step 1:

Ho: = 780

Ha: > 780

Step 2: Test statistics

n =

sample mean = sum of terms / no of terms = 12636 / 15= 842.4

sample sd = s

data	data-mean	(data - mean)2
980	137.6	18933.76
764	-78.4	6146.56
798	-44.4	1971.36
760	-82.4	6789.76
796	-46.4	2152.96
760	-82.4	6789.76
798	-44.4	1971.36
980	137.6	18933.76
796	-46.4	2152.96
796	-46.4	2152.96
798	-44.4	1971.36
790	-52.4	2745.76
960	117.6	13829.76
960	117.6	13829.76
900	57.6	3317.76

Assuming data is normally distributed and as population sd is not given, we will calculate t stat.

t = 2.808

Step 3:

The t-critical value for a right-tailed test, for a significance level of α=0.01 is

tc=2.624

As t stat is does not fall in the rejection area, we fail to reject the Null hypothesis.

Hence we do not have sufficient evidence to believe that the average math test scores of students graduating from a particular high school is greater than 780.

2. New sample of 14 (excluding one 980 score)

new sample mean = 11656 / 14 = 832.5714

new sample deviation = s

data	data-mean	(data - mean)2
764	-68.5714	4702.03689796
798	-34.5714	1195.18169796
760	-72.5714	5266.60809796
796	-36.5714	1337.46729796
760	-72.5714	5266.60809796
798	-34.5714	1195.18169796
980	147.4286	21735.19209796
796	-36.5714	1337.46729796
796	-36.5714	1337.46729796
798	-34.5714	1195.18169796
790	-42.5714	1812.32409796
960	127.4286	16238.04809796
960	127.4286	16238.04809796
900	67.4286	4546.61609796

t stat = 2.456

t criitcal

The t-critical value for a right-tailed test, for a significance level of α=0.01 is tc = 2.65

As the t stat does not fall in the rejection area, we fail to reject the Null hypothesis.

Hence we do not have sufficient evidence to believe that the average math test scores of students graduating from a particular high school is greater than 780.

The cheater did not affect the accreditation.