A RANDOM SAMPLE OF 87 EIGHTH GRADE STUDENTS SCORES ON A NATIONAL MATH TEST HAS A

Question

A RANDOM SAMPLE OF 87 EIGHTH GRADE STUDENTS SCORES ON A NATIONAL MATH TEST HAS A MEAN SCORE OF 283 WITH A STANDARD DEVIATION OF 32. THIS TEST RESULT PROMPTS A STATE SCHOOL ADMIN TO DECLARE THAT THE MEAN SCORE FOR THE STATES EIGHTH GRADERS ON THI SEXAM IS MORE THAN 280. AT a=0.12, is there enough evidence to suppar the claim
Find the Standardized test statistic z, and its corresponding area.
z=_______round to two decimal places
area = _____________round to 4 decimal places
FIND THE P- VALUE
DECIDE WHETHER TO REJECT OR FAIL THE NULL HYPOTESIS

Explanation / Answer

STEP 1: Null Hypothesis H0: P1 = P2 Alternate Hypothesis Ha: P1 ? P2 STEP 2: Analysis Plan for significance level: 0.05 STEP 3: Analyze Sample Data n1: Group 1 sample size = 28 n2: Group 2 sample size = 280 Proportion p1: = 0.0675 Proportion p2: = 0.1 Pooled Sample Proportion p = (p1 * n1 + p2 * n2) / (n1 + n2) = [(0.0675 * 280) + (0.1 * 280)] / (280 + 280) = 67 / 800 = 0.0838 Where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 2, and n2 is the size of sample 2. Standard Error: SE = sqrt{p * (1 - p ) * [(1/n1) + (1/n2)]} = sqrt{0.0838 * (1 - 0.0838 ) * [(1/400) + (1/400)]} = sqrt{0.000384} = 0.0196 STEP 4: Test Statistic z-score: z = (p1 - p2) / SE = (0.0675 - 0.1)/0.0196 = -1.659 For two-tailed test, the p-value is the probability that the z-score is less than -1.659 and more than 1.659 Use the Normal Distribution Table to find P(z < -1.659) = 0.049, and P(z >1.659) = 0.049 The Table for Standard Normal Distribution is organized as a cummulative 'area' from the LEFT corresponding to the STANDARDIZED VARIABLE z. The Standard Normal Distribution is also symmetric (called a 'Bell Curve') which means its an interpretive procedure to Look-Up the 'area' from the Table. For STANDARDIZED VARIABLE z = -1.659 the corresponding LEFT 'area' = 0.049 And due to Table's cummulative nature, the corresponding RIGHT 'area' = 0.049 For a two-tailed test: The P-value = 0.049 + 0.049 = 0.098 STEP 5: Interpret Results Since the P-value (0.098) is more than the significance level (0.05) the Null Hypothesis H0: P1 = P2 must be accepted.

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