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T TEST. DEFINITION OF TERMS. T – Test – the statistical test for comparing a mean with a norm or for comparing two means with small sample sizes (n≤30). Formula: __ X - µ

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## T TEST

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**DEFINITION OF TERMS**• T – Test – the statistical test for comparing a mean with a norm or for comparing two means with small sample sizes (n≤30). Formula: __ X - µ t = ——— SD/√n**_**• Mean ( X ) – the most common measure of central tendency, denoted by µ in the population and by in the sample. Formula: _ X = ΣX n Σ – summation or to add µ - population mean X – individual observations n – number of observations**Population – the entire collection of observations**or subjects that have something in common and to which conclusions are inferred. • Standard Deviation – the most common measure of dispersion or spread, denoted by σ in the population and SD or s in the sample. It can be used with mean to describe the distribution of observations.**Formula:**_ SD = √Σ (X – X)² n – 1 Hypothesis Test – an approach to statistical inference resulting in a decision to reject or not to reject the null hypothesis.**STEPS IN HYPOTHESIS TESTING**STEP 1: State the research question in terms of statistical hypotheses or formulate the hypothesis Null Hypothesis (Ho) – a statement claiming that there is no difference between the assumed or hypothesized value and the population mean. Alternative Hypothesis (Hı)- a statement that disagrees with the null hypothesis.**Step 2: Decide on the appropriate test statistic.**__ X - µ t = ——— SD/√n**STEP 3: Select the level of significance for the statistical**test. Level of Significance – the probability of incorrectly rejecting the null hypothesis in a test of hypothesis, denoted by α when it is actually true (and concluding there is difference when there is nor. - traditional values used for α are 0.05, 0.01 and 0.001.**STEP 4: Determine the value the test statistic must**attain to be declared significant. Determine the rejection region or the critical region. Rejection Region – is a set of values of the test statistic for which the null hypothesis is rejected in a hypothesis test. Acceptance Region – non-critical region Critical Value – the value that a test statistic must exceed for the null hypothesis to be rejected. - refer to the critical values for t distribution table.**Degrees of Freedom (df) – the precise size of the standard**deviation depending on a complex concept related to the sample size which is related to the number of times sample information is used. - n -1 (df) STEP 5: Perform the calculation. STEP 6: Interpret and state the conclusion.**EXAMPLE: YLO Company, manufacturer of Yummy**Oats is claiming that their product contains 8 grams of fiber per 35 gram-pack. VIN Company, its direct competitor conducts an independent test of the product. They test 15 samples of Yummy Oats and it found out to have 8.00, 8.25, 8.16, 7.92, 7.81, 7.93, 8.01, 8.02, 8.50, 8.16, 7.71, 7.81, 7.62, 7.81 and 7.86 of fiber content. If 0.01 level of significance is appropriate, would VIN Company agree on the claim of YLO Co.?**STEP 1: Ho: µ = 8 grams : Yummy**Oats contains 8 grams of fiber/pack Hı: µ ≠ 8 grams : Yummy Oats does not contain 8 grams of fiber/pk STEP 2: __ X X-X (X-X)² 8.00 0.03 0.0009 8.25 0.28 0.0784 8.16 0.19 0.0361 7.92 0.05 0.0025 7.81 0.16 0.0256 7.93 0.04 0.0016 8.01 0.04 0.0016 8.02 0.05 0.0025 8.50 0.53 0.2809 8.16 0.19 0.0361**7.71 0.26 0.0676**7.81 0.16 0.0256 7.62 0.35 0.1225 7.81 0.16 0.0256 7.86 0.11 0.0121 _ X = 119.57 15 = 7.97 Σ = 0.7196**SD =√0.7196/15-1 = √0.7196/14 = √0.0514 = 0.2267**7.97 – 8 -0.03 -0.03 T = ÷ = ÷ = ÷ 0.2267/√15 0.2267/3.8729 0.0585 T = -0.5128**STEP 3: Level of Significance: 0.01**STEP 4: Degrees of Freedom: 15 – 1 = 14 Critical Value : -2.624 RR AR __________________________________ -2.624 -0.5128 0**STEP 6: The computed value of t is -0.5128 is at**the acceptance region. Accepts Null Hypothesis CONCLUSION: VIN Company would agree to the claim of YLO Company that Yummy Oats has 8 grams per 35 gram/pack of fiber content.**EXAMPLE: It was indicated in the label of Vita C**that it contains 235mg of ascorbic acid. An independent test was done to prove whether the claim of the manufacturer is true or not. Ten samples were tested and the results are as follows: 230, 232, 228, 231, 218, 225, 221, 220, 223, and 228. If 0.01 level of significance is appropriate, determine whether to accept or reject the claim of Vita C manufacturer.**STEP 1: Ho: µ = 235 mg : Vita C contains**235 mg ascorbic acid Hı: µ ≠ 235 mg : Vita C does not contain 235 ascorbic a. __ X X-X (X-X)² 230 4.4 19.36 232 6.4 40.96 228 2.4 5.76 231 5.4 29.16 218 7.6 57.76 225 0.6 0.36 221 4.6 21.16 220 5.6 31.36 223 2.6 6.76 228 2.4 5.76**_ X = 2256/10 = 225.6 Σ = 218.4**SD = √218.4/10-1 = √218.4/9 = √24.27 = 4.93 T = 225.6-235/4.93/√10 = -9.4/4.93/3.16 = -9.4/1.56 = -6.03 STEP 3: Level of Significance: 0.01 STEP 4: Degrees of Freedom: 10-1 = 9**RR AR**_______________________________________ -6.03 -2.821 0 STEP 6: The computed value of t is -6.03 is at the rejected region. Rejects Null Hypothesis. CONCLUSION: The independent test for Vita C containing 235 mg of ascorbic acid is rejected.

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