Statistics - Univariate Inferential Tests - Quick Guides

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Study GuideStatistics–Univariate Inferential Tests1.One-Sample z-Test1.1Requirements (When can we use this test?)Aone-sample z-testis used when:•The population isnormally distributed, and•The population standard deviationσis knownThis test is mainly used to make decisions about apopulation mean (μ).1) Testing a Population Mean (Hypothesis Test)In a one-sample z-test, the test statistic is:Formula:Meaning of each symbol•x̄= sample mean•Δ=the value you are testing (the hypothesized mean)•σ=population standard deviation (known)•n= sample sizeAfter you compute the z-value, you use thestandard normal table(Table in Appendix B) to find theprobability value for that z-score.

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Study GuideExample 1: Steer Weight Gain (Right-Tailed Test)A herd of1,500 steerwas fed a special high-protein grain for one month.A random sample of29steer was weighed and had an average gain of:•x̄ = 6.7 poundsThe standard deviation of weight gain for the entire herd is known:•σ = 7.1You want to test this claim:The average monthly weight gain is more than 5 pounds.Step 1: Write the hypothesesSince the question asks “more than 5,” this is aright-tailed test.null hypothesis:H0:μ = 5alternative hypothesis:Ha:μ > 5Step 2: Compute the z statisticUsing the formula, the z-value is calculated as approximately:z = 1.289Then we look up the z-value in the z-table:•Tabled value forz ≤ 1.28is0.8997•So the right-tail probability is:p = 1 − 0.8997 = 0.1003

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Study GuideStep 3: DecisionThis p-value means:The probability of getting a sample mean of6.7 or higher(assuming the null hypothesis is true) isp = 0.1003.Should we reject the null hypothesis?That depends on your chosen significance level.•If you decided beforehand thatα = 0.05,then:p = 0.1003 > 0.05, so youcannot reject H₀.So the null hypothesiswould not be rejectedat the0.05 significance level.Example 2: Vocabulary Test (Two-Tailed Test)A vocabulary test is known to have:•Population mean:μ = 68•Population standard deviation:σ = 13A class of19 studentstakes the test and has:•Sample mean:x̄ = 65Question:Is this class typical of others who have taken the test?Assumeα = 0.05.1.2Why is this a two-tailed test?There are two ways this class could be different:•They could scorelowerthan average, or•They could scorehigherthan averageSo we use atwo-tailed test.

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Study GuideStep 1: State the hypotheses•Null hypothesis:H₀:μ = 68•Alternative hypothesis:Hₐ:μ ≠ 68Step 2: Find the critical valuesBecauseα = 0.05and this istwo-tailed, the alpha is split into two equal tails:•Upper tail =0.025•Lower tail =0.025From the z-table:•Lower critical value =−1.96•Upper critical value =1.96So we reject H₀if the computed z-score is outside:−1.96 to 1.96Step 3: Compute the test statistic“Now compute the z statistic using the one-sample z-test formula:”The computed value is:z = −1.006Step 4: Make the decisionSince−1.006 is between −1.96 and 1.96, it falls in the acceptance region.

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Study GuideTherefore, the null hypothesiscannot be rejected.Final conclusion (in simple words)There isnot enough evidenceto say this class is different from other students who have taken thetest.2) Confidence Interval for a Population Mean (σKnown)A z-based confidence interval for the mean is written as:Formula:“A z-based confidence interval for the mean is written as:”Meaning of the symbols•a and b= lower and upper limits of the interval•x̄= sample mean•z(α/2)=the positive critical z-value for half of alpha(because confidence intervals are always two-tailed)•σ=population standard deviation•n= sample sizeExample 3: Machine Pin Diameters (99% Confidence Interval)A sample of12 machine pinshas:•x̄ = 1.15 inches•σ = 0.04inches(known)

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Study GuideFind a99% confidence intervalfor the population mean diameter.Step 1: Find the z-value for 99% confidenceA 99% confidence level corresponds to:•α = 0.01Since confidence intervals are two-tailed:•α/2 = 0.005The z-value corresponding to an area of0.005in the upper tail is:z = 2.58Step 2: Compute the interval“Now plug the values into the confidence interval formula:”So the confidence interval is:(1.12, 1.18)Correct interpretationWe are99% confidentthat the population mean pin diameter lies between:1.12 inches and 1.18 inchesImportant note:This doesNOTmean that99% of the machine pinshave diameters between 1.12 and 1.18 inches.That conclusion would be incorrect because the interval is estimating thepopulation mean, notindividual values.

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Study Guide3) Sample Size Needed for a Desired Confidence IntervalSurveys and studies can be expensive.So researchers often want to know:“How many subjects do we need to get a confidence interval with a certain width?”The sample size formula is:“To plan ahead, we can use this formula to calculate the required sample size:”Meaning of the symbols•n= required sample size•z(α/2)=critical z-value for the chosen significance level•σ=population standard deviation•w= desired total confidence interval widthImportant reminder:The confidence interval widthwis alwaysdoublethe “plus or minus” amount.Example 4: Average Age at Fisher College (Finding n)How many students are needed to estimate the average age at Fisher Collegewithin ±1 year, with:•95% significance level•σ = 3.5Here, “±1 year” means the total width is:w = 2

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Study Guide“Now substitute the values into the sample size formula:”Final AnswerThe calculation gives:n = 47.06Since sample size must be a whole number, weround up:n = 48 studentsSo, a sample of48 studentswould be sufficient to estimate the mean agewithin ±1 yearat thegiven confidence level.2.Quiz: One-Sample z-test1. QuestionA test is conducted for (H0:μ= 34),with (σ= 5).A sample of size 100 is selected. The standard errorof the sampling distribution is:Answer Choices• 0.05• 0.5• 5Correct Answer0.5Why This Is CorrectStandard error for a one-sample z-test is:

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Study Guide2. QuestionA test is conducted for (H0:μ= 20),with (σ= 4).A sample of size 36 has (x̄= 21.4). The test statisticis:Answer Choices• 0.35• 2.1• 12.6Correct Answer2.1Why This Is CorrectFirst find the standard error:Then compute z:3. QuestionA test is conducted for (H0:μ= 50)vs. (Ha:μ> 50).The test statistic is (z = 2.46). The p-value is:Answer Choices• 0.0069• 0.0138• 0.9931

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Study GuideCorrect Answer0.0069Why This Is CorrectThis is aright-tailed test, so:4. QuestionA test is conducted for (H0:μ= 9.0)vs. (Ha:μ< 9.0).The test statistic is (z =-1.44). Correctconclusions are:Answer Choices• reject (H0) at both (α= 0.05)and (α= 0.10)• reject (H0) at (α= 0.05)but do not reject at (α= 0.10)• reject (H0) at (α= 0.10)but do not reject at (α= 0.05)Correct Answerreject (H0) at (α= 0.10)but do not reject at (α= 0.05)Why This Is CorrectLeft-tailed critical values:•at (α= 0.10): (zc≈-1.28)•at (α= 0.05): (zc≈-1.65)Since (-1.44 <-1.28), reject at 0.10.But (-1.44) isnotless than (-1.65), so do not reject at 0.05.5. QuestionThe null hypothesis is:

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