Skip to content
Quick Guides

Statistics - Univariate Inferential Tests

Preview 10 of 103 pages

See the first 10 pages exactly as they appear in the document.

  1. Page 1
    Read page 1 as text

    Study Guide Statistics – Univariate Inferential Tests 1. One - Sample z - Test 1.1 Requirements (When can we use this test?) A one - sample z - test is used when: • The population is normally distributed , and • The population standard deviation σ is known This test is mainly used to make decisions about a population mean ( μ) . 1) Testing a Population Mean (Hypothesis Test) In a one - sample z - test, the test statistic is: For mula: Meaning of each symbol • x̄ = sample mean • Δ = the value you are testing (the hypothesized mean) • σ = population standard deviation (known) • n = sample size After you compute the z - value, you use the standard normal table (Table in Appendix B) to find the probability value for that z - score.

  2. Page 2
    Read page 2 as text

    Study Guide Example 1: Steer Weight Gain (Right - Tailed Test) A herd of 1,500 steer was fed a special high - protein grain for one month. A random sample of 29 steer was weighed and had an average gain of: • x̄ = 6.7 pounds The standard deviation of weight gain for the entire herd is known: • σ = 7.1 You want to test this claim: The average monthly weight gain is more than 5 pounds. Step 1: Write the hypotheses Since the question asks “more than 5,” this is a right - tailed test . null hypothesis : H 0 : μ = 5 alternative hypothesis : H a : μ > 5 Step 2: Compute the z statistic Using the formula, the z - value is calculated as approximately: z = 1.289 Then we look up the z - value in the z - table: • Tabled value for z ≤ 1.28 is 0.8997 • So the right - tail probability is: p = 1 − 0.8997 = 0.1003

  3. Page 3
    Read page 3 as text

    Study Guide Step 3: Decision This p - value means: The probability of getting a sample mean of 6.7 or higher (assuming the null hypothesis is true) is p = 0.1003 . Should we reject the null hypothesis? That depends on your chosen significance level. • If you decided beforehand that α = 0.05 , then: p = 0.1003 > 0.05 , so you cannot reject H ₀ . So the null hypothesis would not be rejected at the 0.05 significance level . Example 2: Vocabulary Test (Two - Tailed Test) A vocabulary test is known to have: • Population mean: μ = 68 • Population standard deviation: σ = 13 A class of 19 students takes the test and has: • Sample mean: x̄ = 65 Question: Is this class typical of others who have taken the test? Assume α = 0.05 . 1.2 Why is this a two - tailed test? There are two ways this class could be different: • They could score lower than average, or • They could score higher than average So we use a two - tailed test .

  4. Page 4
    Read page 4 as text

    Study Guide Step 1: State the hypotheses • Null hypothesis: H ₀ : μ = 68 • Alternative hypothesis: Hₐ: μ ≠ 68 Step 2: Find the critical values Because α = 0.05 and this is two - tailed , the alpha is split into two equal tails: • Upper tail = 0.025 • Lower tail = 0.025 From the z - table: • Lower critical value = − 1.96 • Upper critical value = 1.96 So we reject H ₀ if the computed z - score is outside: − 1.96 to 1.96 Step 3: Compute the test statistic “Now compute the z statistic using the one - sample z - test formula:” The computed value is: z = −1.006 Step 4: Make the decision Since − 1.006 is between −1.96 and 1.96 , it falls in the acceptance region.

  5. Page 5
    Read page 5 as text

    Study Guide Therefore, the null hypothesis cannot be rejected . Final conclusion (in simple words) There is not enough evidence to say this class is different from other students who have taken the test. 2) Confidence Interval for a Population Mean ( σ Known) A z - based confidence interval for the mean is written as: For mula: “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 size Example 3: Machine Pin Diameters (99% Confidence Interval) A sample of 12 machine pins has: • x̄ = 1.15 inches • σ = 0.04 inches (known)

  6. Page 6
    Read page 6 as text

    Study Guide Find a 99% confidence interval for the population mean diameter. Step 1: Find the z - value for 99% confidence A 99% confidence level corresponds to: • α = 0.01 Since confidence intervals are two - tailed: • α/2 = 0.005 The z - value corresponding to an area of 0.005 in the upper tail is: z = 2.58 Step 2: Compute the interval “Now plug the values into the confidence interval formula:” So the confidence interval is: (1.12, 1.18) Correct interpretation We are 99% confident that the population mean pin diameter lies between: 1.12 inches and 1.18 inches Important note: This does NOT mean that 99% of the machine pins have diameters between 1.12 and 1.18 inches. That conclusion would be incorrect because the interval is estimating the population mean , not individual values.

  7. Page 7
    Read page 7 as text

    Study Guide 3) Sample Size Needed for a Desired Confidence Interval Surveys 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 width Important reminder: The confidence interval width w is always double the “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 College within ±1 year , with: • 95% significance level • σ = 3.5 Here, “±1 year” means the total width is: w = 2

  8. Page 8
    Read page 8 as text

    Study Guide “Now substitute the values into the sample size formula:” Final Answer The calculation gives: n = 47.06 Since sample size must be a whole number, we round up : n = 48 students So, a sample of 48 students would be sufficient to estimate the mean age within ±1 year at the given confidence level. 2. Quiz: One - Sample z - test 1. Question A test is conducted for ( H 0 : μ = 34), with ( σ = 5). A sample of size 100 is selected. The standard error of the sampling distribution is: Answer Choices • 0.05 • 0.5 • 5 Correct Answer 0.5 Why This Is Correct Standard error for a one - sample z - test is:

  9. Page 9
    Read page 9 as text

    Study Guide 2. Question A test is conducted for ( H 0 : μ = 20), with ( σ = 4). A sample of size 36 has ( x̄ = 21.4). The test statistic is: Answer Choices • 0.35 • 2.1 • 12.6 Correct Answer 2.1 Why This Is Correct First find the standard error: Then compute z: 3. Question A test is conducted for ( H 0 : μ = 50) vs. (H a : μ > 50). The test statistic is (z = 2.46). The p - value is: Answer Choices • 0.0069 • 0.0138 • 0.9931

  10. Page 10
    Read page 10 as text

    Study Guide Correct Answer 0.0069 Why This Is Correct This is a right - tailed test , so: 4. Question A test is conducted for ( H 0 : μ = 9.0) vs. (H a : μ < 9.0). The test statistic is (z = - 1.44). Correct conclusions are: Answer Choices • reject ( H 0 ) at both ( α = 0.05) and ( α = 0.10) • reject ( H 0 ) at ( α = 0.05) but do not reject at ( α = 0.10) • reject ( H 0 ) at ( α = 0.10) but do not reject at ( α = 0.05) Correct Answer reject ( H 0 ) at ( α = 0.10) but do not reject at ( α = 0.05) Why This Is Correct Left - tailed critical values: • at ( α = 0.10): ( z c ≈ - 1.28) • at ( α = 0.05): ( z c ≈ - 1.65) Since ( - 1.44 < - 1.28), reject at 0.10. But ( - 1.44) is not less than ( - 1.65), so do not reject at 0.05. 5. Question The null hypothesis is:

93 more pages in the full document

Unlock it once and it stays in your library, ready to chat with or turn into flashcards.

Unlock the full document

More Statistics study materials

See all

Study this document with CramX

Once it is in your library, every tool can work from it.