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## REVIEW Hypothesis Tests of Means

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**REVIEW**Hypothesis Tests of Means**5 Steps for Hypothesis TestingTest Value Method**• Develop null and alternative hypotheses • Specify the level of significance, • Use the level of significance to determine the critical values for the test statistic and state the rejection rule for H0 • Collect sample data and compute the value of test statistic • Use the value of test statistic and the rejection rule to determine whether to reject H0**When to use z and When to use t**• USEz • Large n or sampling from a normal distribution • σ is known • USEt • Large n or sampling from a normal distribution • σ is unknown z and t distributions are used in hypothesis testing. _ These are determined by the distribution of X.**General Form ofTest Statistics for Hypothesis Tests**Depending on whether or not σ is known • A test statistic is nothing more than a measurement of how far away the observed value from your sample is from some hypothesized value, v. • It is measured in terms of standard errors • σ known = z-statistic with standard error = • σ unknown = t-statistic with standard error = • The general form of a test statistic is:**Example**The average cost of all required texts for introductory college English courses seems to have gone up substantially as the professors are assigning several texts. • A sample of 41 courses was taken • The average cost of texts for these 41 courses is $86.15 Can we conclude the average cost: • Exceeds $80? • Is less than $90? • Differs from last year’s average of $95? • Differs from two year’s ago average of $78?**CASE 1: z-tests for σ Known**Assume the standard deviation is $22. • Because the sample size > 30, it is not necessary to assume that the costs follow a normal distribution to determine the z-statistic. • In this case because it is assumed that σ is known (to be $22), these will be z-tests.**Example 1: Can we conclude µ > 80?**1 H0: µ = 80 HA: µ > 80 Select α = .05 TEST: Reject H0 (Accept HA) if z > z.05 = 1.645 z calculation: Conclusion: 1.790 > 1.645 There is enough evidence to conclude µ > 80. 2 3 4 5**Example 2: Can we conclude µ < 90?**1 H0: µ = 90 HA: µ < 90 Select α = .05 TEST: Reject H0 (Accept HA) if z <-z.05= -1.645 z calculation: Conclusion: -1.121 > -1.645 There is not enough evidence to conclude µ < 90. 2 3 4 5**Example 3: Can we conclude µ ≠ 95?**1 H0: µ = 95 HA: µ ≠ 95 Select α = .05 TEST: Reject H0 (Accept HA) if z <-z.025= -1.96 or if z > z.025 = 1.96 z calculation: Conclusion: -2.578 < -1.96 There is enough evidence to conclude µ ≠ 95. 2 3 4 5**Example 4: Can we conclude µ ≠ 78?**1 H0: µ = 78 HA: µ ≠ 78 Select α = .05 TEST: Reject H0 (Accept HA) if z <-z.025= -1.96 or if z > z.025 = 1.96 z calculation: Conclusion: 2.372 > 1.96 There is enough evidence to conclude µ ≠ 78. 2 3 4 5**P-values**• P-values are a very important concept in hypothesis testing. • A p-value is a measure of how sure you are that the alternate hypothesis HA, is true. • The lower the p-value, the more sure you are that the alternate hypothesis, the thing you are trying to show, is true. So • A p-value is compared to α. • If the p-value < α; accept HA – you proved your conjecture • If the p-value > α; do not accept HA – you failed to prove your conjecture Low p-values Are Good!**5 Steps for Hypothesis TestingP-value Method**• Develop null and alternative hypotheses • Specify the level of significance, • Collect sample data and compute the value of test statistic • Calculate p-value: Determine the probability for the test statistic • Compare p-value and : Reject H0 (Accept HA), if p-value < **Calculating p-values**• A p-value is the probability that, if H0 were really true, you would have gotten a value • as least as great as the sample value for “>” tests • at most as great as the sample value for “<” tests • at least as far away from the sample value for “≠” tests • First calculate the z-value for the test. • The p-value is calculated as follows:**P-Value for “<” Test**P-Value for “>” Test P-Value for “≠” Test, With z>0 P-Value for “≠” Test, With z<0 P-value P-value 0 Z 0 Z z z z z P-value = 2*area P-value = 2*area v v v v 0 Z 0 Z**Examples – p-Values**• Example 1: Can we conclude µ > 80? • z = 1.79 • P-value = 1 - .9633 = .0367 (< α = .05). Can conclude µ > 80. • Example 2: Can we conclude µ < 90? • z = -1.12 • P-value = .1314 (> α = .05). Cannot conclude µ < 90. • Example 3: Can we conclude µ ≠ 95? • z = -2.58 • P-value = 2(.0049) = .0098 (< α = .05). Can conclude µ ≠ 95. • Example 4: Can we conclude µ ≠ 78? • z = 2.37 • P-value = 2(1-.9911) = .0178 (< α = .05). Can conclude µ ≠ 78.**=(D4-D7)/(D1/SQRT(D2))**=1-NORMSDIST(D8)**=(D4-D12)/(D1/SQRT(D2))**=NORMSDIST(D13)**=(D4-D17)/(D1/SQRT(D2))**=2*NORMSDIST(D18)**=(D4-D22)/(D1/SQRT(D2))**=2*(1-NORMSDIST(D23))**CASE 2: t-tests for σ Unknown**• Because the sample size > 30, it is not necessary to assume that the costs follow a normal distribution to determine the t-statistic. • In this case because it is assumed that σ is unknown, these will be t-tests with 41-1 = 40 degrees of freedom. Assume s = 24.77.**Example 1: Can we conclude µ > 80?**1 H0: µ = 80 HA: µ > 80 Select α = .05 TEST: Reject H0 (Accept HA) if t >t.05,40 = 1.684 t calculation: Conclusion: 1.590 < 1.684 Cannot conclude µ > 80. 2 3 4 5**Example 2: Can we conclude µ < 90?**1 H0: µ = 90 HA: µ < 90 Select α = .05 TEST: Reject H0(Accept HA) if t<-t.05,40= -1.684 t calculation: Conclusion: -0.995 > -1.684 Cannot conclude µ < 90. 2 3 4 5**Example 3: Can we conclude µ ≠ 95?**1 H0: µ = 95 HA: µ ≠ 95 Select α = .05 TEST: Reject H0 (Accept HA) if t <-t.025,40= -2.021 or if t > t.025,40 = 2.021 t calculation: Conclusion: -2.288 < -2.021 Can conclude µ ≠ 95. 2 3 4 5**Example 4: Can we conclude µ ≠ 78?**1 H0: µ = 78 HA: µ ≠ 78 Select α = .05 TEST: Reject H0 (Accept HA) if t <-t.025,40= -2.021 or if t > t.025,40 = 2.021 t calculation: Conclusion: 2.107 > 2.012 Can conclude µ ≠ 78. 2 3 4 5**The TDIST Function in Excel**• TDIST(t,degrees of freedom,1) gives the area to the right of a positive value of t. • 1-TDIST(t,degrees of freedom,1) gives the area to the left of a positive value of t. • Excel does not work for negative vales of t. • But the t-distribution is symmetric. Thus, • The area to the left of a negative value of t = area to the right of the corresponding positive value of t. • TDIST(-t,degrees of freedom,1) gives the area to the left of a negative value of t. • 1-TDIST(-t,degrees of freedom,1) gives the area to the right of a negative value of t. • TDIST(t,degrees of freedom,2) gives twice the area to the right of a positive value of t. • TDIST(-t,degrees of freedom,2) gives twice the area to the right of a negative value of t.**p-Values for t-Tests Using Excel**P-values for t-tests are calculated as follows:**=(D3-G2)/D4**=TDIST(G3,40,1)**=(D3-G7)/D4**=TDIST(-G8,40,1)**=(D3-G12)/D4**=TDIST(-G13,40,2)**=(D3-G17)/D4**=TDIST(G18,40,2)**Test Value vs P-Value**• Example of one-tailed, positive test value**Review**• When to use z and when to use t in hypothesis testing • σ known – z • σ unknown – t • z and t statistics measure how many standard errors the observed value is from the hypothesized value • Form of the z or t statistic • Meaning of a p-value • z-tests and t-tests • By hand • Excel