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REVIEW Hypothesis Tests of Means

REVIEW Hypothesis Tests of Means. When to use z and When to use t. USE z Large n or sampling from a normal distribution σ is known. USE t Large n or sampling from a normal distribution σ is unknown. z and t distributions are used in hypothesis testing.

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REVIEW Hypothesis Tests of Means

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  1. REVIEW Hypothesis Tests of Means

  2. 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.

  3. 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:

  4. 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?

  5. 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.

  6. 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

  7. 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

  8. 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 ≠ 90. 2 3 4 5

  9. 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

  10. 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!

  11. 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:

  12. 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

  13. 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.

  14. =AVERAGE(A2:A42)

  15. =(D4-D7)/(D1/SQRT(D2)) =1-NORMSDIST(D8)

  16. =(D4-D12)/(D1/SQRT(D2)) =NORMSDIST(D13)

  17. =(D4-D17)/(D1/SQRT(D2)) =2*NORMSDIST(D18)

  18. =(D4-D22)/(D1/SQRT(D2)) =2*(1-NORMSDIST(D23))

  19. 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.

  20. 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

  21. 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

  22. 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

  23. 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

  24. p-Values For t-Tests • Standard tables do not give a comprehensive set of t-values. • For Example 1, t = 1.590. • t-table value for t.05,40 = 1.684 • t-table value for t.10,40 = 1.303 • Since 1.590 falls in between these two values, the best that can be said with this information is that p lies between .05 and .10.

  25. 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.

  26. p-Values for t-Tests Using Excel P-values for t-tests are calculated as follows:

  27. =(D3-G2)/D4 =TDIST(G3,40,1)

  28. =(D3-G7)/D4 =TDIST(-G8,40,1)

  29. =(D3-G12)/D4 =TDIST(-G13,40,2)

  30. =(D3-G17)/D4 =TDIST(G18,40,2)

  31. 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

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