Download Stats Second Semester Final Review Unit 4: Means CLT for a

Stats Second Semester Final Review
Unit 4: Means
CLT for a sample mean:
1) shape: approximately normal
2)center: mean = 
3) spread: standard deviation =
𝜎
√𝑛
Confidence Intervals: Means:
Degrees of freedom (d.f.) = n -1
4-steps of Confidence Interval:
1) Claims: T-Interval; state 𝑥̅ and 𝑠𝑥
2) Conditions: 1) random sample & 2) n ≥30 or the population is normally distributed.
3) Calculations: By Calculator: STAT -> TESTS -> 8:TInterval
𝑠
By hand: 𝑥̅ ± 𝑡 ∗ ( 𝑥⁄ ) , where 𝑡 ∗ is the appropriate critical value from the t-distribution with n-1 degrees of
√𝑛
freedom, n = sample size, 𝑥̅ = sample mean, 𝑠𝑥 = sample standard deviation.
4) Conclusion: “We are ____ % confidence that the actual population mean is between _____________ and
_____________”. (context!)
 Samples with more variability produce wider confidence intervals
 Confidence intervals are use to estimate the value of a population mean. They do not estimate the values of
individual observations in the population or in the sample.
Tests of Significance: Means
4-steps of Inference:
1) Claims: T-Test; hypotheses in symbols and words
Ho: µ = 𝜇0 (𝜇0 is some number)
Ha: µ <, >, 𝑜𝑟 ≠ 𝜇0
2) Conditions: 1) random sample & 2) either sample size is large (n ≥ 30) or the population is normally distributed.
3) Calculations: By calculator: STAT -> TESTS -> 2: T-Test
𝑥̅ −𝜇
By hand: 𝑡 = 𝑠𝑥 0 , where 𝑥̅ is the sample mean, 𝑠𝑥 is the sample standard deviation, and n = sample size.
⁄
√𝑛
3) Conclusions: If p is low, reject Ho (and accept Ha). “We can conclude that µ is significantly ______ than 𝜇0 ”
Otherwise, we fail to reject Ho (so we fail to accept Ha). “We cannot conclude that µ is significantly ______ than
𝜇0 ”
2-Sample T-Interval:
1) Claims: T-Interval; state all givens and confidence interval
2) Conditions: 1) both samples are independently and randomly selected & 2) both samples are large or both are
normal.
3) Calculations: STAT -> TESTS -> 0:2-SampTInt
Pooled: NO
4) Conclusion: We are ____ % confidence that the mean difference is between ______ and ______.
2-Sample T-Test:
1) Claims: State your hypotheses in symbols and words
Ho: 𝜇𝐴 = 𝜇𝐵
Ha: 𝜇𝐴 <, >, 𝑜𝑟 ≠ 𝜇𝐵
2) Conditions: 1) both samples are independently and randomly selected & 2) both samples are large or both are
normal.
3) Calculations: STAT -> TESTS -> 4:2-SampTTest Pooled: NO
4) Conclusion: We can(not) conclude the alternative hypothesis.
Unit 5: Proportions
CLT for a sample proportion:
1) shape: approximately normal
2) center: mean = Θ
3) spread: standard deviation = √
θ(1−θ)
𝑛
Confidence Interval for a population proportion 𝚯
1) Claims: 1-Prop Z Interval and state all the given with the c-level
2) Conditions: 1) random & 2) n𝑝̂ ≥ 10 𝑎𝑛𝑑 𝑛(1 − 𝑝̂ ) ≥ 10
3) Calculations: By Calculator: STAT ->TESTS ->A: 1-PropZInt
𝑝̂(1−𝑝̂)
,
𝑛
By hand: 𝑝̂ ± 𝑧 ∗ √
where p̂ is the sample proportion, n is the sample size an z* is the critical value from
the standard normal distribution.
3) Conclusion: We are ___ % confident that between ___ % and ___ % of the population _____ .”
Test of Significance: Proportions
1) Claims: 1-Prop Z Test and the hypotheses in symbols and words
Ho: 𝜃 = 𝜃0
Ha: 𝜃 <, >, 𝑜𝑟 ≠ 𝜃0
2) Conditions: 1) random & 2) n𝜃0 ≥ 10 𝑎𝑛𝑑 𝑛(1 − 𝜃0 ) ≥ 10
3) Calculations:
By hand: 𝑧 =
𝑝̂−𝜃0
By Calculator: STAT ->TESTS ->5: 1-PropZTest
𝜃 (1−𝜃0 )
√ 0
𝑛
4) Conclusion: Same as usual, if p is low, reject the Ho. State your conclusion in terms of Ha and use %’s.
2-Prop Z Test:
1) Claims: 2-Prop Z Test and the hypotheses in symbols and words
Ho: 𝜃1 = 𝜃2
Ha: 𝜃1 <, >, 𝑜𝑟 ≠ 𝜃2
2) Conditions: 1) random and independent &
2) 𝑛1 𝑝̂𝑐 ≥ 5 𝑎𝑛𝑑 𝑛1 (1 − 𝑝̂𝑐 ) ≥ 5 and 𝑛2 𝑝̂𝑐 ≥ 5 𝑎𝑛𝑑 𝑛2 (1 − 𝑝̂𝑐 ) ≥ 5
3) Calculations:
By hand: 𝑧 =
𝑝̂1 −𝑝̂2
𝑥 +𝑥
1
1
√𝑝̂𝑐 (1−𝑝̂𝑐 )(𝑛 +𝑛 )
1
2
, 𝑝̂𝑐 = 𝑛1 +𝑛2
1
2
By Calculator: STAT ->TESTS ->6: 2-PropZTest
4) Conclusion: Same as usual, if p is low, reject the Ho. State your conclusion in terms of Ha and in context!
2-Proportions Confidence Interval:
1) Claims: 2-Prop Z Int. and state all givens and c-level
2) Conditions: 1) random and independent &
2) 𝑛1 𝑝̂1 ≥ 5 𝑎𝑛𝑑 𝑛1 (1 − 𝑝̂1 ) ≥ 5 and 𝑛2 𝑝̂ 2 ≥ 5 𝑎𝑛𝑑 𝑛2 (1 − 𝑝̂2 ) ≥ 5
3) Calculations:
𝑝̂1 (1−𝑝̂1 )
𝑝̂ (1−𝑝̂
+ 2𝑛 2
𝑛1
2
By hand: (𝑝̂1 − 𝑝̂2 ) ± 𝑧 ∗ √
Calculator: STAT ->TESTS ->B: 2-PropZInt
4) Conclusion: “We’re ___ % confident that the difference in proportions between the two groups (context!) is
between ___ % and ___ %.”
Topic 26: Chi-Square Test for Goodness of Fit and Chi-Square Test for Two-Way Table
 Chi-Square test for Goodness of Fit: (the chi-square statistic X2 is a measure of the distance of the observed
counts from the expected counts. Like any distance, it’s always zero or positive, and it’s zero only when the
observed counts are exactly equal to the expected counts. Large values of X2 are evidence against Ho.
Ho: the _________ is ____________ (uniformly, likely, …) distributed
Ha: the _________ is not ____________ (uniformly, likely, …) distributed

Chi-Square test for two-way table: (the chi-square statistic is a measure of how far the observed counts in a
two-way table are from the expected counts).
Ho: there is no association between _______________
Ha: there is an association between _______________