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Stats Final Cheat Sheet
Probability And Stat For Engr (ENGR 2090)
University of Georgia
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Statistics Final Cheat Sheet | Camille Deguzman | Fall 2022 6, 6, 6, 6, 6, 6.
Steps in Performing a Hypothesis Test Define 𝐻 0 and𝐻 1
- Assume 𝐻 0 to be true
- Compute test statistic (assess the strength of the evidence against 𝐻 0
- Compute P-value (assuming 𝐻 0 to be true) P-value also called observed significance level.
- State a conclusion about strength of the evidence against 𝐻 0
— — — — — — — — — — — — — — — — Right Tailed Test:
𝑧 = 𝑋−μ σ | used when >
( 𝑛 )
P-value: 1 − 𝑃(𝑧 < #)
Left Tailed Test:
𝑧 = 𝑋−μ σ | used when >
( 𝑛 )
P-value: 𝑃(𝑧 < #) — — — — — — — — — — — — — — — — Test Statistic, t: 𝑡 𝑜𝑟 𝑧 = 𝑋−μ σ 𝑛 (for t) 𝑑. 𝑓. = 𝑛 − 1 — — — — — — — — — — — — — — — — Alternate Hypothesis ⇔P-value 𝐻 1 : μ > μ 0 ⇔ Area to the right of𝑧 𝐻 1 : μ < μ 0 ⇔ Area to the left of𝑧 𝐻 1 : μ ≠ μ 0 ⇔ Sum of the areas in the tails cut off by 𝑧 and− 𝑧 — — — — — — — — — — — — — — — — P-value μ ≠ #→ 𝑃(𝑧 <− #) + (1 − 𝑃(𝑧 <+ #) μ > # → (1 − 𝑃(𝑧 <+ #) μ < # → 𝑃(𝑧 <− #) — — — — — — — — — — — — — — — — C. Z - Score 50% → 0 65% → 0. 75% → 1 68% → 0. 80% → 1 85% → 0. 90% → 1 95% → 1. 99% → 2 99% → 2. 97% → 2 99% → 3. 92% → 1 98% → 2. 96% → 2. — — — — — — — — — — — — — — — — n<30 → t-table n>30 → z-table — — — — — — — — — — — — — — — —
The smaller the P-value, the more certain we can be 𝐻 0 is false. The larger the P-value, the more plausible 𝐻 0 becomes, but we can never be certain that 𝐻 0 is true A rule of thumb suggests suggest to reject 𝐻 0 whenever P ≤0. While this rule is convenient, it has no scientific basis. — — — — — — — — — — — — — — — — Let αbe any value between 0 and 1. Then if 𝑃 ≤ α ,
- The result of the test is said to be statistically significant at 100 αlevel.
- The null hypo. Is rejected at the 100α level.
- When reporting the result of hypo. Test, report the P-value rather than just comparing it to 5% or 1%. — — — — — — — — — — — — — — — —
— — — — — — — — — — — — — — — —
chi-square statistics formula
𝑋 2 = 𝑖=
𝑘 ∑
(𝑂𝑖−𝐸𝑖) 2 𝐸𝑖 — — — — — — — — — — — — — — — — To conduct a fixed-level test:
- Choose a number α, where 0 < α < 1 . This is called significance level, or the level, of the test.
- Compute the P-value in the usual way.
- If 𝑃 ≤ α, reject 𝐻 0. If 𝑃 > α, do not reject 𝐻 0. — — — — — — — — — — — — — — — —
— — — — — — — — — — — — — — — —
When conducting a fixed-level test at significance level α, there are 2 types of errors that can be made. These are: - Type I error: Reject 𝐻 0 when it is false. - Type II error: Fail to reject 𝐻 0 when it is false. The probability of type I error is never (> α).
A hypothesis test results in a type II error if 𝐻 0 is not rejected when it is false. The power of a test is the probability of rejecting 𝐻 0 when it is false. 𝑃𝑜𝑤𝑒𝑟 = 1 − 𝑃(𝑡𝑦𝑝𝑒 𝐼𝐼 𝑒𝑟𝑟𝑜𝑟) — — — — — — — — — — — — — — — — Example: Find the power of the 5% level test of 𝐻 0 : μ ≤ 80 vs. 𝐻 1 : μ > 80for the mean yield of the new process under the alternative μ = 82 assuming 𝑛 = 50 and σ = 5.
We have completed the first step of the solution which is to compute the rejection region. We will reject 𝐻 0 of 𝑋 ≥ 81. 16.
This presents the alternate and null distributions on the same plot. The z-score for the critical point of 81 under the alternate hypothesis is 𝑧 = (81−810 =− 1. 19. The area to the right is 0. This is the power. — — — — — — — — — — — — — — — — Example: A test has power 0 when μ = 15. True or false: a. The probability of rejecting 𝐻 0 when μ = 15 is 0. True b. The probability that 𝐻 0 when μ = 15is 0. False c. The probability of making a correct decision when μ = 15is 0. True d. The probability of making a correct decision when μ = 15is 0. False — — — — — — — — — — — — — — — — Example: If the sample size remains the same, and the level αincreases, then the power will increase. — — — — — — — — — — — — — — — — Example: If the level αremains the same, and the sample size increases, then the power will increase. — — — — — — — — — — — — — — — — Example: A power calculation has shown that if μ = 8, the power of a test 𝐻 0 : μ ≥ 10vs. 𝐻 1 : μ < 10 is 0. If instead μ = 7Which of the statements are true? Ans: The power of the test will be greater than 0.
Statistics Final Cheat Sheet | Camille Deguzman | Fall 2022 6, 6, 6, 6, 6, 6.
Example: The specification is that the flow rate be more than 5 gal/min. In an initial study, 8 runs were made. The avg. flow rate was 6 and the SD was 1 gpm. If the mean flow rate is found to meet the specification, the pump will be put into service.
𝑋 = 6. 5 𝑔𝑝𝑚 𝑠 = 1. 9 𝑔𝑝𝑚 μ = 5 𝑛 = 8 𝐻 0 : μ ≤ 5 𝐻 1 : μ > 5
- State the appropriate null and alternate hypotheses Null Hypothesis: 𝐻 0 : μ = 5
Alt. Hypothesis: 𝐻 1 : μ > 5← the flow rate
Ans: 𝐻 0 : μ ≤ 5 versus 𝐻𝑎: μ > 5 2. Find the P-value.
𝑡 = 𝑋−μ 𝑠 𝑛
⇒ 𝑡 = 6−5 1.
8
⇒ 𝑡 = 0.671751 ⇒ 2. 2329
(right-tailed test) 𝑛 − 1 ⇒ 8 − 1 = 7 P-value: 𝑃(𝑡 𝑤𝑖𝑡ℎ 𝑑𝑓 = 7 > 2. 232968783) ⇒ 0. 0305 (from student t-table) → (n<30) Ans: P-value = 0. 3. Should the pump be put into service? Explain. Assume the significant level a = 0. Since the P-value is less than 0, we reject the null hypothesis, concluding the pump should be put in service.
Example: State the most appropriate null hypothesis regarding the population mean. A. A new type of battery will be installed if it can be shown to have a mean lifetime greater than 8 yrs. Ans: This is expecting the mean life to be greater than 8 years. So the null hypothesis will be opposite to the assumption that is lesser than or equal to 8 years. 𝐻 0 : μ ≤ 8
B. A new material for manufacturing tires will be used if it can be shown that the mean lifetime of tires will be more than 60,000 miles. Ans: There is no change in null hypothesis which means the means cannot be more than 60,000 miles. 𝐻 0 : μ ≤ 60, 000
C. A quality control inspector will recalibrate a flow meter if the mean flow rate differs from 10 mL/s. Ans: The alternate hypothesis is showing the difference of 10mL/s 𝐻𝑎: μ ≠ 10 𝐻 0 : μ = 10
This is opposite to the expected assumption of no change.
Example: EPA standards require that the amount of lead in drinking water be less than 15 ppb. 12 samples of water from a particular source have the following concentration, in ppb. 11 13 11 14 15 8 12 8. 10 17 9 15. A hypothesis test will be performed to determine whether the water from this source meets the EPA standard.
𝑋 =
∑𝑋 𝑛 ⇒
- 12 = 12. 383
𝑠 =
∑(𝑥−𝑋) 2 𝑛−1 ⇒
(11−12) 2 +... 12−1 = 2. 932
- State the appropriate null and alternate hypotheses 𝐻 0 : μ = 15 𝐻 1 : μ < 15 (left tailed test) Test statistic 𝑡 = 𝑠/ 𝑛𝑋−μ ⇒ 2/ 1212−15 =− 3. 092 𝑑𝑓 = 𝑛 − 1 ⇒ 12 − 1 = 11
- Compute the P-value P-value: 𝑃(𝑡 <− 3. 092)→ P-value: 0. 0005125
- Can you conclude that the water from this source meets the EPA standard? Explain. (Assume 𝑥 = 0. 005) Since we assume the p-value < 𝑥 = 0. 005, we reject the null hypothesis. There is sufficient evidence that the water from this source meets the EPA standard.
At an assembly plant for light trucks, routine monitoring of the quality of welds yields
(1) Null and Alternate Hypothesis 𝐻 0 : The 2 variables are independent 𝐻 1 : The 2 variables are dependent
(2) Rejection Region Based on the info, the significance level is α = 0. 02, the d. Is (3 − 1) × (3 − 1) = 4, so rejection for this test R= {𝑋 2 : 𝑋 2 > 11. 668}.
(3) Test Statistics
𝑋 2 = 𝑖=
𝑛 ∑
(𝑂𝑖𝑗−𝐸𝑖𝑗) 2 𝐸𝑖𝑗 =. 001 +. 328 +.... = 5. 76
(4) Decision about null hypothesis Since it is observed that 𝑋 2 = 5. 76 ≤ 𝑋𝑐 2 = 11. 668it is concluded that the null hypothesis is not rejected.
(5) Conclusion It’s is concluded that the null hypothesis 𝐻 0 is nor rejected. There is not enough evidence to claim that 2 variables are dependent, at the α = 0. 02significance level. The corresponding p-value for the test is 𝑝 = 𝑃𝑟(𝑋 4 = 0.
2 > 5. 76)
2-sample (z) test: population SD
𝑧 =
(𝑋 1 −𝑋 2 )−(μ 1 −μ 2 )
σ 12 𝑛 1 +
σ 22 𝑛 2 2-sample (t) test: sample SD
𝑡 =
(𝑋 1 −𝑋 2 )−(μ 1 −μ 2 )
𝑠 12 𝑛 1 +
𝑠 22 𝑛 2
Stats Final Cheat Sheet
Course: Probability And Stat For Engr (ENGR 2090)
University: University of Georgia
