![EXAMPLE 4.7
Two independent normal populations with means 𝜇1 and 𝜇2 and unknown equal
variances 𝜎12 and 𝜎22. Two samples drawn one of each population with small
sample sizes 𝑛1 = 12 and 𝑛2 = 10, means 𝑥= 3.11 and 𝑦̅ = 2.04 and variances
𝑠1
2
= 0.594, 𝑠2
2
= 0.201, respectively. Find 95% confidence interval estimation of the
unknown difference 𝜇1 − 𝜇2 [use
𝑡0.975(20)= 2.086]](https://image.slidesharecdn.com/questionsfile-250118210234-0e7d120c/85/Questions-file-for-a-revision-on-probability-and-statistics-31-320.jpg)
Questions file for a revision on probability and statistics
- 2. PART 1
- 3. EXAMPLE 1.1 An electronics company manufactures resistors that have a mean resistance of 100 ohms and a standard deviation of 10 ohms. The distribution of resistance is normal. Find the probability that a random sample of n=25 resistors will have an average resistance less than 95 ohms.
- 4. EXAMPLE 1.2 If the serum iron values for healthy men are approximately normally distributed with a mean 120 and standard deviation 15 micrograms per 100 ml respectively, what is the probability that a random sample of 50 normal men will yield a mean between 115 and 125 micrograms per 100 ml?
- 5. EXAMPLE 1.3 Given a normally distributed population with a mean of 100 and a standard deviation of 20, find the following probabilities based on a sample of size 16:P(X̅≤ 110)
- 6. EXAMPLE 1.4 Suppose that the amount of cosmic radiation to which a person is exposed when flying by jet across in specific country is a random variable having a normal distribution with a mean of 4.5 m, for a random sample of size 12 of this persons having the standard deviation of 0.9 m. What is the probability that the mean of this sample greater than 5.3 m of cosmic radiation?
- 7. EXAMPLE 1.5 The outcome of a chemical process is measured. The outcome should be μ=500 g/ml (supposed population expectation). The outcome was measured in n=25 batches. What is the probability that the sample mean greater than 516 g/ml when the sample standard deviation is s=40 g/ml.
- 8. EXAMPLE 1.6 Suppose it has been established that for a certain type of client the average length of a home visit by a public health nurse is 45 minutes with a standard deviation of 15 minutes, and that for a second type of client the average home visit is 30 minutes long with a standard deviation of 20 minutes. If a nurse randomly visits 35 clients from the first and 40 from the second population. What is the probability that the average length of home visit will differ between the two groups by 20 or more minutes?
- 9. EXAMPLE 1.6 The study cited gives the following data on serum cholesterol levels in two ages males, we select a random sample of size 40 independently from each population with variances 66 and 49 respectively what is the probability that the difference between sample means will be more than 25? By assuming the two populations are normally distributed with means 190 and 180 respectively.
- 10. EXAMPLE 1.7 Given two normally distributed populations with μ1=45 and μ2=30. what is the probability that samples of size 𝑛1 = 15 and 𝑛2 = 10 and with variances of S1=15 ,S2 =20 will yield a value of 𝑥1 − 𝑥2 greater than or equal to 20 (we consider here the case whe𝝈1 2 = 𝝈2 2 =𝝈𝟐 )?
- 11. PART 2
- 12. EXAMPLE 1.1 Suppose that 𝑿𝟏, 𝑿𝟐, … , 𝑿𝒏 is a random sample from an exponential distribution with parameter 𝝀 in the form 𝒇 𝒙 = 𝟏 𝝀 𝒆 − 𝒙 𝝀, 𝒙 > 𝟎 , (𝝀 > 𝟎) Find moment estimator of 𝝀 ?
- 13. EXAMPLE 2.1 Suppose that 𝑿𝟏, 𝑿𝟐, … , 𝑿𝒏 is a random sample from a normal distribution with parameters μ and 𝝈𝟐 . Find moment estimators of μ and 𝝈𝟐 ?"
- 14. EXAMPLE 1.1 Suppose that 𝑿𝟏, 𝑿𝟐, … , 𝑿𝒏 is a random sample from an exponential distribution with parameter 𝝀 in the form 𝒇 𝒙 = 𝝀𝒙 𝒙! 𝒆−𝝀 , 𝒏 = 𝟎, 𝟏, … … , 𝝀 > 𝟎 Find moment estimator of 𝝀 ?
- 15. EXAMPLE 2.2 Suppose that 𝑿𝟏, 𝑿𝟐, … , 𝑿𝒏 is a random sample from an exponential distribution with parameter 𝝀 in the form 𝒇 𝒙 = 𝝀𝒙 𝒙! 𝒆−𝝀 , 𝒏 = 𝟎, 𝟏, … … , 𝝀 > 𝟎 Find maximum likelihood estimator of 𝝀 ?
- 16. EXAMPLE 2.3 Let 𝑿 be a normal random variable with parameter 𝝁, 𝝈𝟐 and pdf is given, for − ∞ < 𝝁 < ∞, 𝝈 > 𝟎 ,by 𝒇 𝒙; 𝝁, 𝝈𝟐 = ൞ 𝟏 𝝈 𝟐𝝅 𝒆 − 𝟏 𝟐𝝈𝟐 𝒙−𝝁 𝟐 , −∞ < 𝒙 < ∞ 𝟎 , 𝒐. 𝒘 Find the maximum likelihood estimators of 𝝁 and 𝝈𝟐 ?
- 17. PART 3
- 18. EXAMPLE 3.1 Suppose a researcher, interested in obtaining an estimate of the average level of some enzyme in a certain human population, takes a sample of 10 individuals, determines the level of the enzyme in each, and computes a sample mean of xˉ=22. Suppose further it is known that the variable of interest is approximately normally distributed with a variance of 45. Find 95% confidence interval for the population mean μ. 𝑧0.975= 1.96
- 19. EXAMPLE 3.2 The yield of a chemical process is being studied. From previous experience yield is known to be normally distributed and σ = 3. The past five days of plant operation have resulted in the following percent yields: 91.6, 88.75, 90.8, 89.95, and 91.3. Find the 95% confidence interval for the true mean yield 𝑧0.975= 1.96
- 20. EXAMPLE 3.3 Ten randomly selected people were asked how long they slept at night. The mean time was 7.1 hours, and the standard deviation was 0.78 hour. Find the 95% confidence interval of the mean time. Assume the variable is normally distributed. 𝑡0.975(9)=2.262
- 21. EXAMPLE 3.4 A research team is interested in the difference between serum uric acid levels in patients with and without Down’s syndrome. In a large hospital for the treatment of the mentally retarded, a sample of 12 individuals with Down’s syndrome yielded a mean of x̅1 = 4.5 mg /100 ml. In a general hospital a sample of 15 normal individuals of the same age and sex were found to have a mean value of x̅2 = 3.4 . If it is reasonable to assume that the two populations of values are normally distributed with variances equal to 1 and 1.5, find the 95% confidence interval for μ1 − μ2 .
- 22. EXAMPLE 3.5 A diversity index was measured in two locations monthly. The measurements lasted one year (n1 = 12) in location 1 and for ten months (n2 = 10) in location 2. The obtained statistics were x̄1 = 3.11; s1 = 0.771; x̄2 = 2.04 and s2 = 0.448. Find the 95% confidence interval for μ1 - μ2. Assume normally distributed populations with equal variances 𝑡0.975(20)=1.725
- 23. MID TERM 2023
- 24. EXAMPLE 4.1 Given 𝜇 = 50, 𝜎 = 16, 𝑛 = 64 and 𝑃(0≤𝑍≤2.5)=0.4938, then 𝑃(45≤𝑋̅≤55)= a) 0.8796 b)b) 0.9876 c) c) 0.8574 d) d) non of the above
- 25. EXAMPLE 4.2 Given 𝜇 = 500, 𝑠=40, 𝑛 =25 and 𝑃(𝑇516)= a) 0.1043 b) b) 0.1142 c) c) 0.025 d) d) non of the above
- 26. EXAMPLE 4.3 The brightness of a television picture tube can be evaluated by measuring the amount of current required to achieve a particular brightness level. A sample of 10 tubes results in 𝑋̅=317.2 and 𝑆=15.7. Then 99% confidence interval on mean current required is (use 𝑡0.995,9=3.25) a) (301.064,333.336) b) (256.743,260.342) c) (243.674,248.236) d) non of the above
- 27. EXAMPLE 4.4 The sample standard deviation, denoted by 𝑆 (random variable) or 𝑠 (realized value), is the positive square root of the population variance . a) True b) b) False
- 28. EXAMPLE 4.5 If the expectation of the random variable of the population distribution is 𝜇 and its variance is 𝜎2, then the expectation of the sample mean is 𝜇. a) True b) b) False
- 29. EXAMPLE 4.5 The rule that tells us how to compute a single numerical value that is used to estimate the corresponding population parameter is referred to as point estimate. a) True b) False
- 30. EXAMPLE 4.6 Estimation: involves estimating a sample parameter. a) True b) b) False
- 31. EXAMPLE 4.7 Two independent normal populations with means 𝜇1 and 𝜇2 and unknown equal variances 𝜎12 and 𝜎22. Two samples drawn one of each population with small sample sizes 𝑛1 = 12 and 𝑛2 = 10, means 𝑥= 3.11 and 𝑦̅ = 2.04 and variances 𝑠1 2 = 0.594, 𝑠2 2 = 0.201, respectively. Find 95% confidence interval estimation of the unknown difference 𝜇1 − 𝜇2 [use 𝑡0.975(20)= 2.086]
- 32. THANK YOU By: Eng/ Youssef Zayed