stat
2poor
assignment_4.xlsx7.9
| 7.9 Solution | |||||||
| a) | A stratified sampling should be taken to ensure that samples from each of the four strata will be proportionately included. | ||||||
| b) | Here the number of sales invoices in each of the four strata differs significantly. | ||||||
| So the stratified sampling should be done by selecting samples proportional to the stratum size. | |||||||
| For example, | |||||||
| from stratum 1 select 500(50/5000) = | 5 | invoices, | |||||
| from stratum 2 select 500(500/5000) = | 50 | invoices, | |||||
| from stratum 3 select 500(1000/5000)= | 100 | invoices, and | |||||
| from stratum 4 select 500(3450/5000)= | 345 | invoices to get a representative sample form each stratum. | |||||
| c) | In simple random sampling, each unit in the population has an equal chance of being selected into the sample. | ||||||
| In stratified sampling, unlike the simple random sampling, proportionate representation across the entire population | |||||||
| is ensured by selecting samples from each stratum. That’s why the sampling in (a) is not simple random sampling. | |||||||
7.21
| 7.21 Solution | |||||||||||||||
| Let Xbar denote the sample mean. | |||||||||||||||
| Then, Xbar is distributed as a normal distribution with mean 8 and standard deviation 2/√25 = 2/5 = 0.4. | |||||||||||||||
| Therefore, Z = (Xbar - 8)/0.4 is distributed as a standard normal distribution. | |||||||||||||||
| a) | The probability the sample mean is between 7.8 and 8.2 minutes is given by, | xbar | z | P(Z < z) | |||||||||||
| P(7.8 < Xbar < 8.2) = P[(7.8 - 8)/0.4 < Z < (8.2 - 8)/0.4] | 7.8 | -0.5 | 0.3085 | ||||||||||||
| = | P(-0.5 < Z < 0.5) | 8.2 | 0.5 | 0.6915 | |||||||||||
| = | P(Z < 0.5) - P(Z < -0.5) | 7.5 | -1.25 | 0.1056 | |||||||||||
| = | 0.3829 | 8 | 0 | 0.5000 | |||||||||||
| b) | The probability that the sample mean is between 7.5 and 8 minutes is given by, | ||||||||||||||
| P(7.5 < Xbar < 8) = P[(7.5 - 8)/0.4 < Z < (8 - 8)/0.4] | |||||||||||||||
| = | P(-1.25 < Z < 0) | ||||||||||||||
| = | P(Z < 0) - P(Z < -1.25) | ||||||||||||||
| = | 0.3944 | ||||||||||||||
| c) | If n = 100, Xbar is distributed as a normal distribution with mean 8 and standard deviation 2/√100 = 2/10 = 0.2. | ||||||||||||||
| Therefore, Z = (Xbar - 8)/0.2 is distributed as a standard normal distribution. | xbar | z | P(Z < z) | ||||||||||||
| Now, the probability the sample mean is between 7.8 and 8.2 minutes is given by, | 7.8 | -1 | 0.1587 | ||||||||||||
| P(7.8 < Xbar < 8.2) = P[(7.8 - 8)/0.2 < Z < (8.2 - 8)/0.2] | 8.2 | 1 | 0.8413 | ||||||||||||
| = | P(-1 < Z < 1) | ||||||||||||||
| = | P(Z < 1) - P(Z < -1) | ||||||||||||||
| = | 0.6827 | ||||||||||||||
| d) | When the sample size increases from n = 25 to n = 100, the standard error of the sample mean reduces from 0.4 to 0.2. | ||||||||||||||
| Therefore more samples will be closer to the distribution mean as the sample size increases from n = 25 to n = 100. | |||||||||||||||
| Hence the likelihood that the sample mean will fall within 0.2 minutes of the mean is much higher for samples of size 100 than for samples of size 25. |
7.23
| 7.23 Solution | |||
| a) | It is given that, n = 64, x = 48 | ||
| The sample proportion, p, of “successful” people is given by, | |||
| p = 48/64 = | 0.75 | ||
| b) | It is given that, the population proportion, π = 0.70 | ||
| Now, the standard error of the proportion is given by, | |||
| √{π(1-π)/n} = √{0.70(1-0.70)/64} | |||
| = | 0.0573 | ||
7.25
| 7.25 Solution | |||
| a) | It is given that, n = 40, | ||
| x = Number of Y's in the data = 14 | |||
| The sample proportion, p, of college students who own shares of stock is given by, | |||
| p = 14/40 = | 0.35 | ||
| b) | It is given that, the population proportion, π = 0.30 | ||
| Now, the standard error of the proportion is given by, | |||
| √{π(1-π)/n} = √{0.30(1-0.30)/40} | |||
| = | 0.0725 | ||
7.29
| 7.29 Solution | |||||||||||||||||
| It is given that, n = 100, π = 0.25 | n | 100 | |||||||||||||||
| Let p denote the sample proportion. | π | 0.25 | |||||||||||||||
| Thus, the sample proportion is normaly distributed with mean π =0.25 and standard deviation √{π(1-π)/n} = 0.0433 | √{π(1-π)/n} | 0.0433 | |||||||||||||||
| Therefore, Z = (p - 0.25)/0.0433 is distributed as Standard Normal. | |||||||||||||||||
| a) | The probability that 25% or fewer male employees will indicate that they have to pick up the slack for moms working flextime is given by, | ||||||||||||||||
| P(p ≤ 0.25) = P[Z ≤ (0.25 - 0.25)/0.0433] | |||||||||||||||||
| = | P(Z ≤ 0) | p | z | P(Z < z) | |||||||||||||
| = | 0.5 | 0.25 | 0 | 0.5000 | |||||||||||||
| 0.2 | -1.1547005384 | 0.1241 | |||||||||||||||
| b) | The probability that 20% or fewer will indicate that they have to pick up the slack for moms working flextime is given by, | ||||||||||||||||
| P(p ≤ 0.20) = P[Z ≤ (0.20 - 0.25)/0.0433] | |||||||||||||||||
| = | P(Z ≤ -1.1547) | ||||||||||||||||
| = | 0.1241 | n | 500 | ||||||||||||||
| π | 0.25 | ||||||||||||||||
| c) | If n =500, the sample proportion is normaly distributed with mean π =0.25 and standard deviation √{π(1-π)/n} = 0.0194 | √{π(1-π)/n} | 0.0194 | ||||||||||||||
| Therefore, Z = (p - 0.25)/0.0194 is distributed as Standard Normal. | |||||||||||||||||
| Now, the probability that 25% or fewer male employees will indicate that they have to pick up the slack for moms working flextime is given by, | |||||||||||||||||
| P(p ≤ 0.25) = P[Z ≤ (0.25 - 0.25)/0.0194] | |||||||||||||||||
| = | P(Z ≤ 0) | p | z | P(Z < z) | |||||||||||||
| = | 0.5 | 0.25 | 0 | 0.5000 | |||||||||||||
| The probability that 20% or fewer will indicate that they have to pick up the slack for moms working flextime is given by, | 0.2 | -2.5820 | 0.0049 | ||||||||||||||
| P(p ≤ 0.20) = P[Z ≤ (0.20 - 0.25)/0.0194] | |||||||||||||||||
| = | P(Z ≤ -2.5820) | ||||||||||||||||
| = | 0.0049 |
