# A random sample of 100 observations from a population has a mean of 60 and a population standard deviation of 18.

- A random sample of 100 observations from a population has a mean of 60 and a population standard deviation of 18.

- What is the point estimate of the population mean? …..60…………………………….

The point estimate of the population mean is 60. The point estimate of the population is equivalent to the point estimate of the sample.

- What is the point estimate of the population variance? ………………………..

The point estimate of the population variance is equivalent to the sample variance. In this case the variance of the population is 18^{2 }= 324

- What is the point estimate of the standard deviation of the population? ……………………

The point estimate of the standard deviation of the population is 18.

- What is the point estimate of the standard deviation of the sample mean? ……………………

The point estimate of the standard deviation of the sample mean is 18.

- Small manufacturers of medical devices have an average production rate of 2200 nationwide. A random sample of 20 manufacturers shows the following production records:

Output: 2100, 2125, 2680, 2200, 2160, 2155, 2055, 2023, 1980, 1860, 1950, 2055, 2160, 2450, 2800, 2500, 1950, 2050, 1980, 2000.

- Write a hypothesis test comparing the average output of sampled manufacturers to national average.

Total = 2125 + 2680 + 2200 + 2160 + 2155 + 2055 + 2023 + 1980 + 1860 + 1950 + 2055 + 2160 + 2450 + 2800 + 2500 + 1950 + 2050 + 1980 + 2000

= 43233

Mean = 43233/20

= 2161.65

The sample mean is 2161.65 while the population mean is 2200. Therefore, the average output of the sampled manufacturers is different from the national average.

- Do a hypothesis test at the .05 level of significance.

Population estimate = sample estimate ± Z-score table value * Standard error of Estimate

Standard error of estimate = standard deviation / √sample number

= 245.7249/4.472= 54.95

Population estimate=2161.65 ± 1.96*54.95

= 2053.948 – 2269.352

The hypothesis test is significant at .05 level of significance. From Table 1 it is evident that the lower and upper boundaries of the sample lie within the same range as the population mean. We fail to reject the hypothesis since the sample mean is within the population estimate.

- Is the sampled manufacturers’ output different from the national average? If yes, how?

The sampled manufacturers’ output is not different from the national average because the sample mean is within the estimated population mean estimate.

**3**.

- Define the following concepts: bias, efficiency, and consistency.

Bias- when used in statistics, it refers to the possibility of an item in a population having a higher probability of being chosen than the other items.

Efficiency-this refers to the stability of a statistic from one sample to another such that there are minimum or no variations of the statistic in the two samples.

Consistency-this concept refers to the sample statistic becoming near in value to the population statistic. The larger the sample size, the near the sample estimate will be to the corresponding parameter.

- What is central limit theorem? What is the importance of the central limit theorem in statistical analysis?

The central limit theorem (CLT) states that as the sample size grows bigger and bigger, the sample statistic approaches the population statistic. This theorem is important in statistical analysis since it proposes that a big sample is better than a small one in reducing errors. Therefore, sample statistics from a larger sample is more reliable than that from a smaller sample.

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**4****.**The manufacturer of YUGO cars claims that their cars can be driven 160000 miles or more with no major repairs. Cheap Rental Inc. has purchased 50 cars from YUGO. Cheap Rental believes that their data does not support YUGO’s claim of 160000 mile. The company’s data shows a no-repair mean mileage of 145000 with a standard deviation of 10000 miles.

- Write a hypothesis test on the mean mileage of the YUGO cars that the mean mileage is not different from 160000miles, as claimed by the company.

Population estimate = sample estimate ± Z-score table value * Standard error of Estimate

Standard error of estimate = standard deviation / √sample number

=10,000/ 7.0711= 1414.2071

Population estimate = 145,000 ± 1.96*1414.2071

=147,771.846 – 142,228.154

The hypothesis by the YUGO Company is rejected because the average speed of the population lies beyond the sample limits. Therefore, the claim that the cars for YUGO Company can be driven 160,000 miles or more with no major repairs is false.

- Write a 95% confidence interval for the mean mileage based on Cheap Rental data.

A 95 % confidence interval for the mean mileage based on Cheap Rental data is 147,771.846 miles to 142,228.154 miles.

- Do a hypothesis test at the .05 significance level that the mean mileage is not different from 160000miles.

The average miles that the YUGO Company provide (160,000) is lying beyond the sample limits (147,771.846 – 142,228.154) indicating that the hypothesis is rejected. Therefore, the mean mileage provided by YUGO is different from 160000 miles.

- Do a hypothesis test at the .05 significance level that the mean mileage of the YUGO cars are less than 160000, as experienced by the Cheap Rental Inc.

The hypothesis that the mean mileage of the YUGO cars is less than 160,000 is accepted.

- Is the Cheap Rental Inc’s experience different from that claimed by the manufacturer at the .05 significance level? Interpret your test results.

The experience obtained by Cheap Rental Inc is different from that claimed by the manufacturer at the .05 significance level. The average miles provided by YUGO lie beyond the estimated range that was found out by Cheap Rental data.

**5****.**The manufacturer of YUGO cars claims that their cars can be driven 160000 miles or more with no major repairs. Cheap Rental Inc. has purchased 25 cars from YUGO. Cheap Rental believes that their data does not support YUGO’s claim of 160000 mile. The company’s data shows a no-repair mean mileage of 135000 with a standard deviation of 10000 miles.

- Write a hypothesis test on the mean mileage of the YUGO cars that the mean mileage is not different from 160000miles, as claimed by the company.
- Write a 95% confidence interval for the mean mileage based on Cheap Rental data.
- Do a hypothesis test at the .05 significance level that the mean mileage is not different from 160000miles.
- Do a hypothesis test at the .05 significance level that the mean mileage of the YUGO cars are less than 160000, as experienced by the Cheap Rental Inc.
- Is the Cheap Rental Inc’s experience different from that claimed by the manufacturer at the .05 significance level? Interpret your test results.

- A sample of 40 observations from a population with a known variance of 242 has a mean of 94. Write a 95% confidence interval for the population mean.

Standard error of estimate = standard deviation / √sample number

= 15.5563 / 6.3246

= 2.4597

Population estimate = 94 ± 1.96* 2.4597

= 94 ± 4.8209

= 89.1791 – 98.8209

- A sample of 40 observations from a population has a mean of 94 and standard deviation of 68. Write a 95% confidence interval for the population mean.

Standard error of estimate = standard deviation / √sample number

= 68 / 6.3246

= 10.7517

Population estimate = 94 ± 1.96*10.7517

= 94 ± 21.0733

= 72.9267 – 105.9267

- A sample of 40 observations from a population has a mean of 94. The standard deviation of the sample mean is 3.4. Write a 95% confidence interval for the population mean.

Standard error of estimate = standard deviation / √sample number

= 3.4 / 6.3246

= 0.5376

Population estimate = 94 ± 1.96*0.5376

= 94 ± 1.0537

= 92.9463 – 95.0537

- A sample of 25 observations from a population has a mean of 94. The standard deviation of the sample mean is 3.4. Write a 95% confidence interval for the population mean.

Standard error of estimate = standard deviation / √sample number

= 3.4 / 5

= 0.68

Population estimate = 94 ± 1.96*0.68

= 94 ± 1.3328

= 92.6672 – 95.3328

- A sample of 25 observations from a population with a known variance of 140 has a mean of 94. Write a 95% confidence interval for the population mean.

Standard error of estimate = standard deviation / √sample number

= 11.8322 / 5

= 2.3664

Population estimate = 94 ± 1.96*2.3664

= 94 ± 4.6382

= 89.3617935 – 98.6382

- National Geographic claims that the average weight of turtles living in the east cost of Maui Island is 58 pounds. A researcher’s sample of 60 turtles from the location has a mean of 72 and variance of 186 pounds. Does researcher’s sample supports the hypothesis that the population mean is no different from 58 pounds (% = .05)?

Standard error of estimate = standard deviation / √sample number

= 13.6382 / 7.7460

= 1.7607

Population estimate = 58 ± 1.96*1.7607

= 58 ± 3.4509

= 54.5491 – 61.4509

The researcher’s sample does not support the hypothesis that the population is no different from 58 pounds. The hypothesis is rejected because the sample mean lies beyond the population estimates.

- National Geographic claims that the average weight of turtles living in the east cost of Maui Island is 58 pounds. A researcher’s sample of 20 turtles from the location has a mean of 72 and variance of 186 pounds. Does researcher’s sample supports the hypothesis that the population mean is no different from 58 pounds (% = .05)?

Standard error of estimate = standard deviation / √sample number

= 13.6382/ 4.4721

= 3.0496

Population estimate = 58 ± 1.96*3.0496

= 58 ± 5.9773

= 52.0227 – 63.9773

The researcher’s sample does not support the hypothesis that the population is no different from 58 pounds. The hypothesis is rejected because the sample mean lies beyond the population estimates.

- National Geographic claims that the average weight of turtles living in the east cost of Maui Island is 58 pounds. A researcher’s sample of 60 turtles from the location has a mean of 72 and variance of 186 pounds. Does researcher’s sample supports the hypothesis that the population mean is greater than 58 pounds (% = .05)?

Standard error of estimate = standard deviation / √sample number

= 13.6382 / 7.7460

= 1.7607

Population estimate = 58 ± 1.96*1.7607

= 58 ± 3.4509

= 54.5491 – 61.4509

The researcher’s sample supports the hypothesis that the population mean is greater than 58 pounds. The population estimate lies within the provided mean of 58 and this helps us support/accept the hypothesis.

- National Geographic claims that the average weight of turtles living in the east cost of Maui Island is 58 pounds. A researcher’s sample of 20 turtles from the location has a mean of 72 and variance of 186 pounds. Does researcher’s sample supports the hypothesis that the population mean is greater than 58 pounds (% = .05)?

Standard error of estimate = standard deviation / √sample number

= 13.6382/ 4.4721

= 3.0496

Population estimate = 58 ± 1.96*3.0496

= 58 ± 5.9773

= 52.0227 – 63.9773

The researcher’s sample supports the hypothesis that the population mean is greater than 58 pounds. The population estimate lies within the provided mean of 58 and this helps us support/accept the hypothesis.

** **

- A manufacturer believes that the average rate of productivity in its manufacturing plant is 45 units per day. A sample of productivity for two weeks shows the following rate a day.

P |
P-P |
(P-P)^{2} |
f(P-P)^{2} |

36 | -8 | 64 | 64 |

48 | 4 | 16 | 16 |

42 | -2 | 8 | 8 |

59 | 15 | 225 | 225 |

52 | 8 | 64 | 64 |

40 | -4 | 16 | 16 |

48 | 4 | 16 | 16 |

53 | 9 | 81 | 81 |

52 | 8 | 64 | 64 |

50 | 6 | 36 | 36 |

Σ(P-P)^{2} = 590 |
ΣF(P-P)^{2 }=590 |

Variance = F(P-*P*)^{2 }/ n

=590/ 10

^{ }= 59

Standard deviation = √59

= 7.6811

^{ }

^{ }

- Write a 95% confidence interval for the average productivity in the manufacturing plant.

Standard error of estimate = standard deviation / √sample number

= 7.6811/ 3.1623

= 2.4284

Population estimate = 44 ± 1.96*2.4284

= 44 ± 4.7597

= 39.2403 – 48.7597

- b. Is the manufacturer right in claiming that the average productivity is 45 units a day?

Yes. The claim that the average productivity is 45 units a day is correct because the mean (45) lies within the population estimate at 95% CI.

- c. Test the hypothesis that the average productivity is more than 45 units a day.

Standard error of estimate = standard deviation / √sample number

= 7.6811/ 3.1623

= 2.4284

Population estimate = 44 ± 1.96*2.4284

= 44 ± 4.7597

= 39.2403 – 48.7597

The hypothesis that the average productivity is more than 45 units a day is correct. The boundaries in the population estimate are within the average of 45 units and hence the claim is correct.

**1**Bozo T. Clown manages a car dealership where the average daily sales for the past ten years has been 25 cars a day. Bozo is concerned with the current downturn in the car sales and would like to compare the current car sales to daily average for the past ten years. He has the following daily car sales for the past two weeks.

Daily Sales |
P-P |
(P-P)^{2} |
f(P-P)^{2} |

18 | -2 | 4 | 4 |

22 | 2 | 4 | 4 |

26 | 6 | 36 | 36 |

28 | 8 | 64 | 64 |

17 | -3 | 9 | 9 |

16 | -4 | 16 | 16 |

20 | 0 | 0 | 0 |

15 | -5 | 25 | 25 |

17 | -3 | 9 | 9 |

21 | 1 | 1 | 1 |

Σ(P-P)^{2} = 168 |
ΣF(P-P)^{2} = 168 |

Variance = F(P-*P*)^{2 }/ n

=168/ 10

= 16.8

Standard deviation = 4.0988

- Write a 95% confidence interval for the average car sales for the past two weeks.

Standard error of estimate = standard deviation / √sample number

= 4.0988/ 3.1623

= 1.2961

Population estimate = 20 ± 1.96*1.2961

= 20 ± 2.5404

= 17.4596 – 22.5404

- b. Test the hypothesis that the car sales in the recent weeks are no different from the average car sales of the past ten years.

Car sales in the recent weeks are different from the average car sales of the past ten years. This is because the average car sales in the past ten years has been 25 but the population estimate ranges from 17.4596 to 22.5404.

- c. Test the hypothesis that the average car sales in recent weeks is less than 25 units per day.

The hypothesis that the average car sales in the recent weeks is not less than 25 units per day is rejected. This is because the average car sales in the past ten years has been 25 but the population estimate ranges from 17.4596 to 22.5404.

- 12. National average SAT score of students accepted to business programs is 1240. A sample of 20 business students in a university has an average SAT score of 1125 and a standard deviation of 116.

- a. Test the hypothesis that the average SAT score in this university is no different from the national average.

- b. Test the hypothesis that the average SAT score in this university is less than the national average.

- 13. Ted Bizarre is managing two plants in a manufacturing industry. A sample of 45 observations from plant one has a mean production of 110 units a day with a population variance of 144 units. A sample of 60 observations from plant two has a mean production of 132 units a day and a population variance of 169. Ted believes that the two plants have the same productivity.

- a. Test the hypothesis that the mean production in two plants are the same.
- b. Test the hypothesis that plant two has a higher productivity than plant one.

- 14. Bozo T. Clown divides his daily working time between standing comic and clowning. His average earning for the past 50 days from standing comic is $120 with a population variance of $165. His average earning from clowning for the same time period is $95 with a population variance of $142. Bozo believe that, on average, his earning from the two jobs are statistically the same.

- a. Test the hypothesis that Bozo’s earning in two jobs are the same.
- b. Test the hypothesis that Bozo earns more income as standing comic than as a clown.

- 15. A researcher believes that the average per capita income in two countries of Banana Republic ans Sahara Republic is the same. A sample of per capita income for the past 15 years for the Banana Republic shows that the country has an average per capita income of $1250, with a sample variance of $1800. A sample of per capita income for the past 10 years for the Sahara Republic shows that the country has an average per capita income of $1010, with a sample variance of $1890.

- a. Test the hypothesis that the average per capita income in two countries are the same.
- b. Test the hypothesis that the average per capita income in Banana Republic is higher than the one in Sahara Republic.

- 16. A manufacturer believes that the average rate of productivity in its two manufacturing plant are the same. A sample of productivity for two weeks in two plants shows the following rate a day.

P1 |
P2 |
||

36 | 38 | ||

48 | 52 | ||

42 | 38 | ||

59 | 62 | ||

52 | 42 | ||

40 | 43 | ||

48 | 49 | ||

53 | 55 | ||

52 | 53 | ||

50 | 58 | ||

- a. Do a hypothesis test that the average productivity in two plants are the same.
- b. Do a hypothesis test that the average productivity in plant two is greater than plant 1.

- 17. Bozo T. Clown manages two car dealership where the average daily sales for the past ten years is given below. Given that the costs in both dealerships are the same, Bozo wonders whether the average sale in two dealership are the same.

- a. Given the data, test the hypothesis that the average daily sales in both dealerships are the same.
- b. Test the hypothesis that the average daily sales in dealership 2 is less than the average sales in dealership 1.

Dealership 1 |
Dealership 2 |
||

18 | 22 | ||

22 | 19 | ||

26 | 12 | ||

28 | 35 | ||

17 | 13 | ||

16 | 15 | ||

20 | 9 | ||

15 | 10 | ||

17 | 14 | ||

21 | 11 | ||

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