## STATISTICS Homework 8.1 Assignment 2016

Question

[3 marks]- 1 of 3 ID: MST.FET.SD.SDM.01.0020A

A company provides tuition for mathematics students across the country. It provides two levels of tuition: tuition for general-level students and tuition for advanced-level students. All students who receive tuition at this company must first complete a test, to assess their initial comprehension of mathematics.

For general-level students, the population mean score in this test is 49.20, with a population standard deviation of 11.81. For advanced students, the population mean score in the test is 81.48, with a population standard deviation of 15.56.

A statistician wants to study the results of a random selection of general students and a random selection of advanced students. The statistician intends to draw a sample of 50 general-level students and a sample of 200 advanced-level students.

a)From the following list, select the statements that are true. For each statement, assume that all other measures remain constant in that instance.

The larger a population mean, the greater the variation in sample means for samples from that population.

The smaller a population mean, the greater the variation in sample means for samples from that population.

The larger a population variance, the greater the variation in sample means for samples from that population.

The smaller a population variance, the greater the variation in sample means for samples from that population.

The larger the sample size for samples from a population, the greater the variation in sample means.

The smaller the sample size for samples from a population, the greater the variation in sample means.

b)For the samples described in the scenario above, the sample mean scores for samples of general-level students will have a HIGHER OR LOWER level of variation than the variation in sample mean scores for samples of advanced-level students.

[1 mark]- 2 of 3 ID: MST.FET.SD.SDM.02.0030A

Loony Park has a new adult ride that holds 15 people. The maximum weight that the ride can carry is 2,850.75 lb. The weight of adults has a mean of 175.28 lb and a standard deviation of 21.58 lb.

Calculate the probability that a random group of 15 people will overload the maximum weight that the ride can carry. You may find thisstandard normal tableuseful. Give your answer as a decimal to 4 decimal places. (Hint).

The probability of 15 people exceeding a maximum weight of 2,850.75 lb is equivalent to the mean weight of the individuals exceeding 190.05 lb.

p =

[1 mark]- 3 of 3 ID: MST.FET.SD.SDM.03.0010A

A company records the number of angry calls made to a particular department every working day. It then takes the average of these numbers over 10 working days. This average is based in the assumption that the distribution of the number of angry calls on a particular day is unaffected by the day on which the call is received.

The probability that this average is greater than 29 is 5%. Calculate the mean number of angry calls on any particular day if you know that the standard deviation of the number of angry calls per day is 14. You may find thisstandard normal tableuseful. Give your answer to 3 decimal places.

Mean number of angry calls per day =

ATTACHMENT 2

[3 marks]- 1 of 2 ID: MST.FET.SD.SDI.01.0040A

It is known that, across the adult population, the average number of words read per minute is 266 and the population standard deviation is 20. You work for a company with many employees across the country, and you would like to know what the average words per minute is for company employees.

Your colleague proposes that he thinks the average will be the same as the average in the general population: 266. You decide to survey 100 employees, and find that in this sample the average words per minute is 269.84.

a)Based on the assumption that the population mean is 266 and that the population standard deviation is 20, calculate the z-score of the sample mean in your survey. Give your answer as a decimal to 2 decimal places.

z =

b)Determine the proportion of the standard normal distribution that lies to the right of this z-score. That is, determine the area to the right of this z-score in the standard normal distribution. You may find thisstandard normal tableuseful. Give your answer as a percentage to 2 decimal places.

Area = %

c)Denote by x% the percentage proportion you calculated in part b). Consider the following five potential conclusions:

A: There is a chance of x% that your friend is correct, that the true population mean is 266.

B: If your colleague is correct and the true population mean is 266, then x% of all samples will produce a sample mean of 269.84 or lower.

C: If your colleague is correct and the true population mean is 266, then x% of all samples will produce a sample mean of 269.84 or higher.

D: There is a chance of x% that the true population mean is 266 or lower.

E: There is a chance of x% that the true population mean is 266 or higher.

Select the statement that can be inferred from your findings:

A

B

C

D

E

[3 marks]- 2 of 2 ID: MST.FET.SD.SDI.02.0010A

You work on a traffic management team for the city. There has been a proposal to add a new lane to a road near a busy intersection. As part of the research into whether or not to build the new lane, the team would like to know how many cars, on average, pass through this intersection at peak hour each day (5 pm to 6 pm). For the purposes of this research, the team is assuming that the standard deviation in the number of cars passing through each hour is 27.

You randomly select 49 days on which to monitor the traffic going through this intersection.

a)Complete the following statement by filling in the correct number. Give your answer to the nearest whole number of cars.

Whatever the true population mean is, for 90% of all samples of size n = 49, the sample mean will be within cars of the population mean.

b)Over the 49 days, you find that an average of 868 cars pass through the intersection. Therefore you can be 90% confident that the true population mean is somewhere between and cars.

ATTACHMENT 3

[1 mark]- 1 of 2 ID: MST.FET.HT.PHT.01.0030A

The proportion of people in the town of Buysville that have credit card debt has consistently been 0.4. Recent consumer trends in Buysville suggest that this figure may have risen. A hypothesis test is conducted in order to find out. Select the correct null and alternative hypotheses:

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H0: = 0.4

HA: ? 0.4

H0: = 0.4

HA: > 0.4

H0: = 0.4

HA: ? 0.4

H0: = 0.4

HA: = 0.4

H0: = 0.4

HA: ? 0.4

H0: = 0.4

HA: < 0.4
[1 mark]- 2 of 2 ID: MST.FET.HT.PHT.02.0040A
Kathy is studying the salinity (level of salt) in a sea near her city. The sea has historically had an average level of salinity of ? = 3.8%.
Kathy would like to test if recent climate change has increased the salinity of the sea. Kathy intends to conduct a hypothesis test with the following hypotheses:
H0: ? = 3.8
HA: ? > 3.8

Before Kathy conducts this test, her friend Craig tells her that he strongly suspects the salinity won’t have increased. He thinks a lot of evidence that has already been collected and studied suggests that the salinity has instead decreased from its old level of 3.8%. Craig believes that Kathy should conduct the following hypothesis test instead:

H0: ? = 3.8

HA: ? < 3.8
Craig believes that:
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instead of conducting a one-tail test, Kathy should conduct a two-tail test.

instead of conducting a two-tail test, Kathy should conduct a one-tail test.

instead of conducting the particular two-tail test that was proposed, Kathy should conduct a different kind of two-tail test.

instead of conducting the particular one-tail test that was proposed, Kathy should conduct a different kind of one-tail test

ATTACHMENT 4

[2 marks]- 1 of 3 ID: MST.FET.HT.CMHT.01.0020A

A hypothesis test is to be performed in order to test the proportion of people in a population that have some characteristic of interest. Select all of the pieces of information that are needed in order to calculate the test statistic for the hypothesis test:

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the size of the sample selected

the actual population proportion

the level of significance used

the sample proportion calculated

the characteristic of interest

the proposed population proportion

[1 mark]- 2 of 3 ID: MST.FET.HT.CMHT.03.0020A

You may find thestandard normal tableuseful in answering this question. Consider the hypothesis test with the following null and alternative hypotheses:

H0: = 0.4

HA: > 0.4

If the level of significance used is ? = 0.025, the region of rejection is the set of numbers:

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greater than 1.96

whose absolute value is greater than 2.24

whose absolute value is less than 1.96

whose absolute value is greater than 1.96

less than -1.96

greater than 2.24

whose absolute value is less than 2.24

less than -2.24

[2 marks]- 3 of 3 ID: MST.FET.HT.CMHT.02.0020A

Painadol is a well trusted, popular pain-killer used by millions throughout the country. Recently an epidemic of headaches has hit the nation and the manufacturers of Painadol have created a new, stronger painkiller called Painadene. The manufacturers are interested in testing whether the speed of pain relief is different with Painadene. It is known that the mean time taken by Painadol tablets to relieve pain is 33 minutes and the standard deviation is 6 minutes.

The manufacturers would like to construct a hypothesis test for the mean time (?) taken for Painadene tablets to relieve pain assuming the same population standard deviation (?) as the Painadol tablets. A random sample of 32 people with headaches tried the new Painadene tablets and the mean time taken to relieve the pain was calculated as 31.85 minutes. The hypotheses that will be used by the manufacturers are H0: ? = 33 and HA: ? ? 33.

You may find thisstandard normal tableuseful throughout the following questions.

a)Calculate the test statistic (z) that corresponds to the sample and hypotheses. Give your answer as a decimal to 3 decimal places.

z =

b)Using the test statistic for Painadol’s hypothesis test and a level of significance of ? = 0.1, Painadol should ACCEPT/ REJECT/ NOT REJECT the null hypothesis.

ATTACHMENT 5

[2 marks]- 1 of 3 ID: MST.FET.HT.CHT.03.0010A

Heather is a student in a class that has been asked to conduct a hypothesis test for a population proportion. The null and alternative hypotheses are:

H0: = 0.5

HA: ? 0.5

Heather is given a sample and calculates the test statistic. Using a level of significance of ? = 0.05, she rejects the null hypothesis. Heather’s friends, Alan and Bill, also use the sample to calculate a test statistic. However, they differ in the level of significance they use. Alan uses a level of significance of ? = 0.1, while Bill uses ? = 0.01. All three students perform their calculations correctly.

For each of Heather’s friends, select whether or not they reject the null hypothesis, or whether it is impossible to tell from the information given.

[1 mark]- 2 of 3 ID: MST.FET.HT.CHT.01.0020A

April is a statistics student who has extensive computer programming skills. She programs a random-number generator that produces numbers that follow the normal distribution with mean ? = 11 and standard deviation ? = 4.

April’s friend Augustus does not know that the population mean is 11. He knows that April’s favorite number is 18, and decides to conduct a hypothesis test with the following hypotheses:

H0: ? = 18

HA: ? ? 18

The result of the test is that Augustus rejects the null hypothesis.

Augustus has:

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committed a Type I error

committed a Type II error

not committed a Type I error or Type II error

committed both a Type I error and a Type II error

[3 marks]- 3 of 3 ID: MST.FET.HT.CHT.02.0020A

Select whether the following changes to a hypothesis test will cause the value of ? to increase, decrease or remain unchanged. In each case, assume all other aspects of the test remain the same.

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a) A lower level of ? is selected for the test.

b) There is a decrease in the magnitude of the critical value(s).

c) A larger sample size is selected for the test.

ATTACHMENT 6

[3 marks]- 1 of 3 ID: MST.FET.HT.HTPP.01.0030A

The Mean Corporation would like to invest in the booming health food industry. It is considering the creation of a health-drink franchise called Goose Juice. The investment department of the Mean Corporation wants to investigate the feasibility of this venture by examining the profits of similar franchises. It is risk averse and does not want to move into the venture if an average annual profit of less than $86,000 can be expected from each Goose Juice that is opened. It is known that the annual profits earned by health-drink franchises has a population standard deviation of $8,900.

The Mean Corporation’s statisticians would like to construct a hypothesis test for the mean annual profit (?) earned by health-drink franchises. A random sample of 28 franchises were chosen and their annual profit for the previous financial year was recorded. The mean annual profit for the sample was calculated as $84,250. The hypotheses that will be used by the statisticians are H0: ? = 86,000 and HA: ? < 86,000. You may find thisstandard normal tableuseful throughout the following questions. a)Calculate the test statistic (z) that corresponds to the sample and hypotheses. Give your answer to 3 decimal places. z = b)Using the test statistic for the Mean Corporation's hypothesis test and level ? = 0.05, the Mean Corporation should ACCEPT/ REJECT/ NOT REJECT the null hypothesis. c)If the sample size is increased to 107 (but the sample mean remains unchanged), the Mean Corporation should ACCEPT/ REJECT / NOT REJECT the null hypothesis. [2 marks]- 2 of 3 ID: MST.FET.HT.HTPP.02.0010A The CEO of the Myths chip company is concerned about the net weight of chips in their 2 ounce chip bags. He decides that he wants to be fairly sure that the mean weight of these chip bags (?) is greater than 2 ounces. A hypothesis test is conducted with the following hypotheses: H0: ? = 2 HA: ? > 2

The level of significance used in this test is ? = 0.05. A random sample of 26 chip bags are collected and weighed. The sample mean weight is calculated to be x = 2.10 and the sample standard deviation is s = 0.33. You may find thisStudent’s t distribution tableuseful throughout this question.

a)Calculate the test statistic (t) for this test. Give your answer to 4 decimal places.

t =

b)The result of this test is that the null hypothesis IS/ IS NOT rejected.

[2 marks]- 3 of 3 ID: MST.FET.HT.HTPP.03.0060A

The government of Preon (a small island nation) was voted in at the last election with 63% of the votes. That was 2 years ago, and ever since then the government has assumed that their approval rating has been the same. Some recent events have affected public opinion and the government suspects that their approval rating might have changed. They decide to run a hypothesis test for the proportion of people who would still vote for them.

The null and alternative hypotheses are:

H0: = 0.63

HA: ? 0.63

The level of significance used in the test is ? = 0.1. A random sample of 114 people are asked whether or not they would still vote for the government. The proportion of people that would is equal to 0.711. You may find thisstandard normal tableuseful throughout this question.

a)Calculate the test statistic (z) for this hypothesis test. Give your answer to 3 decimal places.

z =

b)According to this test statistic, you should ACCEPT/ REJECT / NOT REJECT the null hypothesis.

ATTACHMENT 7

[2 marks]- 1 of 2 ID: MST.FET.HT.PAHT.01.0030A

The Mean Corporation manufactures a line of unassembled furniture. Based on historical evidence, the customer service manager knows that the proportion of customers that have had problems with assembling the furniture at home is 0.43.

Improvements have been made to the instructions provided with the furniture, and the customer service manager believes that this will decrease the proportion of customers that have problems with the furniture () to below 0.43. A hypothesis test is conducted in order to find out.

The null and alternative hypotheses are:

H0: = 0.43

HA: < 0.43 To run the test, the manager randomly selects a sample of 102 people and asks them to assemble a piece of furniture produced by the Mean Corporation. The proportion of people in this sample that have problems is 0.39. You may find thisstandard normal tableuseful throughout this question. a)Calculate the test statistic (z). Give your answer to 2 decimal places. z = b)Calculate the P-value. Give your answer as a decimal to 4 decimal places. P = [3 marks]- 2 of 2 ID: MST.FET.HT.PAHT.02.0020A A psychologist has developed an aptitude test which consists of a series of mathematical and vocabulary problems. They want to test the hypothesis that the mean test score is 65.1. A random sample of 40 people have taken the test and their results recorded: .909px;="" collapse;="" 12px="">

.909px;=”” collapse;=”” 12px=””>

92 75 86 82 65 44 68 55 60 81

57 87 93 95 82 64 60 54 62 61

55 76 54 48 52 77 62 70 74 78

68 52 66 75 46 62 45 97 78 76

You may find thisStudent’s t distribution tableuseful throughout this question.

a)Calculate the test statistic (t) for the hypothesis test. Give your answer to 4 decimal places.

t =

b)A level of significance of ? = 0.05 is to be used for the test. The P-value for this test statistic is LESS/ GREATERthan the level of significance.

c)The result of this test is that the null hypothesis is REJECTED/ NOT REJECTED.

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