stats chapter 7

which of the following is NOT an equivalent expression for the confidence interval given by 161.7 < u < 189.5?

161.7 +/- 27.8

salaries of 32 college graduates who took a stats course in college have a mean of $69,800. assuming a standard deviation of $15,069, construct a 95% confidence interval for estimating the population mean u.

64579 < u < 75021

which of the following is NOT a true statement about error in hypothesis testing

a type I error is making the mistake of rejecting the null hypothesis when it is actually false

In a Harris​ poll, adults were asked if they are in favor of abolishing the penny. Among the​ responses,1293 answered​ "no," 469 answered​ "yes," and 387 had no opinion. What is the sample proportion of yes​ responses, and what notation is used to represent​ it?

choose the correct answer p^= 0.218, the symbol p^ is used to represent a sample proportion

the number of _____ for a collection of sample data is the number of sample data is the number of sample values that can vary after certain restrictions have have been imposed on all data values.

degrees of freedom

find the sample size, n needed to estimate the percentage of adults who have consulted fortune tellers. use a 0.09 margin of error, use a confidence level of 98%, and use results results from a prior poll suggesting that 17% of adults have consulted fortune tellers

n = 94

many states are carefully considering steps that would help them collect sales taxes on items purchased through the internet. how many randomly selected sales transactions must be surveyed to determine the percentage that transpired over the internet? Assume that we want to be 99% confident that the sample percentage is within four percentage points of the true population percentage for all sales transactions.

n= 1037

use the given data to find the minimum sample size required to estimate a population proportion or percentage: Margin of error: 0.04; confidence level 90%; p^ and q^ unknnown

n= 423

the _____ hypothesis is a statement that the value of a population parameter is equal to some claimed value.

null hypothesis

which of the following groups has terms that can be used interchangeably with the others?

percentage, probability, and proportion

a _______ is a single value used to approximate a population parameter

point estimate

the ____ states that if, under a given assumption, the probability of a particular observed event is extremely small, we conclude that the assumption is probably not correct

rare event rule

The ______ is the best point estimate of the population mean

sample mean

which of the following is NOT needed to determine the minimum sample size required to estimate a population proportion?

standard deviation

the _____ is a value used in making decision about the null hypothesis and if found by converting the sample statistic to a score with the assumption that the null hypothesis is true

test statistic

which of the following is NOT a requirement for constructing a confidence interval for estimating a population mean with standard deviation known

the confidence level is 95%

which of the following is NOT required to determine minimum sample size to estimate a population mean?

the size of the population, N

Claim: a coin favors heads when tossed, and there are 13 heads in 28 tosses

there is not sufficient evidence to support the claim because there are not substantially more heads than tails

A ______ is a procedure for testing a claim about a property of a population

hypothesis test

an IQ test is designed so that the mean is 100 and the standard deviation is 10 for the population of normal adults. find the sample size necessary to estimate the mean IQ score of stat students such that it can be said with a 95% confidence that the sample mean is within 3 IQ points of the true mean.

(a) the required sample size is 43 (b) would it be reasonable to sample this number of students yes. this number of IQ test scores is a fairly small number

EXAMPLE: the brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. he wants to first determine the percentage of adults who have heard of the brand. how many adults must he survey in order to be 90% confident that his estimate is within nine percentage points of the true population percentage?

EXAMPLE: (a) E= 0.09 Zα/2= Z0.1/2 = Z0.05 = 1.645 n= [ 1.645]^2(0.25)/0.09^2 = 84 (b) 78% p^= 0.78 n= [1.645]^2(0.78)(1-0.78)/0.09^2 = 58

An IQ test is designed so that the mean is 100 and the standard deviation is 18 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 95​% confidence that the sample mean is within 3 IQ points of the true mean. Assume that σ=18 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation.

(a) 139 (b) would it be reasonable to sample this number of students? yes, this number of IQ test scores is fairly small number

a research institute poll asked respondents if they felt vulnerable to identity theft. in the poll, n=945 and x= 595 who said yes. use a 99% confidence level

(a) find the best point estimate of the population proportion p .063 (b) identify the value of the margin of error E E= .0404 (c)construct a confidence interval .590 < p < .670 (d)write a statement one has 99% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion

in the week before and the week after a holiday, there were 12,000 total deaths, and 5958 of them occurred in the week before the holiday (a) construct a 95% confidence interval estimate of the proportion of deaths in the week before the holiday to the total deaths in the week before and the week after the holiday (b) based on the results, does there appear to be any indication that people can temporarily postpone their death to survive the holiday?

(a) 0.488 < p < 0.505 (b) no, because the proportion could easily equal 0.5. the interval is not less than 0.5 the week before the holiday

a programmer plans to develop a new software system. in planning for the operating system that he will use, he needs to estimate the percentage of computers that use a new operating system. how many computers must be surveyed in order to be 99% confident that his estimate is in error by no more than four percentage points?

(a) n = 1037 (b) assume 87% n= 469 (c) does the additional survey information from part (b) have much of an effect on the sample size that is required? yes, using the additional survey information from part (b) dramatically reduces the sample size

the brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. he wants to first determine the percentage of adults who have heard of the brand. how many adults must he survey in order to be 90% confident that his estimate is within eight percentage points of the true population percentage?

(a) n= 106 (b) 82% n=63 (c) given that the required sample size is relatively small, could he simply survey the adults at the nearest college? no, a sample of students at the nearest college is a convenience sample, not a simple random sample, so it is very possible that the results would not be representative of the population of adults

an IQ test is designed so that the mean is 100 and the standard deviation is 20 for the population of normal adults. find the sample size necessary to estimate the mean IQ score of stat students such that it can be said with a 95% confidence that the sample mean is within 8 IQ points of the true mean. assume that the standard deviation is 20 and determine the required sample size using technology. then determine if this is a reasonable sample size for a real world calculation

(a) 25 (b) yes, this number of IQ test score is fairly small number

In a recent poll of 745 randomly selected adults, 587 said that it is morally wrong to not report all income on tax returns. Use a 0.01 significance level to test the claim that 70% of all adults say that it is morally wrong to not report all income on tax returns. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the p value method. Use the normal distribution as an approximation of the binomial distribution.

(a) Identify the correct null and alternative hypotheses. H0: p = 0.7 H1: p ≠ 0.7 (b) What is the test statistics? z = 5.24 (c) What is the p value? 0.0000 (d) Identify the null conclusion reject H0. there is sufficient evidence to warrant rejection of the claim that 70% of adults say that it is morally wrong not to report all income on tax returns.

In 1997, a survey of 860 households showed that 135 of them still use e-mail. Use those sample results to test the claim that more than 15% of households use e-mail. Use a 0.05 significance level.

(a) What are the null and alternative hypotheses? H0: p = 0.15 H1: p ≠ 0.15 (b) What is the test statistics? z = .57 (c) What is the p value? 0.283 d) What is the conclusion? There is not sufficient evidence to support the claim that more than 15% of households use email. (e) Is the conclusion valid today No, the conclusion is not valid today because the population characteristics of the use of emails are changing rapidly.

A genetic experiment involving peas yielded one sample of offspring consisting of 440 green peas and 145 yellow peas. Use a 0.05 significance level to test the claim that under the same circumstances 24% of offspring peas will be yellow. Identify the null hypothesis, alternative hypothesis, test statistic, p value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the p value method and the normal distribution as an approximation to the binomial distribution.

(a) What are the null and alternative hypotheses? H0: p = 0.24 H1: p ≠ 0.24 (b) What is the test statistics? z = .45 (c) What is the p value? p value = 0.6561 (d) What is the conclusion of the null hypothesis? fail to reject the null hypothesis because the p value is greater than the significance level (e) What is the final conclusion? there is not sufficient evidence to warrant rejection of the claim that 24% of offspring peas will be yellow

the brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. he wants to first determine the percentage of adults who have heard of the brand. how many adults must he survey in order to be 90% confident that his estimate is within six percentage points of the true population percentage?

(a) assume that nothing is known about the percentage of adults who have heard of the brand n= 188 (b) assume that a recent survey suggests that about 77% of adults have heard of the brand n= 133 (c) given that the sample size is relatively small, could he simply survey the adults at the nearest college? no, a sample of students at the nearest college is a convenience sample, not a simple random sample, so it is very possible that the results would not be representative of the population of adults

a physician wants to develop criteria for determining whether a patients pulse rate is atypical, and she wants to determine whether there are significant differences between males and females. use the sample pulse rates males: females: 96 76 94 60 72 72 76 72 68 84 60 72 84 88 72 68 56 84 72 124

(a) construct a 95% confidence interval estimate of the mean pulse rate for males. 63.7 < u < 80.3 (b) construct a 95% confidence interval estimate of the mean pulse rate for males. 67.4 < u < 92.6 (c) compare the preceding results. can we conclude that the population means for males and females are different? no, because the two confidence intervals overlap, we cannot conclude that the two population means are different

In a study designed to test the effectiveness of magnets for treating back​ pain, 40 patients were given a treatment with magnets and also a sham treatment without magnets. Pain was measured using a scale from 0​ (no pain) to 100​ (extreme pain). After given the magnet​ treatments, the 40 patients had pain scores with a mean of 8.0 and a standard deviation of 2.7.After being given the sham treatments, the 40 patients had pain scores with a mean of 8.3 and a standard deviation of 2.9.

(a) construct the 99% confidence interval estimate of the mean pain score for patients given the magnet treatment. what is the confidence interval estimate of the population mean u? 6.8 < u < 9.2 (b) construct the 99% confidence interval estimate of the mean pain score for patients given the sham treatment. what is the confidence interval estimate of the population mean u? 7.1 < u < 9.5 (c) compare the results. does the treatment with magnets appear to be effective? since the confidence intervals overlap, it appears that the magnet treatments are no more effective than the sham treatments.

Claim: at most 28% of internet users pay bills online. a recent survey of 357 internet users indicated that 26% pay their bills online.

(a) express the original claim in symbolic form p ≤ 0.28 (b) identify the null and alternative hypotheses. H0: p = 0.28 H1: p > 0.28

claim: the mean weight of beauty winners is 109 pounds. a study of 18 randomly selected beauty pageants resulted in a mean winner weight of 106 pounds.

(a) express the original claim in symbolic form u = 109 (b) identify the null and the alternative hypotheses H0: u = 109 H1: u ≠ 109

claim: high school teachers have incomes with a standard deviation that is less than $20,000. a recent study of 130 high school teacher incomes showed a standard deviation of $16,000

(a) express the original claim in symbolic form σ < $20,000 (b) identify the null and the alternative hypotheses that should be used to arrive at a conclusion that supports the claim. σ = $20,000 σ < $20,000

claim: the mean weight of beauty pageant winners is 105 pounds. a study of 17 selected beauty pageants resulted in a mean winner weight of 107 pounds.

(a) express the original claim in symbolic form. u = 105 (b) identify the null and alternative hypothesis H0: u = 105 H1: u ≠ 105

a research institute poll asked respondents if they acted to annoy a ad drier. in the poll, n=2344 and x= 1011 who said that they honked. use a 99% confidence level.

(a) find the best point estimate of the population proportion p. ANSWER: .431 (b) identify the value of the margin of error E ANSWER: 0.0264 (c) construct a confidence interval ANSWER: 0.405 < p < 0.457 (d) write a statement ANSWER: one has 99% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population

use the sample data and confidence level given below to complete parts (a) through (d) research institute poll asked respondents if they felt vulnerable to identity theft. in the poll, n=1058 and x=548 who said yes. use a 99% confidence level

(a) find the best point estimate of the population proportion p. ANSWER: 0.518 (b) identify the value of the margin of error E ANSWER: 0.0396 (c) construct the confidence interval ANSWER: 0.478 < p < 0.558 (d) write a statement that correctly interprets the confidence interval. choose the correct answer below ANSWER: one has 99% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion

A survey of 1,630 randomly selected adults showed that 567 of them have heard of a new electronic reader. The accompanying technology display results from a test of the claim that 35​% of adults have heard of the new electronic reader. Use the normal distribution as an approximation to the binomial​ distribution, and assume a 0.05 significance level to complete parts​ (a) through​ (e).

(a) is the test two tailed, left tailed, or right tailed two tailed (b) what is the test statistic? z= -0.18 (c) what is the p value? p value= 0.8558 (d) what is the null hypothesis and what do you conclude about it H0: p = 0.35 (e) choose the correct answer below fail to reject the null hypothesis because the p value is greater than the significance level (f) what is the final conclusion? there is not sufficient evidence to warrant rejection of the claim that 35% of adults have heard of the new electronic reader

A certain drug is used to treat asthma. In a clinical trial of the​ drug, 20 of 270 treated subjects experienced headaches​ (based on data from the​ manufacturer). The accompanying calculator display shows results from a test of the claim that less than 12​% of treated subjects experienced headaches. Use the normal distribution as an approximation to the binomial distribution and assume a 0.05 significance level to complete parts​ (a) through​ (e) below

(a) is the test two-taile, left tailed, or right tailed? left tailed (b) what is the test statistic? z= -2.32 (c) what is the p-value? p-value= 0.0101 (d) what is the null hypothesis, and what do you conclude about it? identify the null hypothesis H0: p = 0.12 (e) decide whether to reject the null hypothesis reject the null hypothesis because the P-value is less than or equal to the significance level (f) what is the final conclusion? there is sufficient evidence to support the claim that less than 12% of treated subjects experienced headahes

identify the type I error and type II error the percentage of college students who own cars is less than 35%

(a) reject the null hypothesis that the percentage of college students who own cars is equal to 35% when that percentage is actually equal to 35% (b) fail to reject the null hypothesis that the percentage of college students who own cars is less than 35% when that percentage is actually less than 35%

In order to estimate the mean amount of the time computer users spend on the internet each month, how many computer users must be surveyed in order to be 95% confident that your sample mean is within 12 minutes of the population mean. assume that the standard deviation of the population of monthly time spent on the internet is 195 minutes. what is a major obstacle to getting a good estimate of the population mean?

(a) the minimum sample size required is 1015 (b) what is a major obstacle to getting a good estimate of the population mean? it is difficult to precisely measure the amount of time spent on the internet, invalidating some data values

The accompanying data table lists the magnitudes of 50 earthquakes measured on the Richter scale. Test the claim that the population of earthquakes has a mean magnitude greater than 1.00. Use a 0.05 significance level. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, and conclusion for the test. Assume this is a simple random sample

(a) what are the hypotheses H0: u = 1.00 H1: u > 1.00 (b) identify the test statistic t= 2.22 (c) identify the p value p-value= 0.016 (d)choose the correct answer below reject H0. there is sufficient evidence to conclude that the population of earthquakes has a mean magnitude greater than 1.00

Assume that a simple random sample has been selected and test the given claim. Identify the null and alternative​ hypotheses, test​ statistic, P-value, and state the final conclusion that addresses the original claim. Listed below are brain volumes in cm3 of unrelated subjects used in a study. Use a 0.01 significance level to test the claim that the population of brain volumes has a mean equal to 1100.9 cm3. 962 1026 1272 1080 1071 1174 1068 1346 1099 1203

(a) what are the hypotheses H0: u = 1100.9 cm3 H1: u ≠ 1100.9 cm3 (b) identify the test statistic t= .787 (c) identify the P- value .4515 (d) state the final conclusion that addresses the original claim fail to reject H0. there is insufficient evidence to warrant rejection of the claim that the population of brain volumes has a mean equal to 1100.9 cm3

Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Identify the null and alternative​ hypotheses, test​ statistic, P-value, and state the final conclusion that addresses the original claim. In a manual on how to have a number one​ song, it is stated that a song must be no longer than 210 seconds. A simple random sample of 40 current hit songs results in a mean length of 218.4 sec and a standard deviation of 54.24 sec. Use a 0.05 significance level and the accompanying Minitab display to test the claim that the sample is from a population of songs with a mean greater than 210 sec. What do these results suggest about the advice given in the​ manual?

(a) what are the hypotheses? H0: u = 210 sec H1: u > 210 sec (b) identify the test statistic t= x-u/s/n x= 218.40 se mean= 8.58 s= 54.24 95% confidence t= 0.98 p=0.167 n= 40 t= .98 (c) identify the P value p-value= 0.167 (d) state the final conclusion fail to reject H0. there is insufficient evidence to support the claim that the sample is from a population of songs with a mean length greater than 210 sec (e) what do the results suggest about the advice given in the manual? the results do not suggest that the advice of writing a song that must be no longer than 210 seconds is not sound advice

Assume that a simple random sample has been selected and test the given claim. Use the​ P-value method for testing hypotheses. Identify the null and alternative​ hypotheses, test​ statistic, P-value, and state the final conclusion that addresses the original claim. The ages of actresses when they won an acting award is summarized by the statistics n=81​ x=35.8 ​years, and s=11.7 years. Use a 0.05 significance level to test the claim that the mean age of actresses when they win an acting award is 34 years.

(a) what are the hypotheses? H0: u = 34 years H1: u ≠ 34 years (b) identify the test statistic *use t test stat* t= 1.384 (c) identify the P-value p-value= .1700 (d)state the final conclusion that addresses the original claim fail to reject H0. there is insufficient evidence to warrant rejection of the claim that the mean age of actresses when they win an acting award is 34 years

A simple random sample of 60 adults is obtained from a normally distributed​ population, and each​ person's red blood cell count​ (in cells per​ microliter) is measured. The sample mean is 5.22 and the sample standard deviation is 0.56. Use a 0.01 significance level and the given calculator display to test the claim that the sample is from a population with a mean less than 5.4, which is a value often used for the upper limit of the range of normal values. What do the results suggest about the sample​ group? t-test u< 5.4 t= -2.489775008 p= 0.0078100119 x= 5.22 sx= 0.56 n=60

(a) what are the null and alternative hypotheses H0: u = 5.4 H1: u < 5.4 (b)identify the test stat -2.490 (c)identify the P-value 0.0078 (d) state the final conclusion reject H0. there is sufficient evidence to support the claim that the sample is from a population with a mean less than 5.4 (e) what do the results suggest about the sample group? there is enough evidence to conclude that the sample is from a population with a mean less than 5.4, so it is unlikely that the population has counts that are too high

Randomly selected students participated in the experiment to test their ability to determine when one minute has passed. forty students yielded a sample mean of 61.6 seconds. assuming that the standard deviation is 10.2 seconds. construct and interpret a 99% confidence interval estimate of the population mean of all students.

(a) what is the 99% confidence interval for the population mean u? 57.4 < u < 65.8 (b) based on the result, is it likely that the students estimates have a mean that is reasonably close to sixty seconds? yes, because the confidence interval includes sixty seconds

a data set includes 106 body temperatures of healthy humans for which x= 98.7F and s= 0.63F

(a) what is the best point estimate of the mean body temperature of all healthy human? 98.7 (b) using the sample stats, construct a 99% confidence interval estimate of the mean body temperature of all healthy humans. do the confidence interval limits contain 98.6F? what does the sample suggest about the use of 98.6F as the mean body temp? 98.539 < u < 98.861 (c) do the confidence interval limits contain 98.6F yes (d) what does this suggest about the use of 98.6F as the mean body temp? this suggests that the mean body temperature could very possibly be 98.6F

a data set includes 104 body temperatures of healthy adult humans for which x= 98.7 F and s= 0.66F

(a) what is the best point estimate of the mean body temperature of all healthy humans? 98.7F (b) what is the confidence interval estimate of the population mean u? 98.53 < u < 98.87 (c) do the confidence interval limits contain 98.6F? yes (d) what does this suggest about the use of 98.6F as the mean body temperature? this suggests that the mean body temperature could very possibly be 98.6F

in a sample of seven cars, each car was tested for nitrogen oxide emissions and the following results were obtained. 0.09, 0.50, 0.14, 0.06, 0.18, 0.07, 0.16 Assuming that this sample is representative of the cars in use, construct a 98% confidence interval estimate of the mean amount of nitrogen oxide emissions for all cars. if the EPA requires that nitrogen oxide emissions be less than 0.165, can we safely conclude that this requirement is being met?

(a) what is the confidence interval estimate of the mean amount of nitrogen oxide emissions for all cars? .045 < u < .169 (b) can we safely conclude that the requirement that nitrogen oxide emissions be less than 0.165 is being met? no, it is possible that the requirement is being met, but it is also very possible that the mean is not less than 0.165

use technology and the given confidence level and sample data to find the confidence interval for the population mean u. assume that the population does not exhibit a normal distribution, weight lost on a diet: 99% Confidence n= 61 x= 3.0 s= 4.5

(a) what is the confidence interval for the population mean u? 1.5 kg < u < 4.5 kg (b) is the confidence interval affected by the fact that the data appear to be from a population that is not normally distributed? no, because the sample size is large enough

EXAMPLE: during a period of 11 years 2650 of the people selected for grand jury duty were sampled, and 18% of them were immigrants. use the sample data to construct a 99% confidence interval estimate of the proportion of grand jury members who were immigrants. given that among the people eligible for jury duty, 68.3% of them were immigrants, does it appear that the jury selection process was somehow biased against immigrants

- p^= 0.18 - Zα/2 = Z0.01/2 Z0.01/2 = Z0.005 Z0.005 = 2.575 - 2.575 √(0.18)(0.82)/2650 = 0.01922 - 0.18 - 0.01922 = 0.161 - 0.18 - 0.01922 = 0.199 the confidence interval for the percentage of immigrants selected for grand jury duty is (0.161, 0.199)

during a period of 11 years 2000 of the people selected for grand jury duty were sampled, and 23% of them were immigrants. use the sample data to construct a 99% confidence interval estimate of the proportion of grand jury members who were immigrants. given that among the people eligible for jury duty, 48.3% of them were immigrants, does it appear that the jury selection process was somehow biased against immigrants

ANSWER: (a) 0.206 < p < 0.254 (b) yes, the confidence interval does not include the true percentage of immigrants

An online site presented this question, " would the recent norovirus outbreak deter you from taking a cruise?" Among the 34,684 people who responded, 64% answered yes. use the sample data to construct a 90% confidence interval estimate for the proportion of the population of all people who would respond yes to that question. does the confidence interval provide good estimate of the population proportion?

ANSWER: (a) 0.636 < p < 0.644 (b) no, the sample is a voluntary sample and might not be representative of the population

during a period of 11 years 1242 of the people selected for grand jury duty were sampled, and 43% of them were immigrants. use the sample data to construct a 99% confidence interval estimate of the proportion of grand jury members who were immigrants. given that among the people eligible for jury duty, 59.3% of them were immigrants, does it appear that the jury selection process was somehow biased against immigrants

ANSWER: .394 < p < .466 (b) yes, the confidence interval does not include the true percentage of immigrants

many states are carefully considering steps that would help them collect sales taxes on items purchased through the internet. how many randomly selected sales transactions must be surveyed to determine the percentage that transpired over the internet? Assume that we want to be 99% confident that the sample percentage is within three percentage points of the true population percentage for all sales transactions.

ANSWER: n= 1842

during a period of 11 years 851 of the people selected for grand jury duty were sampled, and 61% of them were immigrants. use the sample data to construct a 99% confidence interval estimate of the proportion of grand jury members who were immigrants. given that among the people eligible for jury duty, 65.1% of them were immigrants, does it appear that the jury selection process was somehow biased against immigrants

ANSWER: (a) .567 < p < .653 (b) does it appear that the jury selection process was somehow biased against immigrants no, the confidence interval includes the true percentage of immigrants

find the critical value zα/2 that corresponds to α=0.17

ANSWER: 1.37

express the confidence interval 0.222 < p < 0.666 in the form p^ +/- E.

EXAMPE: * 0.444 < p < 0.888 - p^= (up confed limit) + (low conf limit)/2 - E= (up confed limit) - (low confed limit)/2 p^ (0.888 + 0.444)/2 = 0.666 E= (0.888-0.444)/2 = 0.222 therefore p^+/-E= 0.666+/- 0.222 Answer: p^=/-E= 0.444 +/- 0.222

EXAMPLE: a research institute poll asked respondents if they acted to annoy a bad driver. in the poll, n=2564 and x=1027 who said that they honked. use a 95% confidence level

EXAMPLE: (a) find the best point estimate of the population proportion p ANSWER: 0.401 (b) identify the value of the margin of error E ANSWER: 0.0190 (c) construct the confidence interval ANSWER: 0.382 < p < 0.420

EXAMPLE: a research institute poll asked respondents if they felt vulnerable to identity theft. in the poll n=1111 and x=510 who said yes. use a 95% confidene level

EXAMPLE: (a) find the best point estimate of the population proportion p. - the sample proportion p^ is the best point estimate of the population proportion p find the sample proportion p^, rounding to three decimal places p^= x/n = 510/1111 = 0.459 (b) identify the value of the margin of error E - the margin of error E, for a population proportion is found using the following formula, where Zα/2 is the critical value separating an area of α/2 in the right tail of the standard normal distribution, p^ is the sample proportion, q^= 1 -p^, and n is the sample size E=zα/2√p^q^/n -because a 95% confidence interval is requested, α= 0.05. use the accompanying table of z scores to find the critical value, Zα/2 in the right tail of the standard distribution. Zα/2= Z0.05/2 = Z0.025 = 1.96 -now calculate the margin of error E, rounding to four decimal places, using the formula given below. substitute Zα/2= 1.96, p^= 0.459, q^= 1- 0.459, and n=1111 E=zα/2√p^(q^)/n = 1.96√0.459(1-0.459)/1111 = 0.0293 (c) construct the confidence interval -fist check that the requirements to construct a confidence interval used to estimate a population proportion are met. REQUIREMENTS: 1- the sample is a simple random sample 2- the condition for the binomial distribution are satisfied 3- there are at least 5 successes and at least 5 failures - p^-E < p < p^ +E p^-E= 0.459 - 0.0293 = 0.430 p^+E= 0.459 + 0.0293 = 0.488 therefore, 0.430 < p < 0.488 (d) write a statement that correctly interprets the confidence interval. choose the correct answer below. -the confidence level is the probability 1- α that the confidence interval actually does contain the population parameter, assuming that the estimation process is repeated a large number of times. in this case, if one were to select many different samples of size 1111 and construct the corresponding confidence intervals, 95% of them would actually contain the value of the population proportion p.

Example: in the week before and the week after a holiday, there were 14,000 total deaths, and 6951 of them occurred in the week before the holiday (a) construct a 90% confidence interval estimate of the proportion of deaths in the week before the holiday to the total deaths in the week before and the week after the holiday (b) based on the results, does there appear to be any indication that people can temporarily postpone their death to survive the holiday?

EXAMPLE: - the confidence interval for a population proportion p with confidence level 1 - α is p^+/-E, where E is the margin of error given by E= Zα/2 √p^(q^)/n, Zα/2 is the z score separating an area of α/2 in the right tail of the standard normal distribution, p^ is the sample proportion q^= 1-p^ and n is the number of sample values p^= x/n =6951/14,000 = 0.4965 -because of the 90% confidence interval is requested, α= 0.1. use a standard normal table to find the critical value Zα/2= Z0.1/2 Z0.1/2= Z0.05 Z0.05= 1.645 -Evaluate E= Zα/2√p^(q^)/n = 1.645√ 0.4965 (0.5035)/14000 = 0.00695 -use the value of the calculated margin of error E and the value of the sample proportion p^ to find the values of p^-E and p^+E 0.4965 - 0.00695 0.490 0.4965 + 0.00695 0.503

EXAMPLE: find the sample size, n needed to estimate the percentage of adults who have consulted fortune tellers. use a 0.03 margin of error, use a confidence level of 98% and use results from a prior poll suggesting that 9% of adults have consulted fortune tellers

EXAMPLE: - the sample size required to obtain a (1-α)(100%) confidence interval for p (the population proportion) with a margin of error E, with a previous estimate of p^ ( the sample proportion), is given by the following formula, where Zα/2 is the z score separating an area of α/2 in the right tail of the standard normal distribution, and q^= 1 - p^ n= [Zα/2]^2(p^)(q^)/E^2 - because a 98% confidence interval is requested, α = 0.02. while one can use either technology or a standard normal table to find the critical value, zα/2, for the purpose of this exercise, use technology, rounding to 3 decimal places Zα/2 = Z0.02/2 = Z0.01 = 2.326 n= [2.326]^2(0.09)(1-0.09)/0.03^2 = 493

EXAMPLE: Use the given data to find the minimum sample size required to estimate a population proportion or percentage. Margin of​ error: 0.11​; confidence level 95​%; p^ with and q^ unknown

EXAMPLE: -the sample size required to obtain a confidence interval for a population proportion p with confidence level 1- α, a margin of error E, and without a previous estimate p^ is given by n= [Zα/2]^2 0.25/ E^2, where Zα/2 is the z score separating an area of α/2 in the right tail of the standard normal distribution Zα/2 = Z0.05/2 Z0.05/2 = Z0.025 Z0.025 = 1.96 n= 1.96^2 (0.25)/ 0.11^2 = 79.37

Do one of the following, as appropriate. (a) find the critical value Zα/2, (b) find the critical value tα/2, (c) state that neither the normal nor the t distribution applies: confidence level 99%, n=15; σ is known; population appears to be very skewed Find the critical value.

EXAMPLE: confidence interval 99%; n=29; σ is known; population appears to be very skewed. - to be able to use either the normal or the t distribution, either the sample must come from a normally distributed population or the sample size must be greater than 30 - lets assume that the sample is a simple random sample. the critical value cannot be calculated because the population appears to be very skewed and the sample size is less than 30 -since the second requirement is not met, neither normal or t distribution apples ANSWER: neither normal or t distribution applies

EXAMPLE: many states are carefully considering steps that would help them collect sales taxes on items purchased through the internet. how many randomly selected sales transactions must be surveyed to determine the percentage that transpired over the internet? Assume that we want to be 99% confident that the sample percentage is within nine percentage points of the true population percentage for all sales transactions.

EXAMPLE: α = 0.01 because of the 99% confidence interval Zα/2 = Z0.01/2 Z0.01/2 = Z0.005 Z0.005 = 2.575 n= 2.575^2 (0.25)/0.09^2 = 205

EXAMPLE: a programmer plans to develop a new software system. in planning for the operating system that he will use, he needs to estimate the percentage of computers that use a new operating system. how many computers must be surveyed in order to be 99% confident that his estimate is in error by no more than six percentage points?

EXAMPLE: - express the margin of error, six percentage points, as a decimal E= 0.06 - because 99% confidence interval requested α = 0.01 Zα/2 = Z0.01/2 = Z0.005 = 2.576 n= [2.576]^2(0.25)/0.06^2 = 461 -express the previous estimate, 95% as a decimal p^= 0.05 n= [2.576]^2(0.95)(1-0.95)/0.06^2 = 88

Listed below are the amounts of mercury (in parts per million, or ppm) found in tuna sushi sampled at different stores. the sample mean is 0.891 ppm and the sample standard deviation is 0.234 ppm. use technology to construct a 90% confidence interval estimate of the mean amount of mercury in the population. 0.61 0.76 0.63 1.07 1.18 0.88 1.11

EXAMPLE: 90% confidence 1.263 ppm st dev.= 0.429 1.46 1.16 1.44 0.69 0.88 2.00 1.21 Assume that the population is normally distributed. since σ is not known, a t distribution will be used to solve this problem. tα/2= 1.943 E= tα/2 s/√n = 1.943 * 0.429/√7 = 0.315 1.263 - 0.315 < u < 1.263 + 0.315 0.948 ppm < u < 1.578 PROBLEM SOLUTION: 0.719, 1.063

EXAMPLE: An online site presented this​ question, "Would the recent norovirus outbreak deter you from taking a​ cruise?" Among the 33,118 people who​ responded, 71% answered​ "yes." Use the sample data to construct a 95​% confidence interval estimate for the proportion of the population of all people who would respond​ "yes" to that question. Does the confidence interval provide a good estimate of the population​ proportion?

EXAMPLE: although the poll is not a simple random sample construct the confidence interval. first express the point estimate p^ as a decimal p^=/-E, E= Zα/2 √p^(q^)/n p^=0.71 - the critical value Zα/2 is the positive z value that is the boundary separating an area of α/2 in the right tail of the standard normal distribution. find the area in the right tail of the standard normal distribution α/2= 0.05/2 = 0.025 -while either technology or a standard normal distribution table can be used to find the critical value, for the purpose of the explanation,use a table. use a standard normal table to find the critical value Zα/2. the area to the left of Zα/2 is 0.975 Zα/2 = Z0.025 = 1.96 - evaluate the margin of error E, where q^ = 0.29 1.96√(0.71)(0.29)/33,118 = 0.0049 -use the value of the calculated margin of error E and the value of the sample proportion p^, to find the values of p^-E and p^+E.. first the lower p^-E 0.71 - 0.0049 = 0.707 p^ + E 0.71 + 0.0049 = 0.715 therefore the 95% confidence interval estimate of the population of all who would respond yes is 0.705 < p < 0.715

Do one of the following. (a) Find the critical value Zα/2, (b) find the critical value of tα/2, (c) state that neither the normal nor the t distribution applies. confidence level 90%; n= 23; σ= 25.2; population appears to be normally distributed find the critical value.

EXAMPLE: confidence level 99%; n= 29; σ= 33.1; population appears to be normally distributed - when the standard deviation is known, use the normal distribution. when it is not known use the t distribution. since the standard deviation is known, use the standard normal distribution. thus, use a standard normal distribution table that the area to the left of Zα/2, where: α= 100%-99%= 0.01. remember that the area to the left of Zα/2 is needed to determine the critical value. it is calculated using 1-α/2 0.995. Zα/2= 2.575 PROBLEM SOLUTION: α= 100%-90%= 0.10 ANSWER: Zα/2= 1.645

Use technology and the given confidence level and sample data to find the confidence interval for the population mean u. assume that the population does not exhibit a normal distribution. weight lost on a diet: 98% confidence n= 41 x= 3.0 kg s= 5.1 kg

EXAMPLE: weight lost on a diet: 96% confidence n= 61 x= 2.0 kg s= 4.7 kg x-E < u < x+ E where E= tα/2 s/√n First find tα/2 corresponding to a 96% confidence level, where the number of degrees of freedom is given by n- 1= 60 and α= 100%- 96%= 0.004. while either technology or a table of critical t values can be used to find tα/2, for this problem, use technology tα/2= 2.099 use the formula to calculate the margin of error, rounding to one decimal place. E= tα/2 s/√n = 2.099 * 4.7/√61 = 1.3 use the margin of error to compute the confidence interval for the population mean u. x-E < u < x+E 2.0- 1.3 < u < 2.0 + 1.3 0.7 kg < u < 3.3 kg PROBLEM SOLUTION: (a) 1.1 < U < 4.9 (b) no, because the sample size is large enough

express the confidence interval (0.0790, 0.137) in the form of p^-E<p<p^+E

Example: (0.038,0.086) - the confidence interval can be displayed in the following three equivalent formats. p^-E<p<p^+E P^+/_E (p^-E, p^+E) -the confidence interval (low,high) expressed in the form of p^-E<p<p^+E, is low<p<high -therefore, = 0.038<p<0.086 ANSWER: 0.079 < p < 0.137

Do one of the following, as appropriate. (a) Find the critical Zα/2, (b) find the critical value tα/2, (c) state that neither the normal nor the t distribution applies. Confidence level 90%; n= 17; σ is unknown; population appears to be normally distributed.

Example: confidence level 99%; n=29; σ is unknown; population appears to be normally distributed - when the standard deviation, σ, is known, use the normal distribution. when it is not known use the t distribution. since the standard deviation is not know, USE THE T DISTRIBUTION. thus, use a t distribution table to compute the critical value, tα/2, where: α= 100% - 99%= 0.01 and the number of degrees of freedom is given by n - 1= 28. tα/2= 2.763 PROBLEM SOLUTION: α= 100%- 90%= 0.10 n-1 17 - 1= 16 ANSWER: tα/2 = 1.746

find the critical value zα/2 that corresponds to the given confidence to level. 84%

Example: the confidence level is the probability or area 1-α, where α is the compliment of the confidence level. 82% 1) 1 - 0.82 = 0.18 - the critical value is the positive z value that is at the boundary separating an area of α/2 in the right tail of the standard normal distribution - use a standard normal table to find the critical value, zα/2, round to 2 decimal places. - the area to the left of zα/2 = 0.91 2) zα/2 = z0.18/2 z0.18/2 = z0.09 z0.00 = 1.34 ANSWER: 1.41

the claim is that the proportion of adults who smoke a cigarette in the past week is less than 0.35, and the sample statistics include n= 1830 subjects with 659 saying that they smoked a cigarette in the past week. find the value of the test. round to two decimal places

H0: p < 0.35 q= 1 - p = 1 - 0.35 = .65 nq= (1830)(0.65)= 1189.5 ≥ 5 np= (1830)(0.35)= 640.5 ≥ 5 p^= 659/1830 z= (659/1830- .35)/√(.35)(.65)/1830) = .9066 = .91

the claim is that the proportion of peas with yellow pods is equal to 0.25. the sample stats from one experiment include 410 peas with 120 of them having yellow pods. find the value of the test statistic.

H0: p = 0.25 q= 1- p = 1- 0.25 = 0.75 nq= (410)(0.75) = 307.5 ≥ 5 np= (410)(0.25) = 102.5 ≥ 5 both requirements are satisfied to use the normal distribution. therefore the test statistic is z z= (p ̂-p)/√(pq/n) p^ = 120/410 finally substitute 0.25 for p, 120/410 for p^, 0.75 for q and 410 for n in the formula and claculate z, rounding to two decimal places z= (120/410 - 0.25)/ √(0.25)(0.75)/410 = 1.99 = 2

the claim is that the IQ scores of stats professors are normally distributed, with a mean greater than 130. A sample of 17 professors had a mean IQ score of 135 with a standard deviation of 7. find the value of the test stat. round to three decimal

H0: u = 130 t= x-u/s/√n x= 135 s= 7 n= 17 u=130 t= 135- 130/ 7/√17 = 2.9450 = 2.945

Claim: the mean IQ score of students in a large calculus class is greater than 103. a simple random sample of the students has a mean IQ score of 103.3

The sample is not unusual if the claim is true. the sample is not unusual if the claim is false. therefore, there is not sufficient evidence to support the claim

Claim: the mean pulse rate of students in a large stats class is less than 97. a simple random sample of the students has a mean pulse rate of 96.2

the sample is not unusual if the claim is true. the sample is not unusual if the claim is false. therefore there is not sufficient evidence to support the claim.

identify the type I error and the type II that corresponds to the given hypothesis. the proportion of settled medical malpractice suits is 0.19

type I error is the mistakes of rejecting the null hypothesis when it is actually true type II error is the mistake of failing to reject the null hypothesis when it is actually false (a) reject the claim that the proportion claim that the proportion of settled malpractice suits is 0.19 when the proportion is actually 0.19 (b) fail to regret the claim that the proportion of settled malpractice suits is 0.19 when the proportion is actually different from 0.19

which of the following would be a correct interpretation of a 99% confidence interval such as 4.1 < u < 5.6

we are 99% confident that the interval from 4.1 to 5.6 actually does contain the true value of u

listed below are the amounts of mercury found in tuna sushi sampled at different stores. the sample mean is 1.036 ppm and the sample standard deviation is 0.383 ppm. use the technology to construct a 90% confidence interval estimate of the mean amount of mercury in the population. 0.66 1.35 0.57 0.75 1.02 1.44 1.46

what is the confidence interval estimate of the mean amount of mercury in the population? .755 ppm < u < 1.317 ppm

a critical value Zα/2, denotes the

z score with an area of α to its right