STA 125 Ch 10-13 Problems
A catalog sales company promises to deliver orders placed on the Internet within 3 days. Follow-up calls to a few randomly selected customers show that a 95% confidence interval for the proportion of all orders that arrive on time is 88% +/- 6%. What does this mean? Are the conclusions in parts a-e correct? Explain. a)Between 82% and 94% of all orders arrive on time. b)95% of all random samples of customers will show that 88% of orders arrive on time. c)95% of all random samples of customers will show that 82% to 94% of all orders arrive on time. d)The company is 95% sure that between 82% and 94% of the orders placed by the customers in this sample arrived on time. e)On 95% of the days, between 82% and 94% of the orders will arrive on time.
88% ± 6% → (82%, 94%) a) No, we are 95% confident that the true proportion of orders that arrive on time is between 82% and 94%. b) No, it is not about the % of random samples that arrive on time, but how confident we are that the true proportion falls between 0.82 and 0.94. c) No, it is not about the percentage of random samples that fall between the two values, but about the unknown parameter. d) No, it is not a probability and does not describe the sample, but is used to estimate the population proportion. In this sample we know 88% arrived on time. e) No, it is not a probability, it is about the parameter, not about the days.
A clean air standard requires that vehicle exhaust emissions not exceed specified limits for various pollutants. Many states require that cars be tested annually to be sure they meet these standards. Suppose state regulators double-check a random sample of cars that a suspect repair shop has certified as okay. They will revoke the shop's license if they find significant evidence that they shop is certifying vehicles that do not meet standards. a) In this context, what is a Type I error? b) In this context, what is a Type II error? c) Which type of error would the shop's owner consider more serious? d) Which type of error might environmentalists consider more serious?
H0: Car is okay vs. Ha: Car is not okay Or, H0: Shop is okay vs. Ha: Shop is not okay a)Concluding that the vehicle's exhaust emissions exceed the specified limits when actually the car meets the clean air standards. (Or, wrongly concluding the shop is not meeting the standards when it is.) b)Certifying that the car is okay when actually the car's exhaust emission exceed the specified limits. (Or, wrongly conclude the shop is meeting standards when it isn't). c)The shop's owner would consider a Type I error more serious as the state regulator would revoke the shop's license. d)Environmentalists would consider a Type II error more serious as cars that are polluting the environment are being certified as okay.
One method credit card companies use to encourage their delinquent cardholders to pay their credit card balances is the hiring of a collection agency. To justify the cost of using a collection agency, the agency must collect an average of at least $200 per customer. After a trial period during which the agency attempted to collect from a random sample of 100 delinquent cardholder, the 90% confidence interval on the mean collected amount was reported as ($190.25, $250.75). Given this, what recommendation(s), would you make to the credit card company about using the collection agency?
Since $200 is in the confidence interval and the lower limit is fairly close to the needed value ($190.25) it is possible that the collection agency won't collect at least $200. But, the upper limit of $250.75 indicates that it is possible the collection agency can collect much more than $200 on average. However, if we were to run a hypothesis test, we would fail to reject the hypothesis since $200 is included in the interval. More data is needed.
During the 2014-2015 NHL season the average number of penalty minutes per game was 9.68 minutes. A random sample of 40 Miami University hockey games was obtained and the number of penalty minutes per game was recorded, x = 12.65 minutes/game and s = 9.67 minutes/game. Is there any evidence to suggest the true mean number of penalty minutes for Miami University hockey is greater than the mean number of NHL penalty minutes? Assume the sample size of 40 games is large enough for normality according to the CLT, and use a significance level of 5%.
Since the p-value is less than the significance level of 0.05, we reject the null hypothesis and conclude that the true mean number of penalty minutes per game for Miami University hockey games is greater than 9.68. I am 90% confident that the true mean number of penalty minutes per game for Miami University hockey games is between 10.07 and 15.23.
Many young adults live at home with their parents due to various reasons, such as lack of income, postponing marriage, and saving money for school. A recent survey reports that approximately 36% of all young adults (ages 18-31) live with their parents (Fry, 2013). In order to check this claim, a random sample of 440 young adults was obtained and 182 of them were found to live with their parents. Is there sufficient evidence to conclude that the proportion of young adults who live at home with their parents is different from 0.36? Test the appropriate hypotheses using a significance level of 0.01.
Since the p-value is not less than the significance level of 0.01, we fail to reject the null hypothesis. There is not sufficient evidence to conclude that the true proportion of young adults who live at home with their parents differs from 0.36. Since the result is not statistically significant, no confidence interval is needed.
A real estate agent wants to know how many owners of homes worth over $1 million might be considering putting their home on the market in the next 12 months. He surveys 40 of them and finds that 10 of them are considering such a move. Are all the assumptions and conditions for finding the sampling distribution of the proportion satisfied? Explain.
Yes, assuming the survey is random, they should be independent. We don't know the true proportion, so we cannot check to see if np > 10 and nq > 10; but we have observed 10 successes and 30 failures, which is sufficient.
Gallup reported in 2012 that 53% of American investors are likely to say the price of energy (including gas and oil) is hurting the US investment climate 'a lot,' according to a Wells Fargo/Gallup Investor and Retirement Optimism Index survey. The survey results are based on questions asked February 3-12, 2012, of a random sample of 1022 US adults having investable assets of $10,000 or more. The percentage reported to this same question in September 2011 was 62%. a)Is the percentage in 2012 different from that in 2011? Test the appropriate hypothesis. Check conditions. Does the test provide evidence that the percentage has changed from 2011? b)Find a 95% confidence interval for the true proportion of US adults who think the price of energy is hurting the US investment climate 'a lot.'
a) 1.Let p = true proportion of American investors who are likely to say the price of energy is hurting the US investment climate 'a lot.' 2.H0: p = .62 vs. Ha: p ≠ .62 3. α = 0.05 4.Conditions: We have a random sample (RS) of 1022 US adults, both np0 and nq0 at least 10; 1022(.62) = 633.64 and 1022(.38) = 388.36 5.z = -5.9057 6.p-value < .0001 7.Reject H0 8.Since the p-value is less than our 0.05 significance level there is sufficient evidence to reject H0 and conclude that, in 2012, the true proportion of American investors who are likely to say the price of energy is hurting the US investment climate 'a lot' is not equal to 0.62; it has changed from 2011. b) I am 95% confident that the true proportion of US adults who think the price of energy is hurting the US investment climate 'a lot' is between 0.500 and 0.561.
Several factors are involved in the creation of a confidence interval. Among them are the sample size, the level of confidence, and the margin of error. Which statements are true? a)For a given sample size, higher confidence means a smaller margin of error. b)For a specified confidence level, larger samples provide smaller margins of error. c)For a fixed margin of error, larger samples provide greater confidence. d)For a given confidence level, halving the margin of error requires a sample twice as large.
a) False, higher confidence means larger margin of error b) True, as sample size increases, standard error and margin of error decrease c) True d) False, halving the margin of error requires a sample size 4 times as large
According to the 2010 Census, 16% of the people in the US are of Hispanic or Latino origin. One county supervisor believes her county has a different proportion of Hispanic people than the nation as a whole. She looks at their most recent survey data, which was a random sample of 437 county residents, and found that 44 of those surveyed are of Hispanic origin. a)State the hypotheses. b)Name the model and check appropriate conditions for a hypothesis test. c) Calculate the test statistic and P-value. d)State your conclusion in context.
a) Let p = true proportion of Hispanic people who live in the county H0: p = 0.16, Ha: p ≠ 0.16 b) This is a one-proportion z-test. •We have a random sample of 437 county residents. •It is reasonable to assume the independence assumption has been met since we have a random sample. •It is reasonable to assume 437 is almost certainly less than 10% of the population of a county •The sample size is large enough: np0 = 437(0.16) = 69.92 > 10 and n(q0) = 437(0.84) = 367.08 > 10 c) P-value= 2∙P(z<-3.37)=.0007 d) Since the P-value of 0.0007 is less than any reasonable significance level, there is sufficient evidence to reject the null hypothesis and conclude that the true proportion of Hispanic/Latino people in this county differ from that of the nation as a whole. I am 95% confident that the true proportion of Hispanic/Latino people in this county is between 0.072 and 0.129.
All water pipes in two residence halls must be replaced. Water pipes frequently experience water pressures of up to 100 pounds per square inch (psi). To be safe, any pipes used must have a mean breaking strength greater than 120 psi. A water pipe supplier wants to provide the pipes that will be needed. They claim that their pipes are strong enough to meet the 120 psi standard. Before deciding whether to use the supplier's water pipes, the construction manager wants to test the breaking strength of a random sample of pipes from among those that are currently available. From previous experience, the pipe company reports a standard deviation of 8 psi. a) State the appropriate null and alternative hypothesis to test the supplier's claim. b) The supplier will provide 25 pipes for testing. Find the power of the test to detect pipes that have a mean breaking strength of 125 psi at a 5% significance level. c) What is the probability of committing a Type II error?
a) Let µ =true mean breaking strength of the water pipes H_0:μ=120 psi H_a:μ>120 psi b) The power of the test if the true mean breaking strength of the water pipes is actually 125 psi is 0.91777. c). The probability of committing a Type II error, β=1- power = 1 - 0.91777 = 0.08223
After the political ad campaign described in Practice Problem 1(c), pollsters check the governor's negatives. They test the hypothesis that the ads produced no change against the alternative that the negatives are now below 30% and find a P-value of 0.22. Which of these conclusions is appropriate? Explain. a)There's a 22% chance that the ads worked. b)There's a 78% chance that the ads worked. c)There's a 22% chance that their poll is correct. d)There's a 22% chance that natural sampling variation could produce poll results like these if there's really no change in public opinion.
a) No b) No c) No d) Yes
Let's say you toss a coin 40 times and it lands on heads 16 times. a) What is the point estimate? b) Find and interpret a 90% confidence interval c) Find and interpret a 95% confidence interval d) Find and interpret a 99% confidence interval
a) Point estimate is p ̂= (# successes)/Total=16/40=0.40 Find the mean, μ_p ̂ , and standard error, σ_p ̂ , based on our point estimate •μ_p ̂ =0.40 and σ_p ̂ =√((0.40(0.60))/40)=0.077 b) 90% CI = (0.273, 0.527) c) 95% CI = (0.248, 0.552) d) 99% CI = (0.200, 0.600)
According to the Gallup Poll, 27% of U.S. adults have high levels of cholesterol. Gallup reports that such elevated levels "could be financially devastating to the U. S. healthcare system" and are a major concern to health insurance providers. According to recent studies, cholesterol levels in health U. S. adults average about 215 ml/Dl with a standard deviation of about 30 mg/DL and are roughly Normally distributed. If the cholesterol levels of a sample of 42 healthy adults is taken, a)What shape should the sampling distribution of the mean have? b)What would the mean of the sampling distribution be? c)What would its standard deviation be? d)What is the probability of obtaining a sample mean less than 200 ml/Dl? e)What is the probability of obtaining a sample mean between 208 and 220 ml/Dl? f)What sample mean cholesterol level is at the 80th percentile? g)If the sample size were increased to 100, how would your answers to parts a-c change?
a) Symmetric, bell-shaped b) μ_x ̅ =215 mg/dL c) σ_x ̅ =30/√42≈4.6291 mg/dL d) P(x ̅ < 200) = 0.0006 e) P(208 < x ̅ < 220) = 0.79476 f) P(x ̅ < 218.9) = 0.80 g) Only the standard deviation of the sampling distribution would change from 4.6291 mg/dL to 3 mg/dL
Before lending someone money, banks must decide whether they believe the applicant will repay the loan. One strategy used is a point system. The higher the point total, the more convinced the bank is that it's safe to make the loan. Any applicant with a lower point total than a certain cutoff point is denied a loan. We can think of this process as a hypothesis test. Ho: The applicant will repay the loan and therefore should get the money. Ha: The applicant will not repay the loan and therefore should not get the money. a)When a person defaults on a loan, which type of error did the bank make? b)Which kind of error is it when the bank misses an opportunity to make a loan to someone who would have repaid it? c)Suppose the bank decides to lower the cutoff score from 250 points to 200. Is that analogous to choosing a higher or lower value of α for a hypothesis test. Explain. d)What impact does this change in the cutoff value have on the chance of each type of error?
a) Type II error b) Type I error c) Since H0 is giving the loan, lowering the point total needed to secure a loan makes it more difficult to reject H0; therefore, it is analogous to choosing a lower value for α. d) The lowering of points makes it easier to commit a Type II error and makes it more difficult to commit Type I error.
A survey of 200 students is selected randomly on a large university campus. They are asked if they use a laptop in class to take notes. The result of the survey is that 70 of the 200 students responded "yes." a)What is the value of the sample proportion of p ̂. b)What is the standard error of the sample proportion? c)Construct an approximate 95% confidence interval for the true proportion of p by taking ±2 SEs from the sample proportion.
a) p ̂ = 70/200 = 0.35 b) σ_p ̂ =√((0.35(1-0.35))/200)=0.0337 c) 0.35 ± 2(0.0337) = (0.2826, 0.4174) I am 95% confident that the true proportion of university students who use a laptop to take notes is between 0.283 and 0.417.
In 2003, a major vendor of anti-spam software claimed that the proportion of email consisting of unsolicited spam was 0.40; i.e., 4 out of every 10 email messages are spam. Suppose 50 email messages are selected at random. a)Carefully sketch and label the distribution of the sample proportion p-hat (get mean, SD, and n). b)What is the probability the sample proportion is greater than 0.50? c)Find the probability the sample proportion will be between 0.32 and 0.37.
a) μ_p ̂ =0.4, n=50, σ_p ̂ ≈0.07 b) P(p ̂>0.50)=0.0766 c) P(0.32≤p ̂≤0.37)=0.2076
The investment website of Exercise 1 draws a random sample of 200 investors from their customers. Suppose that the true proportion of smart phone users is 36%. a)What would the standard deviation of the sampling distribution of the proportion of smart phone users be? b)What is the probability that the sample proportion of smart phone users is greater than 0.36? c)What is the probability that the sample proportion of smart phone users is between 0.30 and 0.40? d)What is the probability that the sample proportion of smart phone users is less than 0.28? e)What is the probability that the sample proportion of smart phone users is greater than 0.42?
a) σ_p ̂ =√((0.36(1-0.36))/200)=0.0339 b)P(p ̂ > 0.36) = 0.50 c)P(0.30 < p ̂ < 0.40) = 0.8426 d)P(p ̂ < 0.28) = 0.0091 e)P(p ̂ > 0.42) = 0.0384
Police departments often try to control traffic speed by placing speed-measuring machines on roads that tell motorist how fast they are driving. Traffic safety experts must determine where the machines should be placed. In one recent test, police recorded the average speed clocked by cars driving on one busy street close to an elementary school. For a sample of 25 speeds, it was determined that the average amount over the speed limit for the 25 clocked speeds was 11.6 mph with a standard deviation of 8 mph. The 95% confidence interval estimate for this sample is 8.30 mph to 14.90 mph. a)What is the margin of error for this problem? b)The researchers commented that the interval was too wide. Explain specifically what should be done to reduce the margin of error to no more than +/- 2 mph.
a)3.3 mph b)To reduce the margin of error we need to use a larger sample size. 1.96×8/√n=2 1.96×8=2√n ((1.96×8)/2)^2=(√n)^2 〖7.84〗^2=n 62=n
Most car engines need at least 87 octane to avoid "knocking" or "pinging," terms used to describe the preignition that can happen when a fuel's octane is too low. An engineer is designing an experiment to raise the octane of an ethanol-based fuel. From previous studies, she thinks that with 8 experimental runs, she will have a power of 0.90 to detect a real increase of 3 points in the mean octane. a)If the actual increase is only 1 point, will the power be increased or decreased? b)If she wants the power to be the same, but she is interested in detecting an increase of only 1 point what will she need to do.
a)All other things being equal, the power is reduced when trying to detect a smaller effect size. b)She should use a larger sample size to keep the power the same.
Which of the following are true? If false, explain briefly. a)A very high P-value is strong evidence that the null hypothesis is false. b)A very low P-value proves that the null hypothesis is false. c)A high P-value show that the null hypothesis is true. d)A P-value below 0.05 is always considered sufficient evidence to reject a null hypothesis.
a)False, a very low p-value is strong evidence against the null hypothesis (H0) b)False, we never prove or disprove H0 c)False, a high p-value does not provide strong evidence against H0, it also does not provide strong evidence in favor of H0. d)False, the evidence required to reject H0 depends upon the significance level of the test
An enthusiastic junior executive has run a test of his new marketing program. He reports that it resulted in "significant" increase in sales. A footnote on his report explains that he used an alpha level of 7.2% for his test. Presumably, he performed a hypothesis test against the null hypothesis of no change in sales. a)If instead he had used an alpha level of 5%, is it more or less likely that he would have rejected his null hypothesis. Explain. b)If he chose the alpha level of 7.2% so that he could claim statistical significance, explain why this is not an ethical use of statistics.
a)It is less likely that he would have rejected H0 because it requires more stringent evidence in the form of a smaller p-value. b)One must choose the significance level prior to running any hypothesis test. Otherwise, one could always select a significance level so that we could reject the null hypothesis.
For each of the following situations, write the null and alternative hypotheses in terms of parameter values. a)Is a coin fair? b)Last year, customers spent an average of $35.32 per visit to the company's website. Based on a random sample of purchases this year, the company wants to know if the mean this year has changed. c)A governor is concerned about his "negatives" - the percentage of state residents who express disapproval of his job performance. His political committee pays for a series of TV ads, hoping they can keep the negatives below 30%. They will use follow-up polling to assess the ads' effectiveness. d)Only about 20% of people who try to quit smoking succeed. Sellers of a motivational tape claim that listening to the recorded messages can help people quit. e)Regular card customers have been with the company for an average of 17.3 months. The credit card bank wants to know if their Gold card members have been with the company on average a longer amount of time.
a)Let p = true proportion of Heads H0: p = .50 vs. Ha: p ≠ .50 b) Let µ = true mean amount customers spent per visit to company's website H0: µ = $35.32 vs. Ha: µ ≠ $35.32 c) Let p = true proportion of "negatives" H0: p = .30 vs. Ha: p < .30 d) Let p = true proportion of people who quit smoking using the motivational tapes H0: p = .20 vs. Ha: p > .20 e) Let µ = true mean amount of time Gold card customers have been with the credit card bank H0: µ = 17.3 vs. Ha: µ > 17.3
In 1960, census results indicated that the age at which American men first married had a mean of 23.3 years. It is widely suspected that young people today are waiting longer to get married. We want to find out if the mean age of first marriage has increased since then. a)Write appropriate hypotheses. b)We plan to test our hypothesis by selecting a random sample of 40 men who married for the first time last year. Do you think the necessary assumptions for inference are satisfied? Explain. c)Describe the appropriate sampling distribution model for the mean age in such samples. d)The men in our sample married at an average age of 24.2 years, with a standard deviation of 5.3 years. Find the test statistic and the P-value. e)Explain (in context) what this P-value means. f)What do you conclude at alpha = 0.05?
a)Let µ = true mean age at which first marriage occurs for American men H0: µ = 23.3, Ha: µ > 23.3 b) Check the assumptions: •We are going to select a random sample of 40 men who married for the first time last year. •The age at which first marriage occurs for one American man does not affect the age at which first marriage occurs for other men. •A sample size of 40 men should be large enough for normality according to the CLT. •40 men is surely less than 10% of the population of all American men who were married for the first time last year. c) μ_x ̅ =μ_0=23.3, 〖SE〗_x ̅ =s/√n=5.3/√40=0.838 d) t = 1.0740, df = 39, P-value = P (t_39 > 1.0740) = 0.1447 e) If the mean age at first marriage is still 23.3 years, there is a 14.5% chance of getting a sample mean of 24.2 years or older simply from natural sampling variability. f) Since the P-value = 0.1447 is high, we fail to reject the null hypothesis. There is no evidence to suggest that the true mean age of men at first marriage has changed from 23.3 years, the mean in 1960.
In 2012, a large number of foreclosed homes in the Washington, D.C., metro area were sold. In one community, a sample of 30 foreclosed homes sold for an average of $443,705 with a standard deviation of $196,196. a)What assumptions and conditions must be checked before finding a confidence interval for the mean? How would you check them? b)Find a 95% confidence interval for the mean value per home. c)Interpret this interval and explain what 95% confidence means. d)Suppose nationally, the average foreclosed home sold for $350,000. Do you think the average sale price in the sampled community differ significantly from the national average?
a)Random sample, symmetric, unimodal distribution without outliers. b)($370,444, $516,966) c)I am 95% confident that the true mean amount foreclosed homes in the DC area will sell for is between $370,444 and $516,966. The 95% confidence level means that if we were to take all possible samples of 30 foreclosed homes, 95% of them would capture the true mean selling price. d)Yes, since $350,000 is below the confidence interval. I would conclude that houses in the DC area sell for more money.
The analyst in the last problem finds a 95% confidence interval for the true proportion of vaccinated children to be (0.9713, .9767). a)Explain why she can reject the null hypothesis that p = 0.98 vs. p < 0.98 at α=0.01. b)Explain why the difference may or may not be important.
a)She can reject the null hypothesis p = 0.98 vs p < 0.98 since 0.98 is above the values in the confidence interval. b)Since the estimate for p is between .9713 and .9767, it is very close to .98, less than .01, which while statistically significant may not be practically significant. The true value may be as high as 97.67%. A cost analysis should be done to determine the effect of an extra 0.33% measles cases.
Health researchers at a large HMO base their cost predictions for measles on the belief that 98% of children have been vaccinated against measles. A random survey of medical records at many schools across the country found that, among more than 13,000 children, only 97.4% have been vaccinated. A statistician would reject the 98% hypothesis with a P-value of P< 0.0001. a) Explain what the P-value means in this context. b) The result is statistically significant, but is it important? Comment.
a)The p-value is the probability of getting a sample proportion less than or equal to 97.4% if 98% of the children have been vaccinated against measles, which in this case is less than 0.0001. b)The sample size of 13,000 is so large that we were pretty much assured of finding statistical significance. However, 97.4% is not much less than 98% so it really isn't practically significant. A 95% CI is (0.971, 0.977). While 1 million children a year are vaccinated, even a difference of 0.1% represents 1000 kids, so this may be important.
The manufacturer of a metal stand for home TV sets must be sure that its product will not fail under the weight of the TV. Since some larger sets weigh nearly 300 pounds, the company's safety inspectors have set a standard of ensuring that the stands can support an average of 500 pounds. Their inspectors regularly subject a random sample of the stands to increasing weight until they fail. They test the hypothesis H0: µ = 500 against HA: µ > 500, using the level of significance α = 0.01. If the sample of stands fails to pass this safety test, the inspectors will not certify the product for sale to the general public. a)Is this an upper-tail or lower-tail test? In the context of the problem, why do you think this is important? b)Explain what will happen if the inspectors commit a Type I error. c)Explain what will happen if the inspectors commit a Type II error. d)Which error is more serious? Explain. e)Based on which error is more serious, which significance level (0.01, 0.05, 0.10) should you use for the test? Explain.
a)Upper-tail test. It is important because you want the TV stands to be able to withstand at least 500 pounds. b)H0: µ = 500 vs. HA: µ > 500 Type I error: wrongly concluding µ > 500 when the TV stand cannot support at least 500 pounds. Inspectors will certify the stands are safe when they could fail. c)Type II error: wrongly believing µ = 500 when µ > 500. Inspectors will not certify the TV stands as safe when they actually are safe. d)Type I error is more serious as people could get injured. Also, if your product is defective, you will lose business and could be sued. e)The probability of committing a Type I error is the significance level of the test. Since Type I error is more serious, we need to choose a significance level of 0.01.
An investment website can tell what devices are used to access the site. The site managers wonder whether they should enhance the facilities for trading via "smart phones" so they want to estimate the proportion of users who access the site that way (even if they also use their computers sometimes). They draw a random sample of 200 investors from their customers. Suppose that the true proportion of smart phone users was 36%. a)What would you expect the shape of the sampling distribution for the sample proportion to be? b)What would be the mean of this sampling distribution? c)If the sample size were increased to 500, would your answers change? Explain.
a)np = 200(0.36) = 72 and nq = 200(1 - 0.36) = 128; Normal b)μ_p ̂ = 0.36 c)σ_p ̂ =√((0.36(1-0.36))/200)=0.0339 σ_p ̂ =√((0.36(1-0.36))/500)=0.0215 No, only the standard deviation of the sampling distribution would change; it would decrease from 0.0339 to 0.0215.
A marketing researcher for a phone company surveys 100 people and finds that the proportion of clients who are likely to switch providers when their contract expires is 0.15. a)What is the standard deviation of the sampling distribution of the proportion? b)If she wants to reduce the standard deviation by half, how large a sample would she need?
a)σ_p ̂ =√((0.15(1-0.15))/100)=0.0357 b)400 - In order to cut the SD in half you need to quadruple the sample size
A survey of 25 randomly selected customers found the following ages (in years): The mean was 31.84 years and the standard deviation 9.84 years. a) What is the standard error of the mean? b) How would the standard error change if the sample size had been 100, instead of 25? (Assume the SD does not change) c) How many degrees of freedom does the t-statistic have? d) How many degrees of freedom would the t-statistic have if the sample size had been 100?
a)〖SE〗_x ̅ =9.84/√25≈1.968 b)〖SE〗_x ̅ =9.84/√100=0.984 c)df = n - 1 = 25 - 1 = 24 d)df = n - 1 = 100 - 1 = 99
A survey of 25 randomly selected customers found the mean age was 31.84 years and the standard deviation 9.84 years. e) Construct and interpret a 95% confidence interval for the mean age of all customers. Check to see if the assumptions and conditions for the confidence interval have been met. f) How large is the margin of error? g) How would the confidence interval change if you had assumed that the standard deviation was known to be 10.0 years?
e) The data were randomly obtained and should be independent. The sample size is less than 10% of the population. Slightly left skewed, but unimodal, and no outliers, therefore, the t-inference procedure is appropriate. I am 95% confident that the true mean age of customers is between 27.777 and 35.903 years. f) The margin of error is equal to half the width of the confidence interval E=(35.903-27.777)/2=4.063 g) Only use the z-distribution when the population SD is known. I am 95% confident that the true mean age of customers is between 27.9201 and 35.7599 years. The confidence interval is smaller.
A survey of 25 randomly selected customers found the mean age was 31.84 years and the standard deviation 9.84 years. The 95% confidence interval was (27.777, 35.903). h) For the 95% confidence interval found in part (e), how large would the sample size have to be to cut the margin of error in half? i) How large of a sample size would you need to construct a 90% confidence interval with a margin of error of 2, assuming the population standard deviation is 10? j) How large of a sample size would you need to construct a 99% confidence interval with a margin of error of 2, assuming the population standard deviation is 10?
h) The sample size would need to be 4 times larger, n = 100 i) The sample size would need to be 68 j) The sample size would need to be 166
Suppose you want to estimate the proportion of the traditional college students on your campus who own their own car. You have no preconceived idea of what your proportion might be. a)What sample size is needed if you wish to be 95% confident that your estimate is within 0.02 of the true proportion?
n=((0.5)(0.5)〖(1.96)〗^2)/〖(0.02)〗^2 = 2401
The financial aid office wishes to estimate the mean cost of textbooks per semester for students at Miami University. For the estimate to be useful, it should be within $25 of the true population mean. How large a sample should be used to be 90% confident of achieving this level of accuracy? •Use $100 as a reasonable estimate for σ
n=(σ^2 z^2)/m^2 =(100^2 〖1.645〗^2)/25^2 =43.29...→44
