Stat Chp 9.1 9.2
A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of 1013 people age 15 or older, the mean amount of time spent eating or drinking per day is 1.36 hours with a standard deviation of 0.58 hour. (a) A histogram of time spent eating and drinking each day is skewed right. Use this result to explain why a large sample size is needed to construct a confidence interval for the mean time spent eating and drinking each day. (b) There are more than 200 million people nationally age 15 or older. Explain why this, along with the fact that the data were obtained using a random sample, satisfies the requirements for constructing a confidence interval. (c) Determine and interpret a 99% confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day. (d) Could the interval be used to estimate the mean amount of time a 9-year-old spends eating and drinking each day? Explain.
(A) Since the distribution of time spent eating and drinking each day is not normally distributed (skewed right), the sample must be large so that the distribution of the sample mean will be approximately normal. (B) The sample size is less than 5% of the population. (C) STAT > TESTS> 8 = The nutritionist is 99% confident that the mean amount of time spent eating or drinking per day is between 1.313 and 1.407 hours. (D) No; the interval is about people age 15 or older. The mean amount of time spent eating or drinking per day for 9-year-olds may differ.
The trade magazine QSR routinely checks the drive-through service times of fast-food restaurants. A 95% confidence interval that results from examining 539 customers in Taco Bell's drive-through has a lower bound of 169.0 seconds and an upper bound of 172.2 seconds. (a) What is the mean service time from the 539 customers? (b) What is the margin of error for the confidence interval? (c) Interpret the confidence interval.
(A) The mean service time from the 539 customers is enter your response here seconds = 169.0 + 172.42/ 2 = 170.6 (B) The margin of error is 170.6 - 169.0 = 1.6 seconds (C) One can be 95% confident that the mean drive-through service time of Taco Bell is between 169 seconds and 172.2 seconds
A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x, is found to be 114, and the sample standard deviation, s, is found to be 10. (a) Construct a 96% confidence interval about μ if the sample size, n, is 25. (b) Construct a 96% confidence interval about μ if the sample size, n, is 19. (c) Construct an 80% confidence interval about μ if the sample size, n, is 25. (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed?
(a) Lower bound = 0.04/2= 0.02 = Find 0.02 on the t-distrubution and subtract 1 from the number on the left 24 = 2.172 Then, 114 -2.172 x 10/ square root of 25 = 109.7 Upper bound = 114 + 2.172 x 10/ square root of 25 = 118.3 (B) Lower bound = 2.214 = 114 - 2.214 x 10/ square root of 19 Upper bound = 2.214 = 114 + 2.214 x 10/ square root of 19 As the sample size decreases. the margin of error increases (C) Lower bound 0.20/2 = 0.10 = 1.318 = 114 - 1.318 x 10/square root of 25 = 111.4 Upper bound = 114 + 1.318 x 10/square root of 25 = 116.6 As the level of confidence decreases, the size of the interval decreases. (D) No, the population need to be normally distributed with no outliers and the sample data must come from a population that is normally distributed too
A survey of 2298 adults in a certain large country aged 18 and older conducted by a reputable polling organization found that 412 have donated blood in the past two years. (a) Obtain a point estimate for the population proportion of adults in the country aged 18 and older who have donated blood in the past two years. (B). Verify that the requirements for constructing a confidence interval about p are satisfied. (c) Construct and interpret a 90% confidence interval for the population proportion of adults in the country who have donated blood in the past two years. Select the correct choice below and fill in any answer boxes within your choice.
(a). 412/ 2298 = 0.179 So, P^ = 0.179 (b). The sample can be assumed to be a simple random sample, the value of np^(1 - p^) is 337.712, which is greater than or equal to 10, and the sample size can be assumed to be less than or equal to 5% of the population size. 2298(0.179)(1-0.179) = 337.712 (c). We are90% confident the proportion of adults in the country aged 18 and older who have donated blood in the past two years is between 0.166 and 0.192. Use STAT>TEST> A1-PropZInt..>x=412n=2298
In a survey conducted by a reputable marketing agency, 227 of 1000 adults 19 years of age or older confessed to bringing and using their cell phone every trip to the bathroom (confessions included texting and answering phone calls). (a) What is the sample in this study? Determine the population of interest. Select the correct choice below and, if necessary, fill in the answer box within your choice. (b) What is the variable of interest in this study? Is it qualitative or quantitative? (c) Based on the results of this survey, obtain a point estimate for the proportion of adults 19 years of age or older who bring their cell phone every trip to the bathroom. (d) Explain why the point estimate found in part (c) is a statistic. Explain why it is a random variable. What is the source of variability in the random variable? e) Construct and interpret a 95% confidence interval for the population proportion of adults 19 years of age or older who bring their cell phone every trip to the bathroom. Select the correct choice below and fill in any answer boxes within your choice. (f) What ensures that the results of this study are representative of all adults 19 years of age or older?
(a). The sample is the 1,000 adults 19 years of age or older. The population is all adults 19 years of age or older. (B). The variable of interest is bringing one's phone every trip to the bathroom. This variable is qualitative with two outcomes because individuals are classified based on a characteristic. (c). P^ = 227/1,000 = 0.227 (d). Why is the point estimate found in part (c) a statistic? Its value is based on a sample. Why is the point estimate found in part (c) a random variable? Its value may change depending on the individuals in the survey. What is the source of variability in the random variable? The individuals selected to be in the study (e). Thus, the 95% confidence interval for the population proportion of adults 19 years of age or older who bring their cell phone every trip to the bathroom is between 0.201 and 0.253. (f). Random sampling
Compute the critical value zα/2 that corresponds to a 93% level of confidence.
.93 = (1- α) x 100% .93 = (1- α) x 1 100-93=7 so, 0.07 = α then, 0.07/2= 0.035 Find 0.035 in the negative standard normal distribution chart = 1.81
In a poll, 51% of the people polled answered yes to the question "Are you in favor of the death penalty for a person convicted of murder"? The margin of error in the poll was2%, and the estimate was made with 94% confidence. At least how many people were surveyed?
0.02/2= 0.01 then find the z score that matches that 0.51(1-0.51)(2.05/ 0.02)^2 = 2200
Determine the point estimate of the population proportion, the margin of error for the following confidence interval, and the number of individuals in the sample with the specified characteristic, x, for the sample size provided. Lower bound=0.093, upper bound=0.307, n=1500
0.307 + 0.093 / 2 = 0.2 0.307 - 0.093 / 2 =0.107 Then, 1,500 (0.2) = 300 The point estimate of the population proportion is 0.2, the margin error is 0.107 and the number of individuals in the sample with the specified characteristic is 300.
A doctor wants to estimate the mean HDL cholesterol of all 20- to 29-year-old females. How many subjects are needed to estimate the mean HDL cholesterol within 3 points with 99% confidence assuming s=14.2 based on earlier studies? Suppose the doctor would be content with 90% confidence. How does the decrease in confidence affect the sample size required?
A 99% confidence level requires = 0.01/2 = 0.005 = 2.575 Then, (2.575 x 14.2 / 3)^2 = 149 subjects 90% = 1.645 A 90% confidence level requires (1.645 x 14.2 x 3)^2 = 61 Decreasing the confidence level dereases the sample size needed
Explain what "90% confidence" means in a 90% confidence interval.
If 100 different confidence intervals are constructed, each based on a different sample of size n from the same population, then we expect 90 of the intervals to include the parameter and 10 to not include the parameter.
A simple random sample of size n is drawn. The sample mean, x, is found to be 19.2, and the sample standard deviation, s, is found to be 4.2. (a) Construct a 95% confidence interval about μ if the sample size, n, is 34. (b) Construct a 95% confidence interval about μ if the sample size, n, is 51. (C) (c) Construct a 99% confidence interval about μ if the sample size, n, is 34. (d) If the sample size is 14, what conditions must be satisfied to compute the confidence interval?
STAT > TEST > 8 (a) lower bound = 17.74 Upper bound = 20.67 (b). Lower bound = 18.02 Upper bound = 20.38 Increasing the sample size affects the margin of error by decreasing it (C) Lower bound = 17.23 Upper bound = 21.17 Increasing the level of confidence affects the size of the margin of error (D) The sample data must come from a population that is normally distributed with no outliers.
A survey was conducted that asked 1004 people how many books they had read in the past year. Results indicated that x=13.3 books and s=16.6 books. Construct a 90% confidence interval for the mean number of books people read. Interpret the interval.
STAT > TESTS> 8 There is 90% confidence that the population mean number of books read is between 12.44 and 14.16
A survey was conducted that asked 1005 people how many books they had read in the past year. Results indicated that x=11.2 books and s=16.6 books. Construct a 95% confidence interval for the mean number of books people read. Interpret the interval.
STAT > TESTS> 8 There is 95% confidence that the population mean number of books read is between 10.17 and 12.23
Construct a 95% confidence interval of the population proportion using the given information. x =120, n = 300
STAT>TESTS> A:1-PropZInt..input data The lower bound is 0.345 The upper bound is 0.455
Construct a confidence interval of the population proportion at the given level of confidence. x=540, n=1100, 96% confidence
STAT>TESTS> A:1-PropZInt..input data The lower bound of the confidence interval is 0.460 The upper bound of the confidence interval is 0.522
In a poll, 51% of the people polled answered yes to the question "Are you in favor of the death penalty for a person convicted of murder?" The margin of error in the poll was 2%, and the estimate was made with 94% confidence. At least how many people were surveyed?
The minimum number of surveyed people was: 0.02/2 = 0.01 Use the negative standard normal distribution table to find the number close to 0.01 in the chart = 2.32 0.69 (1-0.69)(2.32/0.02)^2 =
A researcher wishes to estimate the percentage of adults who support abolishing the penny. What size sample should be obtained if he wishes the estimate to be within 4 percentage points with 95% confidence if (a) he uses a previous estimate of 25%? (b) he does not use any prior estimates?
(a) n= using the table of critical values 95% = 1.96 0.25 (1-0.25) (1.96/ 0.04)^2 = 450 =451 (B) 0.25 (1.96/ 0.04) = 600 = 601 The 0.25 in this equation is going to stay the same no matter what's the equation