St. Petersburg College | STA 2023 – ELEMENTARY STATISTICS | Module 4 Test Answered Correctly |

n 1 1 Select the most appropriate response. The purpose of statistical inference is to provide information about the Question options: sample based upon information obtained from the population population based upon information obtained from a sample on Determine the point estimate. A survey of 400 non-fatal accidents showed that 189 involved the use of a cell phone. Determine a point estimate for p, the population proportion of non-fatal accidents that involved the use of a cell phone. Question options: 0.527 0.896 0.473 0.321 on Construct the requested confidence interval. A media poll of 500 air travelers surveyed in national airports found that 370 favor tighter security procedures in boarding planes. Construct a 90% confidence interval for the proportion of all air travelers who are in favor of tighter security procedures. Question options: .74±1.645(.74)(.26)500 .74±1.645(.5)(.5)500 St. Petersburg College | STA 2023 – ELEMENTARY STATISTICS | Module 4 Test | .74±1.96(.74)(.26)500 370±1.645(370)(130)500 on Determine the margin of error. It has been determined with 95% confidence that the proportion of online students at SPC who live in Pinellas County is between 0.73 and 0.77. Determine the margin of error associated with this confidence interval. Question options: ±0.01 ±0.02 ±0.03 ±0.04 ±0.05 on Determine the sample proportion. It has been determined with 95% confidence that the proportion of online students at SPC who live in Pinellas County is between 0.73 and 0.77. Determine the sample proportion of on-line SPC students who live in Pinellas County. Question options: 0.10 0.730.75 0.79 0.90 on Construct the requested confidence interval. A survey of 280 homeless persons showed that 63 were veterans. Construct a 90% confidence interval for the proportion of homeless persons who are veterans. Question options: (0.161, 0.289) (0.184, 0.266) (0.176, 0.274) (0.167, 0.283) on Determine the confidence level. Determine the confidence level for confidence interval shown in the figure below. Question options: 98% 95% 90% 80%99% on Construct the requested confidence interval. A survey of 300 union members in New York State reveals that 112 favor the Republican candidate for governor. Construct a 90% confidence interval for the percentage of all New York State union members who favor the Republican candidate. Question options: (30.1%, 44.5%) (32.7%, 41.9%) (31.9%, 42.8%) (26.7%, 47.9%) (17.8%, 56.8%) on Determine the necessary sample size. What is the approximate sample size required to estimate the percentage of Americans who have earned a college degree to within 2%, with 95% confidence? Recall: the worst case scenario would assume that p = 0.5 Question options: 2401 3382 4145 on Select the most appropriate response. What will happen (other things being equal) if you decrease the confidence level for a given confidence interval? Question options: It will be narrower It will be wider It will not change on Determine the margin of error. In a survey of 500 residents, 300 were opposed to the use of photo-cops for issuing traffic tickets. Find the margin of error that corresponds to a 95% confidence interval for the proportion of residents that are opposed to the use of photo-cops for issuing traffic tickets. Question options: 0.03 0.4 0.05 0.6 0.04 on Select the most appropriate response. What will happen (other things being equal) if you increase the sample size used to construct a given confidence interval?Question options: It will be narrower It will be wider It will not change on Construct the requested confidence interval. An article estimates, with 90% confidence, that the proportion of elementary school teachers who are female is 82%. The article also states the maximum error of their estimate is 3%. Determine the resulting 90% confidence interval for the proportion of elementary school teachers who are female. Question options: (82%, 85%) (79%, 82%) (80.5%, 83.5%) (87%, 93%) (79%, 85%) on Construct the requested confidence interval. In a random sample of 60 computers, the mean repair cost was $150 with a standard deviation of $36. Construct a 90% confidence interval for the population mean. Question options:($537, $654) ($138, $162) ($141, $159) ($142, $158) on Determine the width of the given confidence interval. An environmental journal discussing current water usage within the US reports, with 95% confidence, the average number of household gallons used per person per day is between 82 and 88. The study was based on a random sample of 30 individuals throughout the nation. Question options: 82 88 3 85 6 on Select the most appropriate response. An article in an educational journal discussing a non-conventional high school reading program reports, with 95% confidence, the average score on the reading portion of the SAT is between 520 and 560. The study was based on a random sample of 20 students and it was assumed that SAT scores would be approximately normally distributed. Determine the average SAT reading score of the 20 students sampled. Question options: 520560 530 540 550 on Construct the requested confidence interval. In a recent study of 42 eighth graders, the mean number of hours per week that they watched television was 19.6 with a standard deviation of 5.8 hours. Construct a 98% confidence interval for the population mean. Question options: (14.1, 23.2) (18.3, 20.9) (17.4, 21.8) (19.1, 20.4) on Interpret the confidence interval. An experiment was conducted to estimate the average fat content of a quarter pounder with cheese. A random sample of 30 burgers produced a sample mean of 26 grams which resulted in a 95% confidence interval of ( 25 , 27) grams. Which of the following is the correct interpretation of this interval? Question options: 95% of the quarter pounders with cheese sampled had a fat content between 25 and 27 grams.We are 95% confident that the mean fat content of the sampled quarter pounders with cheese is between 25 and 27 grams We are 95% confident that the mean fat content of a quarter pounder with cheese is between 25 and 27 grams. We are 95% confident that the range of fat content of a quarter pounder with cheese is between 25 and 27 grams. on Select the most appropriate response. In a random sample of 28 families, the average weekly food expense was $95.60 with a standard deviation of $22.50. Determine whether a normal distribution or a t-distribution should be used or whether neither of these can be used to construct a confidence interval. Assume the distribution of weekly food expenses is approximately normal. Question options: Cannot use normal distribution or t-distribution. Use a t-distribution. Use the normal distribution. on Construct the requested confidence interval. A random sample of 10 parking meters in a beach community showed the following incomes for a day. Assume the incomes are normally distributed. $3.60 $4.50 $2.80 $6.30 $2.60 $5.20 $6.75 $4.25 $8.00 $3.00 Construct a 95% confidence interval for the true mean. Hint: s=1.83 Question options: ($2.11, $5.34)($1.35, $2.85) ($4.81, $6.31) ($3.39, $6.01) on Select the most appropriate response. Data collected by child development scientists produced the following 90% confidence interval for the average age (in months) at which children say their first word: 10.4 μ 13.8. In this example, what does the phrase 90% confident mean? Question options: There is a 90% chance that the sample mean is between 10.4 and 13.8 months. If many samples are drawn, the computed confidence intervals will contain the population mean approximately 90% of the time. There is a 90% chance that the population mean is between 10.4 and 13.8 months. If many samples are drawn, the population mean will be between 10.4 and 13.8 months approximately 90% of the time. on Select the most appropriate response. Which of the following is a benefit of increasing the sample size when trying to estimate the mean from a sample average? Question options: A reduction in the bias of the estimate. A reduction in the variability of the estimate. An increase in the width of the resulting confidence interval. There will be no benefit because the population has not changed. onConstruct the requested confidence interval. An electronics company just finished designing a new tablet computer and is interested in estimating its battery-life. A random sample of 20 laptops with a full charge was tested and the battery-life was found to be approximately normal with a mean of 6 hours and a sample standard deviation of 1.5 hours. Which of the following is the correct form for a 99% confidence interval? Question options: CI.99(μ)=6±2.576(0.3354) CI.99(μ)=6±2.576(1.5) CI.99(μ)=6±2.861(0.3354) CI.99(μ)=6±2.861(1.5) on Evaluate the claim. A weight-loss program claims that, on average, participants lose 10 pounds during the first month of participation. An independent organization constructed a 95% confidence interval of (6.63 , 9.15) for the average amount of weight lost during the first month of this program. What conclusion best evaluates the claim made about this program’s effectiveness? Question options: With 95% certainty, the claim of losing 10 pounds on average during the first month of participation is false. With 95% certainty, the claim of losing 10 pounds on average during the first month of participation is correct. With 100% certainty, the claim of losing 10 pounds on average during the first month of participation is false. With 100% certainty, the claim of losing 10 pounds on average during the first month of participation is correct. on Evaluate the claim. An insurance company reported that, on average, claims for a certain medical procedure are $942. An independent organization constructed a 95% confidence interval of ($930 , $950) for the average amount claimed for the particular medicalprocedure. What conclusion best evaluates the truthfulness of the number reported by the insurance company? Question options: With 95% certainty, the average claim for this medical procedure is $942. With 95% certainty, the average claim for this medical procedure is not $942. The confidence interval is consistent with an average claim of $942 for this medical procedure. With 100% certainty, the average claim for this medical procedure is $942.