95%confidence interval estimate of the population standard deviation 64 60 60 56 60 54 61 60 61 70 62 66
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A: Please like my answer..... It supports me a lot...?
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Listed below are speeds mi/h measured from traffic on a busy highway. The simple random sample was obtained at 3:30 p.m. on a weekday. Use the sample data to construct a 95%confidence
64 60 60 56 60 54 61 60 61 70 62 66
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- Researchers were interested in the impact of texting on student learning. A group of 99 college students received texts from the researcher during a prerecorded psychology lecture. At the end of the 20-minute lecture, students answered a 17-question quiz about the material that had just been presented; scores were compared to a population mean of 15 on the test, though the population standard deviation was unknown. Which statistical test should the researchers use to analyze their data? a. dependent-samples t test b. one-sample t test c. independent-samples t test d. z testThe mean tar content of a simple random sample of 25 unfiltered king-size cigarettes is 21.4 mg, with a standard deviation of 3 mg. The mean tar content of a simple random sample of 25 filtered 100-mm cigarettes is 13.0 mg with a standard deviation of 3.8 mg. The accompanying table shows the data. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Let population 1 be unfiltered king-size cigarettes. Complete parts (a) through (c) below. Click the icon to view the data. a. Use a 0.05 significance level to test the claim that unfiltered king-size cigarettes have a mean tar content greater than that of filtered 100-mm cigarettes. What does the result suggest about the effectiveness of cigarette filters? Identify the null and alternative hypotheses. O B. Ho: H1 =H2 O C. Ho: H1 #H2 H1: H1 =H2 O A. Ho: H1 = H2 H:H1 H2 O F. Ho: H1 = H2 H1:H1> H2 H:H1=H2 Hq: H1…Dr. Graham is interested in determining if middle-aged adults use text messaging more or less frequently than the general population. Dr. Graham collects information on text messaging from a random sample of 50 adults ages 25 to 44. Dr. Graham finds that these individuals send or receive an average of 68 text messages per day. Using the population mean (and standard deviation) of 41.5 texts per day (34 texts per day), determine whether adults in this age group use text messaging more than the general public.
- A random sample of 8 adults aged 30 years were asked how much they spent on medical costs in the year 2009. Using the following data, compute the sample mean, the sample standard deviation, the sample median, and the first and third quartiles. 300 140 5600 520 470 700 640 1200A single burger contains a mean of 100 g of cooked beef with a standard deviation of 5 g. The restaurant owner claims that new triple burger contains three times as much beef. The table shows a sample of masses from 30 triple burgers. Does the sample support the claim at a 95% confidence level? Triple Burger Beef Mass (g) 292 297 295 290 290 294 291 296 282 291 289 291 290 292 295 289 289 294 298 292 294 295 285 297 296 287 302 289 290 290A report stated that after obtaining a bachelor's degree, 43% of college students enroll in a graduate degree program. For a random sample of 61 undergraduate students, find the mean, variance and standard deviation for the number of students who will enroll in a graduate degree program after obtaining a bachelor's degree. a.) mean b.) variance c.) standard deviation (round to two decimal places)
- Given two dependent random samples with the following results: Population 1 48 29 32 29 38 31 28 Population 2 45 38 41 16 40 28 38 Use this data to find the 95% confidence interval for the true difference between the population means. Assume that both populations are normally distributed. Copy Data Step 2 of 4 : Calculate the sample standard deviation of the paired differences. Round your answer to six decimal places.Health insurers and the federal government are both putting pressure on hospitals to shorten the average length of stay (LOS) of their patients. The average LOS in the United States is 4.6 days (Healthcare Cost and Utilization Project Statistical Brief, December 2018). A random sample of 22 hospitals in one state had a mean LOS of 3.6 days and a standard deviation of 2.1 days. Assume that the sampled population is approximately normal. Complete parts a through c. Question content area bottom Part 1 a. Use a 99% confidence interval to estimate the population mean LOS for the state's hospitals. enter your response here, enter your response here (Round to three decimal places as needed.)A study was conducted for an aviation safety agency in which they surveyed thousands of passengers at airports across Europe. In this study, the "mass" for a passenger includes the weight of the passenger and all of the passenger's carry-on items, including infants without their own seats. For the 5,904 adult male passengers measured in summer, the sample mean and standard deviation for this mass were 88.8 kg and 15.9 kg, respectively. (a) Construct a 95% confidence interval (in kilograms) for the population mean mass for adult male passengers in summer. (Round your answers to three decimal places.) ___to__ kg (b) Write a sentence or two interpreting the confidence interval you found in part (a). With 95% confidence, we can estimate that the population mean mass of male passengers traveling in the summer is contained within the interval. We can estimate that 95% of the male passengers traveling in the summer in the sample have a mass that is contained within the interval.…
- A food distribution company claims that a restaurant chain receives, on average, 26 pounds of fresh vegetables on a daily basis. The standard deviation of these shipments is known to be 4.4 pounds. The district manager of the restaurant chain decides to randomly sample 35 shipments from the company and finds a mean weight of 24.7 pounds. Test at a 3% level of significance to determine whether or not the food distribution company sends less than 26 pounds of fresh vegetables. a. Check the TWO requirements that are satisfied. The Central Limit Theorem applies. The a distribution is normal since n > 30. The a distribution is normal since the x distribution is normal. The p distribution is normal since np > 5 and nq > 5.A special academic program for skilled students conduct an admission exam to select their new cohort. The admission exam is known to have a mean score of 66. An examiner thinks that the actual mean score for the most recent applicants is lower than 66. He randomly samples the scores of 18 recent applicants and obtains the average scores as 62.0 with a standard deviation of 8.7. He performs a hypothesis test using a 5% level of significance to reach an appropriate conclusion. a. Calculate the value of the the appropriate test statistic for this test. Answer rounded to at least 2 decimal places. b. Determine the tabulated critical value for this test. Only write the absolute value (without +/- sign) Answer rounded to at least 3 decimal places. c. Determine the p-value for this test. Answer in exact fraction, or rounded to at least 4 decimal places.The salaries of professional baseball players are heavily skewed right with a mean of $3.2 million and a standard deviation of $2 million. The salaries of professional football players are also heavily skewed right with a mean of $1.9 million and a standard deviation of $1.5 million. A random sample of 40 baseball players’ salaries and 35 football players’ salaries is selected. The mean salary is determined for both samples. Let represent the difference in the mean salaries for baseball and football players. Which of the following represents the shape of the sampling distribution for ? skewed right since the populations are both right skewed skewed right since the differences in salaries cannot be negative approximately Normal since both sample sizes are greater than 30 approximately Normal since the sum of the sample sizes is greater than 30