A random sample of 14 cigarettes of a certain brand has an average nicotine content of 4.4 milligrams and a standard deviation of 1.9 milligrams. What is the margin of error of the 99% confidence interval for the average nicotine content of the cigarettes.
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- A new drug is being tested to see if it reduces the frequency of migraines. Study participants were divided into two groups. 49 participants in group 1 received the medication; 42 participants in group 2 received a placebo. After a period of six months, group 1 had a mean of 4 migraines. It is known that the population standard deviation for this group is 1.3 migraines. Group 2 had a mean of 5.4 migraines. The population standard deviation for this group is 1.7 migraines. Can we conclude that the medication reduces the population mean number of migraines? Use αα =0.01. Note: If t-test, then unequal variances are assumed. Question a)Let: μ1μ1 = the population mean number of migraines with the medication μ2μ2 = the population mean number of migraines with the placebo (Step 1) State the null and alternative hypotheses by selecting the appropriate symbol, and identify which tailed test: H0:H0: μ1μ1 μ2μ2 H1:H1: μ1μ1 μ2μ2 Which tailed test…Fran is training for her first marathon, and she wants to know if there is a significant difference between the mean number of miles run each week by group runners and individual runners who are training for marathons. She interviews 42 randomly selected people who train in groups and finds that they run a mean of 47.1 miles per week. Assume that the population standard deviation for group runners is known to be 4.4 miles per week. She also interviews a random sample of 47 people who train on their own and finds that they run a mean of 48.5 miles per week. Assume that the population standard deviation for people who run by themselves is 1.8 miles per week. Test the claim at the 0.01 level of significance. Let group runners training for marathons be Population 1 and let individual runners training for marathons be Population 2. Step 2 of 3 : Compute the value of the test statistic. Round your answer to two decimal places.The manager of Tee Shirt Emporium reports that the mean number of shirts sold per week is 1,210, with a standard deviation of 325. The distribution of sales follows the normal distribution. What is the likelihood of selecting a sample of 25 weeks and finding the sample mean to be 1,100 or less?
- A car company claims that its new SUV gets better gas mileage than its competitor's SUV. A random sample of 35 of its SUVs has a mean gas mileage of 16.7 miles per gallon (mpg). The population standard deviation is known to be 0.9 mpg. A random sample of 42 competitor's SUVS has a mean gas mileage of 16.4 mpg. The population standard deviation for the competitor is known to be 0.6 mpg. Test the company's claim at the 0.02 level of significance. Let the car company's SUVs be Population 1 and let the competitor's SUVS be Population 2. Step 1 of 3: State the null and alternative hypotheses for the test. Fill in the blank below. Họ :-1 = M2 Ha М2A car company claims that its new SUV gets better gas mileage than its competitor's SUV. A random sample of 49 of its SUVs has a mean gas mileage of 17.1 miles per gallon (mpg). The population standard deviation is known to be 1.3 mpg. A random sample of 33 competitor's SUVS has a mean gas mileage of 16.3 mpg. The population standard deviation for the competitor is known to be 1.6 mpg. Test the company's claim at the 0.05 level of significance. Let the car company be Population 1 and let the competitor be Population 2. Step 2 of 3: Compute the value of the test statistic. Round your answer to two decimal places.A study by the department of education of a certain state was trying to determine the mean SAT scores of the graduating high school seniors. The study examined the scores of a random sample of 136 graduating seniors and found the mean score to be 474 with a standard deviation of 112. Determine a 95% confidence interval for the mean, rounding all values to the nearest tenth.
- A nurse is interested in the amount of time patients spend exercising per day. According to a recent study, the daily workout time per adult follows an approximately normal distribution with a mean of 94 minutes and a standard deviation of 27minutes. If the nurse randomly samples patients in her office to analyze their exercise time and gets a standard error of 3minutes, how many patients did she sample?A study of a local high school tried to determine the mean number of text messages that each student sent per day. The study surveyed a random sample of 114 students in the high school and found a mean of 195 messages sent per day with a standard deviation of 73 messages. Determine a 95% confidence interval for the mean, rounding all values to the nearest whole number.Scores on a standarized exam are known to follow a normal distribution with a standard deviation of 5. A researcher finds a random sample of 25 exam scores has a mean score of 72. Use this to find a 88% confidence interval for the mean exam score of all students taking the exam.
- A random sample of 25 brand A cigarettes shoed an average nicotine content of 5 milligram, while a sample of 40 brand D cigarette showed an average nicotine of 4.8 milligram. If the standard deviation of nicotine is 1.6 milligrams, would you say that brand D has a lesser nicotine content? Use a 0.01 level of significance. Assume the distribution of nicotine content to be normal.A recent study of 28 randomly selected employees of a company showed that the mean ofdistance they traveled to work was 14.3 miles. The standard deviation of the sample was 2.0miles. Find the 95% confidence interval of the true mean.Rhys is a quality control manager at a facility that manufactures snack foods. He is interested in the number of whole mini- pretzels that are in the 10 oz bags in the latest lot produced. Rhys selects 24 bags of mini-pretzels at random from the latest lot and counts the number of pretzels in each bag His sample has a mean of 158 pretzels with a standard deviation of 1.6 pretzels. What is the margin of error with 95% confidence (2 = 2.07)? O 0.02 O 66.76 O 0.68 O 6.34