A random sample of 29 glass sheets is obtained and their thicknesses are measured. The sample mean is x= 3.12 mm and sample standard deviation is S = 0.14 mm. Construct a 95% two-sided confidence interval for the mean glass
Q: een asked by Forest Hill Elementary School to evaluate the school's drinking water. The lab tested…
A: We have, x¯=sample mean=8.43s=sample standard deviation=1.12n=sample size=24Confidence…
Q: A random sample of 21 Camden County College students had a mean age of 24 years, with a standard…
A:
Q: The average length of an elephant's trunk from a population of African forest elephants is unknown.…
A:
Q: A random sample of 174 products is selected and sample standart deviation are found to be 32 cm and…
A: Let a random variable X denotes the length of the product. Then, the sample mean and the sample…
Q: The Sugar Producers Association wants to estimate the mean yearly sugar consumption. A sample of 16…
A:
Q: In a manufacturing process, a random sample of 9 bolts has a mean length of 3 inches with a sample…
A:
Q: A sample of only 8 adult rhinos had an average weight of 2,500 pounds. The standard deviation for…
A:
Q: The average height of a population is unknown. A random sample from the population yields a sample…
A: From the provided information, the sample mean and the standard deviation of the sampling…
Q: A researcher collected sample data for 18 women ages 18 to 24. The sample had a mean serum…
A:
Q: Lifetimes of AAA batteries are approximately normally distributed. A manufacturer wants to estimate…
A: Given that the lifetime of the AAA batteries is approximately normally distributed.The sample size…
Q: A test to measure the gullibility of college students is designed to have a mean of 70 and a…
A:
Q: Find the sample size necessary for a 99% confidence level with maximal margin of error E = 2.00 for…
A: Solution
Q: A random sample of 100 NBA women players had a mean height of x¯=69.5in and a standard deviation…
A: The sample size is 100, sample mean is 69.5, and sample standard deviation is 3.3.
Q: A data set includes 103 body temperatures of healthy adult humans having a mean of 98.5°F and a…
A: According to the given question A data set includes 103 body temperatures of healthy adult humans…
Q: The firm wants to estimate the average weight of a sugar package with an accuracy of 0.5 gram and a…
A: We want to find the sample size (n)
Q: To determine the average number of minutes it takes to manufacture one unit of a new product, an…
A: Given information Sample size (n) = 15 Sample mean x̅ = 3.92 Standard deviation (s) = 0.45 Taking…
Q: Arandom sample of 25 observations selected from this population gave a mean equal to 143.72. The…
A: We have given that Mean = 143.72, Population standard deviation = 14.8, Sample size n = 25…
Q: A simple random sample of 101 body temperatures have a mean of 98.20°F and a standard deviation of…
A:
Q: The average zinc concentration recovered from a sample of measurements taken in 40 different…
A:
Q: The average metal concentration recovered from a sample of metallic measurements in 36 sites is…
A:
Q: A sample of 5 water speciments selected for treatment found a sample mean of arsenic concentration…
A: Solution: Let X be the arsenic concentration. From the given information, x-bar=24.3, s=4.1 and n=5.…
Q: A simple random sample from a population with a normal distribution of 106 body temperatures has…
A: Given that, A simple random sample from a population with a normal distribution of 106 body…
Q: In a random sample of twelvetwelve people, the mean driving distance to work was 24.624.6 miles and…
A: The given values are:Sample size (n) = 12Sample mean (x̄) = 24.6 milesSample standard deviation (s)…
Q: The length of the skulls of 10 fossil skeletons of an extinct species of bird has a mean of 5.68 cm…
A:
Q: The total of individual weights of garbage discarded by 16 households in one week has a mean of 35.7…
A: Given information- Sample size, n = 16 Significance level, α = 0.01 Sample mean, x-bar = 35.7 lbs…
Q: The diameter of holes for a cable harness is known to have a normal distribution with a standard…
A: Given that The diameter of holes for a cable harness is known to have a normal distribution with a…
Q: The average zinc concentration recovered from a sample of measurements taken in 40 different…
A:
Q: Over a period of months, an adult male patient has taken nine blood test fir uric acid. The mean…
A: The sample mean is given as 5.29. The sample standard deviation is 1.77. The sample size, n is 9.…
Q: Mechanical Engineering
A: Where, n is the sample size E is the margin of error σ is the standard deviation
Q: A simple random sample from a population with a normal distribution of 97 body temperatures has…
A:
Q: In a random sample of 100 male students the average height is 68.2 with standard deviation 2.5 .…
A: The summary of statistics is, x¯=68.2,s=2.5,n=100 The degree of freedom is, df=n-1 =100-1 =99…
Q: a) A sample of 10 measurements of the radius of a marble gave a mean of 4.38mm and a standard…
A:
Q: standard deviation was estimated to be 11.2 µg m-³. a. Determine the 95% confidence interval for the…
A: xbar = 41.5 s = 11.2 n= 30
Q: In a random sample of 21 people, the mean commute time to work was 33.6 minutes and the standard…
A: Given : For n=21 people in the sample, sample mean and sample standard deviation of commute time to…
Q: The average metal concentration recovered from a sample of metallic measurements in 36 sites is…
A: The random variable metallic concentration in the liver follows normal distribution. We have to…
Q: The average weight of 25 chocolate bars selected from a normally distribute population is 200g with…
A: Given: n=25x¯=200s=10α=0.05
Q: A simple random sample from a population with a normal distribution of 104 body temperatures has…
A:
Q: A random sample of n = 100 frames of 100mm x 50mm was selected from a consignment of 500 frames.…
A: 1). The sample mean is given as 106 mm. The sample standard deviation is 3.5 mm. The sample size, n…
Q: A researcher collected sample data for 20 women ages 18 to 24. The sample had a mean serum…
A:
Q: A sample of size 250 has mean 57.1 and standard deviation 11.8. (a) find the standard error of the…
A:
Q: The firm wants to estimate the average weight of a sugar package with an accuracy of 0.7 grams and a…
A: Population standard deviation is 2.5. The z-critical value for 98% confidence interval is 2.33 using…
Q: A beverage company uses a machine to fill half-gallon bottles with juice. The company wants to…
A: In order to construct a confidence interval, the three basic information we needed are as follows,…
Q: A city planner wants to estimate the average monthly residential water usage in the city. He…
A:
Q: In a study of an Oil field, 100 permeability measurements have been usedand resulted in a normal…
A: Level of significance: The level of significance or α is the probability of rejecting H0 when, in…
Q: mple of 20 tortillas made by a specific tortilla maker had a mean diameter of 243mm with a sample…
A:
Q: What is the minimum sample size required to estimate the population mean (mu) with 90% confidence if…
A: The margin of error for population mean with confidence level 90%, sample size n and standard…
Q: 8 holes drilled on a lathe were measured and its variance was found to be 2.81 cm. What is the…
A: Given: s2=2.81n=8 The degrees of freedom is obtained as below: df=n-1=8-1=7 The critical value is…
A random sample of 29 glass sheets is obtained and their thicknesses are measured. The sample mean is x= 3.12 mm and sample standard deviation is S = 0.14 mm. Construct a 95% two-sided confidence interval for the mean glass thickness
Step by step
Solved in 3 steps
- In a random sample of 21 people, the mean commute time to work was 33.9 minutes and the standard deviation was 7.1 minutes. Assume the population is normally distributed and use a t-distribution to construct a 80% confidence interval for the population mean u. What is thr margin of error of u? Interpret the results.We wish to give a 95% confidence interval for the mean time it takes to complete a routine assembly task. to this end we obtain a random sample of 100 assembly times. We obtain a sample mean value of 15 minutes with a sample standard deviation of 5 minutes. Give the point estimate of the population average assembly time.The pH of rain, measured at a weather station in Michigan, was observed for 39 consecutive rain storms. The sample mean is 4.6982 and the sample variance is 0.39623. Obtain a 99% confidence interval for the mean pH of the population of storms at that location.
- The quality control manager at a light bulb factory needs to estimate the mean life of a large shipment of light bulbs. The standard deviation is 98 hours. A random sample of 49 light bulbs indicated a sample mean life of 370 hours. Complete parts (a) through (d) below. ..... a. Construct a 95% confidence interval estimate for the population mean life of light bulbs in this shipment. The 95% confidence interval estimate is from a lower limit of hours to an upper limit of hours. (Round to one decimal place as needed.) b. Do you think that the manufacturer has the right to state that the lightbulbs have a mean life of 420 hours? Explain. Based on the sample data, the manufacturer the right to state that the lightbulbs have a mean life of 420 hours. A mean of 420 hours is standard errors the sample mean, so it is that the lightbulbs have a mean life of 420 hours. c. Must you assume that the population light bulb life is normally distributed? Explain. A. Yes, the sample size is not large…Suppose average pizza delivery times are normally distributed with an unknown population mean and a population standard deviation of six minutes. A random sample of 28 pizza delivery restaurants is taken and has a sample mean delivery time of 36 minutes. Find a 90% confidence interval estimate for the population mean delivery time.A researcher collected sample data for 16 women ages 18 to 24. The sample had a mean serum cholesterol level (measured in mg/100 mL) of 192.4, with a standard deviation of 9.6 Assuming that serum cholesterol levels for women ages 18 to 24 are normally distributed, find a 90% confidence interval for the mean serum cholesterol level of all women in this age group. Then find the lower limit and upper limit of the 90% confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place. Lower level: Upper Level:
- Blood sugars from eight patients in a rural clinic were recorded last summer by the local physician. The blood sugars results were: 120, 145, 200, 250, 79, 100, 894, 255 mg/d1. The standard deviation was 5. Calculate the 95% Confidence Interval (CI) of the mean blood sugars.A recording company wants to estimate the mean length of musical cd’s being recorded. A random sample of 28 cd’s has a mean length of 48.47 minutes and standard deviation of 13.9 minutes. Use this sample data to construct the 95% confidence interval for the population mean length of cd’s. Show lower and upper values for interval using two-decimal accuracy.In a random sample of 50 male students the average height is 68.2 with standard deviation 2.5 . Obtain 99% confidence interval for the mean height of all students.
- The mean number of the sample of 16 bolts produced on a specific machine per day was found to be 42 with a standard deviation of 2. Assume that the number of bolts produced per day on this machine has a normal distribution. Construct a 95% confidence interval for the population mean A tyre manufacturer claims that its tyres have a mean life of 50000 kms. A random sample of 16 of these tyres is tested and the sample mean is 33 000 kms. Assume the populations standard deviation is 3000kms and the lives of tyres are approximately normally distributed. To test the manufacturer’s claim using the 5% level of significance the analyst shouldIn a random sample of twelvetwelve people, the mean driving distance to work was 24.624.6 miles and the standard deviation was 4.54.5 miles. Assume the population is normally distributed and use the t-distribution to find the margin of error and construct a 9999% confidence interval for the population mean muμ. Interpret the results.