Answered: he average weight of 20 randomly…

he average weight of 20 randomly selected cars was 3500 pounds. The sample standard deviation was 315 pounds. Find the 90% confidence interval of the true population mean weight of the c

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Q: A random group of 36 people in Toms River were asked “how much money did you spend at the grocery…

A: Given: Sample mean x¯=149.50 Sample size (n) = 36 Sample standard deviation (s) = 22.50 Confidence…

Q: Find a 90%confidence intervalfor thế populadofn stan

A: (1-α)100% confidence interval for σ2 is First…

Q: y selected textbooks' weights, and find they have a mean weight of 78 ounces. Assume the population…

A: As the population standard deviation is known, we will use z distribution. The value of z at 95%…

Q: You measure 41 watermelons' weights, and find they have a mean weight of 47 ounces. Assume the…

A: Given that: n= 41.0 , x̄= 47.0 s= 2.5 At 90.0 % confidence level , z score= 1.645 So, 90.0 %…

Q: Assume you are going to test the equality of population means in a paired samples design against a…

Q: The standard deviation of the weights of elephants is known to be approximately 15 pounds. We wish…

A: Sample mean = x̅ = 244 Sample size = n = 50 Population S.D = σ = 15 Standard Error = σ/√n =…

Q: The following sample data are from a normal population: 10, 8, 12, 15, 13, 11, 6, 5. What is the…

A: Given data is10,8,12,15,13,11,6,5sample size(n)=8sample mean(x¯)=10standard…

Q: A sample of 56 Charleston County households have a mean income of $27,519 with a standard deviation…

Q: A car repair business randomly selects receipts for parts purchased on 16 different jobs. The mean…

A: given data, n=16x¯=31.32s=30.42CI=0.95α=1-0.95=0.05df=n-1=16-1=15we have to construct the 95%…

Q: Construct a confidence interval for the mean of the paired differences for the two populations with…

A: Solution: From the given information, n=32, d-bar=1.358, α=0.05 and sd=0.635.

Q: 4. A recent study of 19 students of San Rafael College showed that the mean of the distance they…

Q: A recent study shows that out of 5, 000 people who moved to Arizona, 2, 350 did so to take a new…

A: From the given information, sample size n = 1000. Total number of people moved n(S) = 5000. Number…

Q: The weights, in pounds, of a small group of randomly selected dogs are 34, 37, 42, 57, 64, ar…

A: We have given that, The data set are :- 34, 37, 42, 57, 64 Then, We will find the sample standard…

Q: A sample of 47 Charleston County households have a mean income of $32,346 with a standard deviation…

A: The sample size is 47, sample mean is $32346, and sample standard deviation is $8571. The confidence…

Q: Pyramid Lake is on the Paiute Indian Reservation in Nevada. The lake is famous for cutthroat trout.…

A: Given information Hypothesized mean µ = 19 inches Sample size (n) = 49 Mean x̅ = 18.5 inches…

Q: The population standard deviation for the height of college hockey players is 3.4 inches. If we want…

Q: The average cost to attend a particular university is 63,551. Assume the population standard…

A: Approach to solving the question:So, with a sample size of 60 students, we can be 95% confident that…

Q: The average number of hours worked per week for college students is 27, and the standard deviation…

Q: You measure 35 textbooks' weights, and find they have a mean weight of 36 ounces. Assume the…

A: Given information- Population standard deviation, σ = 9.3 ounces Sample size, n = 35 Sample mean,…

Q: A study was commissioned to find the mean weight of the residents in certain town. The study…

A: Given: n=58, x=174 and s=38. Now,Degrees of freedom=n-1=58-1=57 Then, at 95% confidence level, the…

Q: Airline companies are interested in the consistency of the number of babies on each flight, so that…

A: Test for Variance: The χ2 Chi-Square test is used to test the significance difference between…

Q: 2. In a recent study of 35 college students, the mean number of hours per week that they played…

A: According to the given information in this question We need to find the 95% confidence interval

Q: Calculate a 95% upper confidence bound for the mean of a normally distributed population using the…

A: Given that, x = 32289 s = 2846 n=15 For 95% confidence interval,the z value is 1.96

Q: The Statistics Games are played by a large number of teams across the county. Teams raise money to…

A: We have given that, Sample mean (x̄) = $ 375, standard deviation (s) = $ 32 and sample size (n) =…

Q: ou measure 46 dogs' weights, and find they have a mean weight of 32 ounces. Assume the sample…

Q: Suppose we are contracted by the Justice Department to study the extent of juvenile delinquency. We…

Q: You measure 24 textbooks' weights, and find they have a mean weight of 53 ounces. Assume the…

A: Given that, x¯=53,n=24,σ=12.9 Critical value: By using the z-tables, the critical value at 5% level…

Q: You measure 32 textbooks weight and find they have a mean weight of 67 ounces. Assume the standard…

A: Given: Sample size (n) = 32 Mean (x̄) = 67 ounces Population standard deviation (σ) = 8.302…

Q: construct a 99% confidence interval for the true population mean textbook weight.

Q: The daily salaries of security guards for 10 security agencies are as follows: P350, P400, P375,…

A: Solution To find the point estimate we will find the sample mean.

Q: Karen wants to advertise how many chocolate chips are in each Big Chip cookie at her bakery. She…

Q: If the mean and standard deviation of a sample of size 45 is 1050 and 52 respectively, construct a…

A: It is given that Sample mean = 1050Sample standard deviation = 52Sample size = 45Confidence level =…

Q: You measure 37 textbooks' weights, and find they have a mean weight of 59 ounces. Assume the…

Q: the following random sample was selected from a normal distribution: 11, 16, 3, 13, 18, 18, 16, 19,…

A: Given data is11, 16, 3, 13, 18, 18, 16, 19, 17, 11sample size(n)=10

Q: You measure 50 textbooks' weights, and find they have a mean weight of 66 ounces. Assume the…

A: From the provided information, Sample size (n) = 50 Sample mean (x̅) = 66 ounces Population standard…

Q: You measure 41 watermelons' weights, and find they have a mean weight of 79 ounces. Assume the…

A: Mean = 79Standard deviation = 2.4Sample size = 41

Q: The lead levels in a random sample of 35 crows in the region were measured and recorded. The mean…

A: It is given that Sample size n = 35 Sample mean M = 4.50 Sample SD s = 1.12 Confidence level = 99%…

Q: For a random sample of 55 overweight men, the mean of the number of pounds that they were overweight…

A: Given data, n=55 μ=32σ=4.1Point estimate μx=?

Q: 26 randomly selected students took the calculas final. If the sample mean was 90. Construct a 95$…

Q: 73 full time workers were surveyed on how many hours they spend watching television during the week.…

A: Given information, n=73x¯=3.2s=0.2 Confidence level is 95% significance level is 5% df=n-1=73-1=72 t…

Question

The average weight of 20 randomly selected cars was 3500 pounds. The sample standard deviation was 315 pounds. Find the 90% confidence interval of the true population mean weight of the cars

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps

SEE SOLUTION Check out a sample Q&A here

Similar questions

In a sample of 100 newborn elephants, the mean weight of a newborn elephant is 244 pounds with a population standard deviation of 11 pounds. a. Find a 90% confidence interval for the mean weight of newborn elephants. Interpret your answer in a complete sentence. b. Find a 99% confidence interval for the mean weight of newborn elephants. Interpret your answer in a complete sentence.
You measure 50 textbooks' weights, and find they have a mean weight of 79 ounces. Assume the population standard deviation is 12.4 ounces. Based on this, construct a 99% confidence interval for the true population mean textbook weight .
The time patients wait for examination in a hospital fits the normal distribution and the population standard deviation is 30 minutes. The average waiting time of 75 randomly selected patients is 90 minutes. Calculate and interpret the 90% confidence interval for the population average.
You measure 24 turtles' weights, and find they have a mean weight of 34 ounces. Assume the population standard deviation is 11.7 ounces. Based on this, construct a 95% confidence interval for the true population mean turtle weight.
The Iowa Department of Transportation repaired hundreds of bridges in 1993. To check the average cost to repair a bridge, a random sample of n = 55 bridges was chosen. The mean and standard deviation for the sample are $25,788 and $1,540, respectively. Records from previous years indicate an average bridge repair cost was $25,003. Use the sample data to test that the 1993 mean μ is greater than $25,003. Which type of test is this? A. 2 tailed B. 1 tailed right C. 1 tailed left D. 0 tailed
FIND 90% CONFIDENCE INTERVAL FOR THE DIFFERENCE IN POPULATION PROPORTIONS Pf -Pm .
Calculate z scores from two samples The tallest living man at one time had a height of 252 cm. The shortest living man at that time had a height of 30.3 cm. Heights of men at that time had a mean of 170.25 cm and a standard deviation of 8.47 cm. Answer Z score for the tallest man Z-score for the shortest men
Suppose a random sample of 50 basketball players have an average height of 78 inches. Also assume that the population standard deviation is 1.5 inches. Find a 99% confidence interval for the population mean height. Find a 95% confidence interval for the population mean height. Find a 90% confidence interval for the population mean height. Find an 80% confidence interval for the population mean height.
At the local fish hatchery we take a random sample of 21 trout from one of the holding pens. The trout in the sample have a mean length of 11.2 inches with a standard deviation of .73 inches. Construct and interpret a 95% confidence interval for the mean length of all trout in this holding pen. Type your interpretation sentence in the written response box.
In a simple random sample of 144 households in a county in Virginia, the average number of children in these households was 3.62 children. The standard deviation from this sample was 2.40 children. A 90% confidence interval for the mean number of children in all households in this county is?
The size of fish is very important to commercial fishing. A study conducted in 2012 found the length of Atlantic cod caught in nets in Karlskrona to have a mean of 49.9 cm and a standard deviation of 3.74 cm. Assume the length of fish is normally distributed. A sample of 17 fish was taken. d) Find the mean of the sampling distribution of the sample mean. Put the numeric value in the first box and the correct units in the second box. e) Find the standard deviation of the sampling distribution of the sample mean. Put the numeric value rounded to two decimal places in the first box and the correct units in the second box. g) Find the probability that the sample mean length of the 17 randomly selected Atlantic cod is less than 51.5 cm. Round final answer to 4 decimal places. DO NOT use the rounded standard deviation from part e in this computation. Use the EXACT value of the standard deviation with the square root. h) Find the probability that the sample…
At a large university, data were collected on the number of sisters and brothers that each student had. Let therandom variable X represent the number of sisters and the random variable Y represent the number ofbrothers. The distribution of X has mean 1.00 and standard deviation 0.94. The distribution of Y has mean1.07 and standard deviation 1.04. What is the mean of the distribution of X+Y? WHAT IS THE CORRECT ANSWER A. 1.98 B. 2.01 C. 2.04 D. 2.0528 E. 2.07