Dyan Jones kept careful records of the fuel efficiency of his new car. After the first 7 times he filled up the tank, he found the mean was 20.9 miles per gallon (mpg) with a sample standard deviation of 0.7 mpg.  a) Compute the 95% confidence interval for his mpg. (Use T distribution table) (Round your answers to 3 decimal places)  b) How many times should he fill his gas tank to obtain a margin of error below 0.10 mpg? (Use Z distribution table) (Round your answer to the next whole number).

Q: OMPUTE FOR THE MARGIN OF ERROR OF THE FOLLOWING AND SHOW THE STEP-BY-STEP SOLUTION. 1. Confidence…

A: Steps Solution Step1:Write given data The confidence interval is 90%,σ=5,n=50 Step2:Find…

Q: A parenting magazine reports that the average amount of wireless data used by teenagers each month…

A:

Q: The distribution of scores on a standardized test is approximately Normal with a standard deviation…

A: According to the provided information, Mean (x̅) = 76.8 Standard deviation (σ) = 22 Sample size (n)…

Q: AT scores are distributed with a mean of 1,500 and a standard devieation of 287. You are interested…

A: Given that, Critical value: Using the z-table, the critical value at 85% confidence level is ,

Q: A doctor wants to estimate the mean HDL cholesterol of all 20-29 year old females. How many subjects…

A: formula for the sample size needed to estimate a population mean is

Q: An important factor in selling a residential property is the number of people who look through the…

A: Since the sample size is small and population standard deviation is unknown, t distribution can be…

Q: Solve the following. Show your complete solution. Fifty randomly selected jeepneys were stopped, and…

A:

Q: A doctor wants to estimate the mean HDL cholesterol of all​ 20- to​ 29-year-old females. How many…

A: The formula for the sample size is as follows:

Q: A taxi company claims that its drivers have an average of at least 12.4 years' experience. In a…

A:

Q: you want to obtain a sample to estimate a population mean. based on previous evidence, you believe…

A:

Q: #1 The mean cholesterol level of a random sample of 100 adults aged 45-54 was 216 milligrams with a…

A: Givensample size(n)=100Mean(x)=216standard deviation(s)=19.5confidence level=98%

Q: e chips per cookie for Big Chip cookies? Round your answers to one decimal place.

A: Let X be the random variable such that number of chocolate chips per cookie for Big Chip cookies x̅:…

Q: (T/F) To choose a sample size before estimating a proportion, we need to know the desired confidence…

A: The objective of the question is to determine whether the desired confidence level is necessary when…

Q: Noah wants to advertise how many chocolate chips are in each Big Chip cookie at his bakery. He…

A:

Q: estimate the mean HDL cholesterol of all​ 20- to​ 29-year-old females. How many subjects are needed…

A: The sample size can be calculated as: n=z×σE2 Here, z = desired confidence level E = The allowable…

Q: An economist would like to estimate the average household income in Martin county. The mean…

A: We have given that,  Sample mean (x̄) = $ 52590, population standard deviation (σ ) = $ 864 and…

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

A:

Q: Find a 90% confidence interval for the population mean age of the Smurfs if a sample of 52 smurfs…

A: It is given that, There is a random sample of size 52 and has a mean of 49.7 and the standard…

Q: In a certain population, body weights are normally distributed with a mean of 152 pounds and a…

A: The number of people surveyed is 1271 and it is obtained below: Let us define the random variable X…

Q: A random sample of gasoline prices in 30 gas stations gives the following statistics: y =$1.29, s…

A: Given  n=30, s=0.16, ȳ=1.29  Find 95% CI for population mean μ

Q: A study of 77 golfers showed that their average score on a particular course was 93. The st.dev. of…

A: 1). Assume, μ  be the population mean golfers score. x be the random variable representing the…

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

A:

Q: On average, the level of radioactivity was 198.2 gammas. The standard deviation was 3.3 gammas.…

A: Given information Sample size n = 10 Sample mean x̅ = 200 Population standard deviation = 3.3…

Q: You plan to build a 95% confidence interval estimate of the mean of a large population. You want a…

A: Givenconfidence level=95%Margin of error(E)=10standard deviation(σ)=200

Q: Females a.Give 95% Confidence Interval (CI) for the mean males grades? b.Give 95% CI for the…

A: It is given asFor part (a):Sample mean = 7.4Sample variance = 3.3Sample std. deviation = 1.342Sample…

Q: A nutritionist wants to estimate the mean amount of cholesterol in a certain variety of chicken…

A: Given Data: Level of significance=0.01  σ=25mg ME=10   To Find: 'n'

Q: People with high blood pressure suffer from hypertension. Data on fasting blood sugar…

A: Given: Data on fasting blood sugar (milligrams/decilitre) and magnesium (milligrams/decilitre) in…

Q: Heights for teenage boys and girls were calculated. The mean height for the sample of 73 boys was…

A: given data, n1=73x¯1=167s12 =71s1=8.4261n2=47x¯2=180s22 =66s2=8.124

Q: Heights for teenage boys and girls were calculated. The mean height for the sample of 47 boys was…

A: The sample data for boys is given as:Sample mean: Sample variance: Sample size: The sample data for…

Q: Heights for teenage boys and girls were calculated. The mean height for the sample of 50 boys was…

A:

Q: COMPUTE FOR THE MARGIN OF ERROR OF THE FOLLOWING AND SHOW THE STEP-BY-STEP SOLUTION. 1. Confidence…

A: The confidence level is 99%. The standard deviation of the data is 25. The sample size of the data…

Q: Mrs. Gallas wanted to estimate the average weight of an Oreo cookie to determine if the average…

A: In statistical inference, there are two types of estimation,  Point estimation and Interval…

Q: Heights for teenage boys and girls were calculated. The mean height for the sample of 51 boys was…

A: given data x¯1  = 183s12 = 66n1 = 51x¯2 =173s22 = 61n2 = 7187% ci for  μ1 - μ2α = 1-0.87 = 0.13

Estimation

A new framework of data analysis where the new statistical methods are used for interpreting the results and analyzing the data is known as estimation in statistics.