According to the Centers for Disease Control and Prevention, 33.5% of U.S. adults have high LDL, or “bad” cholesterol. Suppose we sample 240 adults, out of which 81 have high LDL. a.  Based on our sample, what is our estimate for the population proportion of adults with high LDL? b. What is the standard error for our estimate? c. Using your answers for parts a and b, construct a 95% confidence interval to estimate the population proportion. Can we conclude that the percentage of those that have high LDL is higher than 33.5%? d. Test the CDC’s claim that the percentage of those that have high LDL is higher than 33.5% at the α = 0.05 level. Report your p-value and conclusion.

Q: A large family-held department store had the business objective of improving its response to…

A: Given that the variable is “Number of days between complaint and resolution of complaint”. The data…

Q: Here are summary statistics for randomly selected weights of newborn​ girls: n=232​, x=32.3 ​hg,…

A: Given, Sample Size(n)=232 Sample Mean=32.3 Standard Deviation=6.5 Confidence Level=90%…

Q: The Centers for Disease Control reported the percentage of people 18 years of age and older who…

A: a) The sample size can be estimated as follows: n=ze2p(1-p)where, z = z scoree = errorp =…

Q: If n = 220,p = 0.4,then what will be the 87.4% confidence interval for the population proportion?

A:

Q: A. Determine the mean and standard deviation of the total thickness of the two halves. B. What is…

A: plastic casing for a magnetic disk is composed of two halves. The thickness of each half is normally…

Q: ool graduates, and ask them their annual incomes. They give you an average income of $35,000 per…

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

Q: If n = 370 and p=0.31p, construct a 90% confidence interval. This should be rounded to three decimal…

A:

Q: Suppose we want to estimate the proportion of teenagers (aged 13-18) who are lactose intolerant. If…

A:

Q: Suppose you want to construct a confidence interval with a margin of error of 0.03 (3%). Find the…

A: The formula for calculating sample size is, n = z2*p^(1-p^)MOE2 Where, z is the z score (default…

Q: If n=18, ¯x=x¯=34, and ss=15, construct a confidence interval at a 98% confidence level. Assume the…

A:

Q: A developmental psychologist would like to determine how much fine motor skill changes for children…

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

Q: A​ 95% confidence interval for the proportion of U.S. adults who subscribe to some sort of streaming…

A: In the above question, we are given,  a confidence interval is

Q: According the the U.S. Census bureau, the percentage of Santa Ana residents living in poverty is…

A: Given: p^1=0.177p^2=0.133n1=1200n2=1500

Q: If α = 0.05, then what is the confidence level?

A: Given: α = 0.05   we know that, Confidence level+ α = 1

Q: On September 14, 2000, a cross-sectional survey of 24,000 residents of a western state revealed that…

A: Given: n = 24000 Number of tuberculosis cases X = 145 Proportion p^ = Xn Formula Used: Confidence…

Q: A sample of 700 from population A has a proportion of interest of 0.24 while a sample of 600 from…

A: Confidence Interval: The form of the confidence interval is, CI = Point estimate ± Margin of error…

Q: ccording to a past poll conducted of Lexington residents, 52% indicated support for a light rail…

A: Given: Sample proportion p^=52%=0.52 Sample size (n) = 400

Q: You are trying to estimate the percent of people in California who have a college degree. You…

A: According to the given information in this question We need to identify that from the given…

Q: What is stated by the alternative hypothesis (H1) for an ANOVA? ​a. There are no differences…

A: We have given that the statement about to the alternative hypothesis of ANOVA.

Q: I am struggling setting this problem up. Any help would be greatly appreciated! For this…

A: As the information is said to be assumed, so I assumed some values.

Q: In a study of government financial aid for college students, it becomes necessary to estimate the…

A: Solution: From the given information, confidence level is 0.99 and E=0.05.

Q: If the alpha level is .05, then the confidence level is ?

A: Given: alpha = 0.05 confidence level / significant level = 1 - alpha

Q: An insurance company collects data on seat-belt use among drivers in a country. 90% confidence…

A: Givenn1=2000p1^=28%=0.28P2^=xn2=4821500=0.32α=1-0.90=0.10α2=0.05zc=zα2=z0.05=1.6449      (from z…

Q: A study compares the difference in proportions of those who can taste Phenylthiocarbamide (PTC)…

A: Note - “Since you have posted multiple questions with multiple sub parts, we will provide the…

Q: According to the Centers for Disease Control and Prevention, 33.5% of U.S. adults have high LDL, or…

A: The hypothesized percentage (or proportion) of US adults with LDL, p0=33.5%.The sample size, i.e.,…

Q: What Z score would be used to construct a 90% confidence level? Is it ... a. 1.65 b. 2.33 c.…

A: Confidence Level = 90 Significance Level = α = (100-90)% = 0.1 Critical value = z* = 1.65   […

Q: Ve want to know how many people plan on buying a cell phone this year. Determine the sample size…

A:

Q: A sample of freshman takes a reading comprehension test and their scores are summarized below.…

A:

Q: An insurance company selected samples of clients under 18 years of age and over 18 and recorded the…

A: Given Information: Under age of 18  n1=480 x1=160 Over age of 18 n2=500 x2=140

Q: a. interpret the corresponding 95% confidence intervals for both Birth weight and Age.  b. What is…

A: Given data: x1 x2 y 135 3 89 120 4 90 100 3 83 105 2 77 130 4 92 125 5 98 125 2 82…