n a survey of 3484 adults aged 57 through 85 years, it was found that 84.7% of them used at least one prescription medication.   Complete each of the following.  (a) Construct a 90 % confidence interval of the percentage of adults aged 57 through 85 years who use at least one prescription medication.  {Hint: You may use the Calc function STAT>Tests>1-PropZintvl; MAKE SURE YOU ENTER X AS A WHOLE NUMBER}  (b) Find the margin of error. {Hint: Right side of the inequality minus the left side = 2E.} Answer: CI = Blank 1  % < p <  Blank 2  % ; ( to one decimal place). (Note: the "units" are %.)   Answer: The margin of error is: Blank 3 _____________________ % . to 1 dec. places.

Q: In general, individuals ages 9 to 50 should have a sodium intake of 1500 mg (milligrams) per day to…

A: The objective of the question is to interpret the given 90% confidence interval for the mean intake…

Q: A random sample of 423 adult women showed that 64% are currently married. Another sample of 310…

A: Solution: From the given information,

Q: PT. “ABC” in 2018 offers a new portfolio for investment. For this new product, companies need to…

A: Given that :  Older people (x1) = 90 Younger people ( x1) = 70 each of which was taken as many as…

Q: The 95% confidence interval for the difference in the proportions of people who make donations if…

A: Given : n1 = 200 x1 = 56 n2 = 200 x2 = 36 confidence level,c=0.95

Q: A random sample of 300 high school students in a particular town showed that 165 smoke on a regular…

A: Given Information: Sample size (n) = 300  Proportion of students who smoke (p) = 165/300 = 0.55

Q: A study was done to determine whether there is a relationship between snoring and the risk of heart…

A: a. From the given information, x1=89, n1=1105, x2=22 and n2=1379.

Q: Is 18* 1/3= greater then 1/3 ?

A: Given : We have to check whether 18*1/3 is greater than 1/3

Q: A random survey of 70 male sophomores showed 16 preferred off campus living and out of 63 female…

A:

Q: Suppose a TA wanted to estimate the proportion of students who failed Stat 119 for the 980 students…

A: Obtain the point estimate for the true proportion of students who failed the course. The point…

Q: Insurance companies are interested in knowing the population percent of drivers who always buckle up…

A:

Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 200…

A: The objective of this question is to construct a 90% confidence interval for the proportion of adult…

Q: confidence interval for the proportion of unemployed adults who have been unemployed for more than…

A: Here given  In a survey of 532 unemployed adults, 225 said that they had been unemployed for more…

Q: I want to estimate the average time it takes a tomato plant to bear fruit. I randomly sample 16…

A: Given,sample size(n)=16mean(x¯)=36standard deviation(σ)=17α=1-0.85=0.15α2=0.075Z0.075=1.4395…

Q: I want to construct a confidence interval (CI) for estimating the population mean of IQ scores. I…

A: As per the solution the sampled observations are n=65

Q: random sample of 500 Jordanian firms, it was found that 420 Jordanian firms made a profit, The…

A: Given that  Sample size (n) = 500 Films made profit (x) = 420 Sample proportion (p) = 420500=0.84…

Q: Ten car drivers were asked to parallel park 2 cars having very different wheel bases and turning…

A: Sample sizen=10 The 90% confidence interval for the mean differences is [-2, 5].   At the 90% level…

Q: A surprising number of young adults (ages 19 to 25) still live in their parents’ homes. A random…

A: Given  X1=986, n1=2253, X2=923, n2=2629

Q: Among a random sample of 400 restaurant workers, we find that 6.25% of workers have been working at…

A:

Q: A recent study of 2700 children randomly selected found 21​% of them deficient in vitamin D. ​…

A: Given that, 2700 children randomly selected found 21​% of them deficient in vitamin D. n=2700p^=0.21…

Q: A survey of 80 recent fatal traffic accidents showed that 50 were alcohol-related.  Find the 84%…

A:

Q: An investigator conducts a study to investigate whether there is a difference in mean PEF in…

A:

Q: people from Tanjung Malim has been given cholesterol tests, and 60 of these people had levels over…

A: We have given that,  Favorable cases (X) = 60 and sample size (n) = 150 Then,  We will find the 95%…

Q: A survey of 100 customers showed that 36 of them preferred brand B of a particular product. Give a…

A: p = proportion of people vote for BSample size : n = 100Number of favorable outcomes : x =…

Q: We wish to estimate what percent of adult residents in a certain county are parents. Out of 600…

A: Obtain the 95% confidence interval for the true proportion of adult residents who are parents in…

Q: Joe’s Pizza Joint found that 64.2% of the 328 pizzas sold last week included pepperoni. Find a98%…

A: Obtain the 95% confidence interval for the proportion of Joe’s customers that order pepperoni. The…

Q: A Hollywood film producer is working on a new action film. She wants to estimate the true mean…

A: Confidence interval for mean gives the probabilit that the parameter lies between the two values…

Q: The drug OxyContin is used to treat pain, but it is dangerous because it is addictive and can be…

A: Given,n=278x=76p^=xnp^=76278=0.2731-p^=1-0.273=0.727α=1-0.90=0.1α2=0.05Z0.05=1.645  (from Z-table)

Q: Random samples were taken of the number of daily visitors to Goldstream Park and French Beach…

A: Here, alternative hypothesis is greater than type.

Q: The Pew Research Center polled n = 1048 U.S. drivers and found that 69% enjoyed driving their…

A: Given Information: Sample size n=1048 69% enjoyed driving their automobiles.

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

A:

Q: mong 1815 randomly selected high school students surveyed by the Centers for Diseases Control, 300…

A: Given : Total number of high school students = 1815 Number of students said they were current…

Q: A drug manufacturer conducted a study to try to determine if the proportion of elderly patients that…

A: The 95% confidence interval for the difference of proportions (elderly less younger adults) was…

Q: A laboratory tested 82 chicken eggs and found that the mean amount of cholesterol was 228 milligrams…

A: Given ,x=228σ=19n=82The z-value at 95 % confidence is 1.96

Q: A random sample of 100 employees at a company shows that 42 drive US made automobiles (the rest…

A: A random sample of 100 employees at a company shows that 42 drives US made automobiles  p = Sample…

Q: A researcher collected sample data for 16 middle-aged women. The sample had a mean serum cholesterol…

A: Given , Sample size , n = 16Sample mean , 191.7standard deviation , s = 8confidence interval  = 95%…

Q: A sample of 421 adults showed that 239 of them are connected to the Internet from home.  Construct…

A: From the provided information, Sample size (n) = 421 From which 239 are connected to the Internet…

Q: A survey found that 611 out of 663 college graduates cannot correctly spell the word aluminum. Find…

A: Given that  A survey found that 611 out of 663 college graduates cannot correctly spell the word…

Q: laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 206…

A:

Q: A data set includes 103 body temperature of healthy adult humans having a mean of 98.1F and standard…

A: It is given that sample mean is 98.1 and the standard deviation is 0.56.

Q: A recent study by the McGraw-Hill company showed that 61% (1621) of college students use their…

A: Given, Proportion of college students use their smartphones for study = 0.61Total number of students…

Q: Does this interval give the quality control inspector reason to believe that the mean amount of…

A: The 95% confidence interval for the true mean amount of juice dispensed per bottle is (296.4,…

Q: A random sample of prices of petrol per litre (in p per litre) from UK filling stations was taken…

A: Given : 1 litre =0.22 gallons 95% CI for mean price of petrol per litre is (131.42 , 142.22)

Q: The American Heart Association is about to conduct in anti smoking campaign and wants to know the…

A: Solution: Out of a sample of 1421 Americans over 35, 412 smoke.  Sample size: n = 1421 The number of…

2-Proportion Test

In a 2-proportion test, two separate proportions are measured at the same time, and their similarity is evaluated. It is also known as a 2-proportion z-test, and a comparison of two different proportions is done in this method.