Given,
p = 35% = 0.35
n = 37
a)
It is possible to use the normal approximation to Binomial distribution because n*p =37*0.35 12.95 , which is > 5 and n*q =37*(1-0.35) = 24.05 , which is > 5
b)
= p = 0.35
32. Do you have a great deal of confidence in the advice given to Suppose a...
1) Workers at a large toxic cleanup project are concerned that their white blood cell counts may have been reduced. Let x be a random variable that represents white blood cell count per cubic millimeter of whole blood in a healthy adult. Then μ = 7500 and σ ≈ 1750.† A random sample of n = 50 workers from the toxic cleanup site were given a blood test that showed x = 6760. Compute P(x ≤ 6760). (Round your answer...
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion p. Round to three decimal places. Of 89 adults selected randomly from one town, 66 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance.
According to a study conducted in one city,
32%
of adults in the city have credit card debts of more than
$2000. A simple random sample of
n=200
adults is obtained from the city. Describe the sampling
distribution of
p,
the sample proportion of adults who have credit card debts of
more than $2000. Round to three decimal places when necessary.
Awnser this A B C D pick the correct awnser and make sure it
correct I put it in...
Calculate the margin of error and construct a confidence interval for the population proportion using the normal approximation to the p̂ p̂ -distribution (if it is appropriate to do so). Standard Normal Distribution Table a. p̂ =0.85, n=140, α =0.2 p̂ =0.85, n=140, α =0.2 E=E= Round to four decimal places Enter 0 if normal approximation cannot be used < p < < p < Round to four decimal places Enter 0 if normal approximation cannot be used b. p̂ =0.3, n=160, α =0.2 p̂ =0.3, n=160, α =0.2...
According to a survey in a country, 34% of adults do not own a credit card. Suppose a simple random sample of 300 adults is obtained. Complete parts (a) through (d) below. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2) the sample proportion of adults who do not own a credit card. Choose the phrase that best describes the shape of the (a) Describe the sampling...
I need help answering this
question
Confidence in banks: A news poll conducted in 2012 asked a random sample of 1268 adults in the United States how much confidence they had in banks and other financial institutions. A total of 144 adults said that they had a great deal of confidence. An economist claims that less than 11% of U.S. adults have a great deal of confidence in banks. Can you conclude that the economist's claim is true? Use both...
A random sample of 320 medical doctors showed that 180 had a solo practice. (a) Let p represent the proportion of all medical doctors who have a solo practice. Find a point estimate for p. (Use 3 decimal places.) (b) Find a 98% confidence interval for p. (Use 3 decimal places.) lower limit upper limit Give a brief explanation of the meaning of the interval. 98% of the all confidence intervals would include the true proportion of physicians with solo...
A random sample of 1002 adults in a certain large country was asked "Do you pretty much think televisions are a necessity or a luxury you could do without?" Of the 1002 adults surveyed, 516 indicated that televisions are a luxury they could do without. Complete parts (a) through (e) below. Click here to view the standard normal distribution table (page 1). LOADING... Click here to view the standard normal distribution table (page 2). LOADING... (a) Obtain a point estimate...
A random sample of 1028 adults in a certain large country was asked "Do you pretty much think televisions are a necessity or a luxury you could do without?" Of the 1028 adults surveyed, 527 indicated that televisions are a luxury they could do without. Complete parts (a) through (e) below. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). (a) Obtain a point estimate for the...
a.) Use the normal distribution to find a confidence interval for a proportion p given the relevant sample results. Give the best point estimate for p, the margin of error, and the confidence interval. Assume the results come from a random sample. Answer all parts please, not just the confidence interval. A 95% confidence interval for p given that p-hat=0.75 and n=100. Round your answer for the point estimate to two decimal places, and your answers for the margin of...