The American Medical Association wishes to have information on the mean cost of Coronovirus test kits....
The American Medical Association wishes to have information on the mean cost of Coronovirus test kits. A random sample of 500 kits reveals a sample mean of $86,450. The standard deviation of this population is $5,050. The AMA would like answers to the following questions:
What is a reasonable range of values for the population mean?
Assume the association decides to use the 95 percent level of confidence(z=1.96).
Homework Answers
Add Answer to:
The American Medical Association wishes to have information on
the mean cost of Coronovirus test kits....
The American Veterinary Medical Association conducted a survey of veterinary clinics to estimate the proportion that...
The American Veterinary Medical Association conducted a survey of veterinary clinics to estimate the proportion that do not treat large animals (cows, horses, etc.). The survey was mailed to a random sample of 120 veterinary clinics throughout the country and of these, 88 responded that they do not treat large animals. (a) (2 points) Calculate a 95% confidence interval for the population proportion p of veterinary clinics that do not treat large animals. (b) (5 points) Typically, 70% of the...
The American Sugar Producers Association wants to estimate the annual mean sugar consumption per capita. A...
The American Sugar Producers Association wants to estimate the annual mean sugar consumption per capita. A sample of 347 individuals had a mean of 61 pounds consumed per year, with a standard deviation of 19 pounds. Construct and interpret a 90 percent confidence interval for the true population mean of annual sugar consumption per capita.
The Sugar Producers Association wants to estimate the mean yearly sugar consumption. A sample of 25...
The Sugar Producers Association wants to estimate the mean yearly sugar consumption. A sample of 25 people reveals the mean yearly consumption to be 27 kg with a sample standard deviation of 9 kg. Assume a normal population. For a 95% confidence interval, what is the critical value needed?
A survey conducted by the American Automobile Association showed that a family of four spends an...
A survey conducted by the American Automobile Association showed that a family of four spends an average of $215.60 per day while on vacation Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per day and a sample standard deviation of $79.50 Devo a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls (to 2 decimals). . Based on the...
Supplementary Exercises 385 A survey conducted by the American Automobile Association (AAA) showed that a fam-...
Supplementary Exercises 385 A survey conducted by the American Automobile Association (AAA) showed that a fam- ily of four spends an average of $215.60 per day while on vacation. Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per 45. day and a sample standard deviation of $74.50 Develop a 95% confidence interval estimate of the mean amount spent per day by a a. family of four visiting Niagara Falls....
A researcher wishes to test the claim that for a particular manufacturer of cereal, the mean...
A researcher wishes to test the claim that for a particular manufacturer of cereal, the mean weight in its boxes of cereal is less than 18 ounces. A sample of 36 boxes yields a sample mean weight of 17.88 ounces. Assume that the population standard deviation is .28 ounces. Let a =.05. a. Conduct a 5-step test of hypotheses: i. H0: Ha: ii. a=_______. iii. Test Statistic (if a t test is used, also report the df): iv. P-value: v....
1. Please test hypothesis for a mean or for a proportion. Please state null and alternative...
1. Please test hypothesis for a mean or for a proportion. Please state null and alternative hypothesis Hypothesized mean = 125, sample mean = 135, standard deviation = 15; n= 500, confidence level = 95% 2. Test Hypotheses of population percent = 33%， sample percent = 31%，n= 120， confidence level = 99%
Recent studies have shown that about 16% of American adults fit the medical definition of being...
Recent studies have shown that about 16% of American adults fit the medical definition of being obese. A large medical clinic would like to estimate what percentage of their patients are obese, so they take a random sample of 500 patients and find that 65 are obese. Suppose that in truth, the same percentage holds for the patients of the medical clinic as for the general population, 16%. Give a numerical value for each of the following concepts. (a) The...
Scores on a certain test are normally distributed with a variance of 84. A researcher wishes...
Scores on a certain test are normally distributed with a variance of 84. A researcher wishes to estimate the mean score achieved by all adults on the test. Find the sample size needed to assure with 95 percent confidence that the sample mean will not differ from the population mean by more than 5 units.
One researcher wishes to estimate the mean number of hours that high school students spend watching...
One researcher wishes to estimate the mean number of hours that high school students spend watching TV on a weekday. A margin of error of 0.29 0.29 hour is desired. Past studies suggest that a population standard deviation of 2.4 2.4 hours is reasonable. Estimate the minimum sample size required to estimate the population mean with the stated accuracy. The required sample size is nothing . (Round up to the nearest whole numbe These are the only numbers I was...
A survey conducted by the American Automobile Association showed that a family of four spends an average of $215.60 per day while on vacation Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per day and a sample standard deviation of $79.50 Devo a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls (to 2 decimals). . Based on the...
Supplementary Exercises 385 A survey conducted by the American Automobile Association (AAA) showed that a fam- ily of four spends an average of $215.60 per day while on vacation. Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per 45. day and a sample standard deviation of $74.50 Develop a 95% confidence interval estimate of the mean amount spent per day by a a. family of four visiting Niagara Falls....
Most questions answered within 3 hours.
-
Find the sample size, n, needed to estimate the percentage of
adults who have consulted fortune...
asked 11 seconds ago -
When the relationship is perfect, the regression of Y on X is a
perfect straight line....
asked 15 minutes ago -
What would be the balanced equation for the addition of
concentrated hydrochloric acid to cobalt chloride?...
asked 20 minutes ago -
Regarding a constructive discharge claim, whether an employee's
working conditions were so intolerable that he/she had...
asked 20 minutes ago -
Use a z -Table to find the specified area.
Between the mean and 3.01 deviations from...
asked 22 minutes ago -
If the rate constant for a particular reaction
triples between 35 °C and
70 °C, calculate...
asked 27 minutes ago -
if a magnetic field of 0.634 T is measured a distance of 0.009 m
away from...
asked 32 minutes ago -
In
the past, Ben has been able to hit a baseball at the batting cage
20%...
asked 35 minutes ago -
"Function does not take 0 arguments". I keep getting this error
for my last line of...
asked 35 minutes ago -
C++
What are the advantages of having a full copy of an object over
just having...
asked 36 minutes ago -
The airline you’re flying on has an on-time rate of 76% for the
route you are...
asked 54 minutes ago -
The return for Target stock next year will depend on the state
of the economy. The...
asked 1 hour ago