Question

In: Statistics and Probability

A random-digit-dialing survey sampled 1750 adults and found that 1140 used some form of prescription medication....

A random-digit-dialing survey sampled 1750 adults and found that 1140 used some form of prescription medication. Of these 1140 adults who used prescription drugs, 940 said they are concerned about the current state of health care; 320 of the 610 nonusers said this, too.

The margin of error in your 90% confidence interval is approximately

  1. Could cancer cells be selectively targeted by using antibodies recognizing a tumor-specific protein marker? Researchers grafted human cancerous cells onto 20 healthy adult mice and then randomly assigned 10 of these mice to be treated with tumor-specific antibodies. They found that only 1 of the 10 mice treated with antibodies developed metastases, whereas all 10 of the 10 mice in the control group developed metastases.

    We want to compare the proportions of mice developing metastases under the two conditions. Here, a large-sample significance  test is    

    A.

    appropriate because a total of 20 mice is large enough for any inference.

    B.

    not appropriate because the respective counts of successes and failures are not large enough.

    C.

    appropriate because the study was randomized.

    D.

    not appropriate because the experiment is not double-blind.

1 points   

QUESTION 7

  1. Refer to question number 6.

    A plus-four 90% confidence interval for pcontrol - ptreatment (the difference in the proportions that develop metastases) is

    A.

    0.53 to 0.97.

    B.

    0.74 to 1.05.

    C.

    0.49 to 1.01.

    D.

    0.17 to 0.92.

Solutions

Expert Solution

first sample size,     n1=   1140
number of successes, sample 1 =     x1=   940
proportion success of sample 1 , p̂1=   x1/n1=   0.825


      
second sample size,     n2 =    610
number of successes, sample 2 =     x2 =    320
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.525

Std error , SE =    SQRT(p̂1 * (1 - p̂1)/n1 + p̂2 * (1-p̂2)/n2) =     0.0231  

α=0.05
Z critical value =   Z α/2 =    1.6449   [excel function: =normsinv(α/2)
          
margin of error , E =   Z*SE =    0.0381 (answer)

-----------------------------------------------------------------------------------------------------------------------------------------------------



Related Solutions

Pharmaceutical companies promote their prescription drugs using television advertising. In a survey of 75 randomly sampled...
Pharmaceutical companies promote their prescription drugs using television advertising. In a survey of 75 randomly sampled television viewers, 12 indicated that they asked their physician about using a prescription drug they saw advertised on TV. Develop a 90% confidence interval for the proportion of viewers who discussed a drug seen on TV with their physician. (Round your answers to 3 decimal places.) Is it reasonable to conclude that 26% of the viewers discuss an advertised drug with their physician?
A survey of 25 randomly sampled judges employed by the state of Florida found that they...
A survey of 25 randomly sampled judges employed by the state of Florida found that they earned an average wage (including benefits) of $59.00 per hour. The sample standard deviation was $6.05 per hour. (Use t Distribution Table.) What is the best estimate of the population mean? Develop a 99% confidence interval for the population mean wage (including benefits) for these employees. (Round your answers to 2 decimal places.) How large a sample is needed to assess the population mean...
A survey of 25 randomly sampled judges employed by the state of Florida found that they...
A survey of 25 randomly sampled judges employed by the state of Florida found that they earned an average wage (including benefits) of $67.00 per hour. The sample standard deviation was $5.75 per hour. What is the best estimate of the population mean? Develop a 95% confidence interval for the population mean wage (including benefits) for these employees. (Round your answers to 2 decimal places.) How large a sample is needed to assess the population mean with an allowable error...
A survey of 20 randomly sampled judges employed by the state of Florida found that they...
A survey of 20 randomly sampled judges employed by the state of Florida found that they earned an average wage (including benefits) of $63.00 per hour. The sample standard deviation was $5.90 per hour. (Use t Distribution Table.) What is the best estimate of the population mean? Develop a 98% confidence interval for the population mean wage (including benefits) for these employees. (Round your answers to 2 decimal places.) Confidence interval for the population mean wage is between   and How...
A survey found that 34​% of 875 randomly sampled teens said that their parents checked to...
A survey found that 34​% of 875 randomly sampled teens said that their parents checked to see what Web sites they visited. The following year the same question posed to 800 teens found 42​% reporting such checks. Do these results provide evidence that more parents are​ checking? need: null and alternative hypothesis, z value, p value, anything else that. might help
In an opinion survey, a random sample of 1000 adults from the U.S.A. and 1000 adults...
In an opinion survey, a random sample of 1000 adults from the U.S.A. and 1000 adults from Germany were asked whether they supported the death penalty. 560 American adults and 590 German adults indicated that they supported the death penalty. Researcher wants to know if there is sufficient evidence to conclude that the proportion of adults who support the deal penally in the US is different than that in Germany. Use a confidence interval (at confidence interval of 98%) to...
Mimi conducted a survey on a random sample of 100 adults. 58 adults in the sample...
Mimi conducted a survey on a random sample of 100 adults. 58 adults in the sample chose banana as his / her favorite fruit. Construct a 95% confidence interval estimate of the proportion of adults whose favorite fruit is banana. Show all work. Just the answer, without supporting work, will receive no credit.
In a random sample of 820 adults in the U.S.A., it was found that 86 of...
In a random sample of 820 adults in the U.S.A., it was found that 86 of those had a pinworm infestation. You want to find the 90% confidence interval for the proportion of all U.S. adults with pinworm. (a) What is the point estimate for the proportion of all U.S. adults with pinworm? Round your answer to 3 decimal places. (b) What is the critical value of z (denoted zα/2) for a 90% confidence interval? Use the value from the...
In a random sample of 810 adults in the U.S.A., it was found that 76 of...
In a random sample of 810 adults in the U.S.A., it was found that 76 of those had a pinworm infestation. You want to find the 99% confidence interval for the proportion of all U.S. adults with pinworm. (a) What is the point estimate for the proportion of all U.S. adults with pinworm? (b) What is the critical value of z for a 99% confidence interval? (c) What is the margin of error (E) for the 99% confidence interval? (d)...
1. A survey about same-sex marriage used a random sample with 1200 adults. The results showed...
1. A survey about same-sex marriage used a random sample with 1200 adults. The results showed that 30% favored legal marriage, 32% favored civil unions, and 38% favored no legal recognition. Computer the right boundary at the 91% C.L. for the proportion of adults who favor the “legal marriage” position. 2. A survey about same-sex marriage used a random sample with 1200 adults. The results showed that 30% favored legal marriage, 32% favored civil unions, and 38% favored no legal...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT