Question

Question 9 of 21

In a random sample of 562 items produced by a machine, the quality control staff found 6.6% to be defective. Based on this sample, the 95% confidence interval for the proportion of defective items in all items produced by this machine, rounded to four decimal places, is:

    A. the lower endpoint is 0.0425, the upper endpoint is 0.0895

    B. the lower endpoint is 0.0455, the upper endpoint is 0.0865

    C. The lower endpoint is 0.0485, the upper endpoint is 0.0835

    D. the lower endpoint is 0.0515, the upper endpoint is 0.0805

Question 10 of 21

A company wants to estimate the mean net weight of all 32-ounce packages of its Yummy Taste cookies at a 95% confidence level. The margin of error is to be within 0.025 ounces of the population mean. The population standard deviation is 0.096 ounces. The sample size that will yield a margin of error within 0.025 ounces of the population mean is:

    A. 57

    B. 65

    C. 71

    D. 89

Question 11 of 21

A company wants to estimate, at a 95% confidence level, the proportion of all families who own its product. The most conservative estimate of the sample size that would limit the margin of error to be within 0.046 of the population proportion is:

    A. 230

    B. 319

    C. 454

    D. Not enough information is given

Question 12 of 21

In a 1997 poll of 261 male, married, upper-level managers conducted by Joy Schneer and Frieda Reitman for Fortune magazine, 31% of the men stated that their wives worked either full-time or part-time (Fortune, March 17, 1997). What are the boundaries for a 99% confidence interval for p, the proportion of all male, married, upper-level managers whose wives work?

    A. The lower limit is 0.1561, The upper limit is 0.4639

    B. The lower limit is 0.1961, The upper limit is 0.4239

    C. The lower limit is 0.2361, The upper limit is 0.3839

    D. The lower limit is 0.2761, The upper limit is 0.3439

Answer 1

9)

the lower endpoint is 0.0455, the upper endpoint is 0.0865

10)

for 95 % CI value of z=	1.960
standard deviation σ=	0.0960
margin of error E =	0.025
required sample size n=(zσ/E)2                                         =	57.0

11)

here margin of error E =	0.046
for95% CI crtiical Z          =	1.960
estimated proportion=p=	0.500
required sample size n =	p*(1-p)*(z/E)2=	454.00

12)

The lower limit is 0.2361, The upper limit is 0.3839