Suppose a large telephone manufacturer has a problem with excessive customer complaints and consequent returns of the phones for repair or replacement. The manufacturer wants to estimate the magnitude of the problem in order to design a quality control program. How many telephones should be sampled and checked in order to estimate the proportion defective to within 9 percentage points with 89% confidence

80 telephones should be sampled

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the z-score that has a p-value of .

The margin of error is of:

89% confidence level

So , z is the value of Z that has a p-value of , so .

How many telephones should be sampled and checked in order to estimate the proportion defective to within 9 percentage points with 89% confidence?

n telephones should be sampled, an n is found when M = 0.09. We have no estimate for the proportion, thus we use

Rounding up(as 79 gives a margin of error slightly above the desired value).

80 telephones should be sampled

answer: yes i agree that you should not put it on/p>

step-by-step explanation: the reason for this is that it wastes other peoples time and also make them fail as well and that is never good.

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;