A national survey of companies included a question that asked whether the customers like the new flavor of a cola from company A. The sample results of 1000 customers, and 850 of them indicated that they liked the new flavor. The 98% confidence interval on the population proportion of people who like the new flavor is .

The 98% confidence interval on the population proportion of people who like the new flavor is (0.8237, 0.8763).

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 zscore that has a pvalue of .

For this problem, we have that:

98% confidence level

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

The lower limit of this interval is:

The upper limit of this interval is:

The 98% confidence interval on the population proportion of people who like the new flavor is (0.8237, 0.8763).

if you are saying purchasing 3 social media ads and one newspaper ad counts as one purchase then there are:

6 purchases

if you are asking how many of each then:

18 social media ads

6 newspaper ads

(0.2)+(-8)=0.3+-3+0.3m is the equation i guess? ?

step-by-step explanation: