Question

Question Help in a survey of 3149 adults, 1498 say they have started paying bils online in the last year Construct a 99% cotidence nterval for he population proportion interpret the results. A 99% confidence interval for the population proportions Round to three decimal places as needed.) Interpret your results. Choose the correct answer below OA W 99% confidence, tcan be said that the sample propori ofauts no say they hive tated paying isoïne heast year s beteen he С B. The endponts ofthe given confidence interval Show rat adts pay bats online 99% of the tre OC. With 99% cottdence, tcan be said that the popaation proportion ofadīts who ay hey have stated paying tiis onine n he ast year between the OD contidence interval confidence interval

Answer 1

We are required to create a 99% confidence interval for the given population proportion. Let's try to understand what we are trying to find out here. We are required to find out an interval which will contain the population proportion of the number of people who have started paying bills online in 99% of the times that we take the samples of this given population.

Since it's a 99% confidence interval, $\alpha$=1-0.99=0.01 where $\alpha$ gives the significance level i.e. the probability of rejecting the null hypothesis when it's true.

The 99% confidence interval is given as [P₀ - Z_α/2 * SE(P₀), P₀ + Z_α/2 * SE(P₀)] where P₀ is the sample proportion, Z_α/2 is the Z-statistic associated with $\alpha$=0.01 and SE(P₀) is the standard error of the proportion.

Now, considering that in a population of 3149 adults, 1498 have started paying bills. Therefore, the required proportion is 1498/3149 = 0.476. Though it's a population proportion, we can substitute this value for the sample proportion. The answers in both the cases will be same.

P0=0.476

1-P₀ = 1-0.476 = 0.524

The standard error of the population will be calculated as $hXuv2dm098A89lV4d6XjgrAAAAABJRU5ErkJggg=$ .

In this case, it will be equal to = 0.0089.

Therefore, the 99% confidence interval will be given as [0.476 - Z_α/2 (0.0089),0.476 + Z_α/2 (0.0089)] = [0.453,0.499].

This can be interpreted in the following manner: With 99% confidence, it can be stated that the population proportion of adults who started paying their bills online in the last year lies between 45.3% and 49.9%. This answers the second part of the question i.e. Option 3.