Question

Suppose that a simple random sample of 145, 91 said they drink coffee in the morning.

Step 1 of 5:

What is the sample proportion for people who drink coffee in the morning?

Step 2 of 5:

Do we have what we need to compute a confidence interval?

Yes, it's binomial. Yes, n≥30.

Yes, we have a simple random sample, it is binomial with n/.05 gives a value less than the number of people than that in the world, and n*p-hat and n*(1-p-hat) are both greater than 5.

No, we do not have a normal population.

No, it's not binomial.

Step 3 of 5:

Find the lower bound for a 98% confidence interval of the true proportion of people who drink coffee in the morning, accurate to 3 decimal places.

Step 4 of 5:

Find the upper bound for a 98% confidence interval of the true proportion of people who drink coffee in the morning, accurate to 3 decimal places.

Step 5 of 5:

Choose the correct interpretation of this confidence interval.

We are 98% confident that the true proportion of people who drink coffee in the morning is between [lower bound as a percentage] and [upper bound as a percentage].

We are 98% confident that between [lower bound] and [upper bound] people drink coffee in the morning.

We are between [lower bound as a percentage] and [upper bound as a percentage] sure that 98% of people drink coffee in the morning.

I used my calculator and did 1-Prop Z-Int.

Answer 1

solution YES, 3 Here N=145, d=91 = = 0.6276 Sample proportion (Þ)= sample simple random 2) This is a here np=9135 ng=5435 At 98% confidence 24/2=2 the 98% (-I for p is B #2 P 8 = 0.6276 +233 0:6276 15 = 0.6276 + 0.0935 = {0.53u1, 0.7211) lower bound=101534) of people ") the upper bound = 50-7211] confident that the true proportion 5 we are 98% morning is between who drink coffee in the (53.4%) and 172.1%