Smartphone Purchases In a survey of 3611 adult Americans 18...

Smartphone Purchases In a survey of 3611 adult Americans 18 years and older conducted in July 2010 by SmartRevenue, it was found that 542 have used their smartphone to make a purchase. (a) Obtain a point estimate for the population proportion of adult Americans 18 years and older who have used their smartphone to make a purchase. (b) Verify that the requirements for constructing a confidence interval about $p$ are satisfied. (c) Construct a $90 \%$ confidence interval for the population proportion of adult Americans who have used their smartphone to make a purchase. (d) Interpret the interval.

Smartphone Purchases In a survey of 3611 adult Americans 18 years and older conducted in July 2010 by SmartRevenue, it was found that 542 have used their smartphone to make a purchase.
(a) Obtain a point estimate for the population proportion of adult Americans 18 years and older who have used their smartphone to make a purchase.
(b) Verify that the requirements for constructing a confidence interval about $p$ are satisfied.
(c) Construct a $90 \%$ confidence interval for the population proportion of adult Americans who have used their smartphone to make a purchase.
(d) Interpret the interval.

Key Concepts

Sample Proportion

The sample proportion is the estimate of the population proportion based on a sample. It is calculated by dividing the number of successes (in this case, individuals who used a smartphone for a purchase) by the total sample size. This serves as a point estimate for the true population proportion.

Confidence Interval for a Proportion

A confidence interval for a proportion provides a range of values within which the true population proportion is likely to fall at a specified level of confidence (such as 90%). The interval is typically constructed using the sample proportion and an estimation of its standard error, along with a critical value derived from the normal distribution.

Checking Normal Approximation Conditions

Before constructing a confidence interval using the normal approximation for a binomial distribution, it is important to verify that the sample size is sufficiently large. This is often done by ensuring that both the number of successes (np) and the number of failures (n(1-p)) are at least 10. These conditions justify the use of a normal model for the sampling distribution of the proportion.

Interpretation of the Confidence Interval

Interpreting a confidence interval involves understanding that the interval provides a range in which we are confident, at a specified level (here, 90%), that the true population proportion lies. It reflects the uncertainty inherent in using a sample to estimate a population parameter, implying that if the survey were repeated numerous times, approximately 90% of the calculated intervals would contain the true population proportion.

Transcript

00:01 There are oftentimes that we want to find a confidence interval for a population proportion, and they'll state what confidence level that we are to use.

00:11 In this problem, we'll take each of the components of forming that confidence interval for a population proportion p, and then also look at what our interpretation is of that confidence interval.

00:23 So here we have a sample of size n equal 3 ,600.

00:30 11 randomly selected and out of those we have 542 had a particular characteristic.

00:40 So this could be embedded within a word problem and then you'd also want to incorporate those specifics of the word problem in with your answer.

00:48 But here we'll just be doing the numerical calculations.

00:52 Now the very first part of the question asks us to find the point estimate for the population proportion and the point estimates notation is p with a hat over it.

01:05 The notation for the population proportion itself is p without the hat and for the point estimate for p it's p with a hat.

01:15 Now we find the point estimate from our information from our sample by taking the number in our sample that have the characteristic x in this case that's 542.

01:29 And dividing it by the number in our sample.

01:34 And in this case, the number in our sample is 3 ,611.

01:39 And when you calculate that, do the division, you get p hat is about 0 .150.

01:51 Part b asks us to verify that we can use the critical value of z sub alf over 2 and our methods that we learned in this section to form the confidence interval for the population proportion.

02:06 And to verify it, there's two things we need to check.

02:09 The first thing we need to check is if n our sample size times p hat times one minus p hat, multiplying that all together, do you get a number that's greater than or equal to 10? well, here our n is 3 ,611.

02:26 Our p hat is 0 .150 that we found in the previous part of this example, and then our 1 minus, and then the p hat of 0 .150 for the third factor.

02:47 Now, multiplying that all together, we get the value 460 .65, which definitely is a number that's greater than or equal to 10.

02:59 So yes, we've met that condition.

03:02 The second condition that we need to verify is to check to see if the sample size n, if that 3 ,611 for this question is less than or equal to 5 % or 0 .05 in decimal form times the population size.

03:25 And this was particularly tied to adult americans 18 years or older.

03:32 And you can find that with the current census, a number of adult americans 18 years or over is 257 ,000, $536 ,000.

03:47 And when you multiply that out, definitely you'll find that 3 ,611 is less than that particular value on the right -hand side.

03:59 So that condition is met as well.

04:03 So next up, let's form the confidence interval.

04:06 So to find your confidence interval for a population proportion p, meaning those conditions, it's p -hat minus e to p -hat plus e...

Please give Ace some feedback