7.4 Using Technology for Confidence Intervals

This is another section to show technology tools that can be used to help with the statistics concepts taught in this chapter. This section will cover tools to help find Confidence Intervals.

Graphing Calculator

Similar to all the other technology sections, the main tool introduced will be the TI-83, 83+, 84, or 84+ Graphing Calculator. The graphing calculator can help to find values to help calculate the error bound formula and even construct the entire confidence interval.

A Confidence Interval with a Known Standard Deviation

We learned in 7.1 A Single Population Mean Using the Normal Distribution, to construct a confidence interval for a single unknown population, mean [latex]\mu[/latex], where the population standard deviation is known, we need [latex]\overline{x}[/latex] as an estimate for [latex]\mu[/latex] and we need the margin of error. Remember, the margin of error ([latex]\text{EBM}[/latex]) is called the error bound for a population mean (abbreviated [latex]\text{EBM}[/latex]). The sample mean [latex]\overline{x}[/latex] is the point estimate of the unknown population mean [latex]\mu[/latex].

The confidence interval estimate will have the form: [latex](\text{point estimate} - \text{error bound}, \text{point estimate} + \text{error bound})[/latex] or, in symbols, ([latex]\overline{x}-\text{EBM},\overline{x}+\text{EBM}[/latex]).

We did by-hand calculations to calculate the error bound [latex]\text{EBM}[/latex] and construct the confidence interval. The error bound formula for an unknown population mean, [latex]\mu[/latex], when the population standard deviation [latex]\sigma[/latex] is known is: [latex]\text{EBM} = {z}_{\frac{\alpha }{2}}\left(\frac{\sigma }{\sqrt{n}}\right)[/latex]. The by-hand work we did had us find the value of z that puts an area equal to the confidence level (in decimal form) in the middle of the standard normal distribution [latex]Z \sim N(0, 1)[/latex]:

• The confidence level, [latex]\text{CL}[/latex], is the area in the middle of the standard normal distribution. [latex]\text{CL} = 1 - \alpha[/latex], so [latex]\alpha[/latex] is the area that is split equally between the two tails.
• Each of the tails contains an area equal to [latex]\frac{\alpha}{2}[/latex].
• The z-score that has an area to the right of [latex]\frac{\alpha }{2}[/latex] is denoted by [latex]{z}_{\frac{\alpha }{2}}[/latex].

For example, when [latex]\text{CL} = 0.95[/latex], [latex]\alpha = 0.05[/latex] and [latex]\frac{\alpha }{2} = 0.025[/latex]; we write [latex]{z}_{\frac{\alpha }{2}} = z_0.025[/latex]. The area to the right of [latex]z_{0.025}[/latex] is 0.025 and the area to the left of [latex]z_{0.025}[/latex] is [latex]1 – 0.025 = 0.975[/latex]. We then used a standard normal probability table to find [latex]{z}_{\frac{\alpha }{2}} = {z}_{0.025} = 1.96[/latex]. It was work, but we got there, right?

Well, the first thing we can do with the graphing calculator is to use the [latex]\text{invNorm}[/latex] function.

Legend

Notice, we are using 0 and 1 since we want a z-score and [latex]Z \sim N(0, 1)[/latex]. We still have to go through taking our [latex]\text{CL}[/latex], finding [latex]\alpha[/latex] and [latex]\frac{\alpha }{2}[/latex]. It is helpful to remember the graph below.

For the example provided above, we would use [latex]\text{invNorm}(0.975, 0, 1)[/latex] and get [latex]z = 1.96[/latex]. We would take that z-score and use it to calculate the error bound [latex]\text{EBM}[/latex] and construct the confidence interval. Let's see another example.

We know the following from our problem:

• [latex]\overline{x} = 68[/latex]
• [latex]\sigma = 3[/latex]
• [latex]n = 36[/latex]
• The confidence level is 90%, so ([latex]\text{CL} = 0.90[/latex])

To find the confidence interval, you need the sample mean, [latex]\overline{x}[/latex], and the [latex]\text{EBM}[/latex], where [latex]\text{EBM} = \left({z}_{\frac{\alpha }{2}}\right) \left(\frac{\sigma }{\sqrt{n}}\right)[/latex]. Since [latex]\text{CL} = 0.90[/latex], [latex]\alpha = 1 – \text{CL} = 1 – 0.90 = 0.10[/latex], and [latex]\frac{\alpha }{2} = 0.05[/latex]. That makes [latex]{z}_{\frac{\alpha }{2}}={z}_{0.05}[/latex]. The area to the right of [latex]{z}_{0.05}[/latex] is 0.05 and the area to the left of [latex]{z}_{0.05}[/latex] is [latex]1 - 0.05 = 0.95[/latex].

Using [latex]\text{invNorm}(0.95, 0, 1)[/latex] on the TI-83, 83+, and 84+ calculator, we would get [latex]z=1.645[/latex].

• [latex]\text{EBM} = (1.645)(\frac{3}{\sqrt{36}}) = 0.8225[/latex]
• [latex]\overline{x} - \text{EBM} = 68 - 0.8225 = 67.1775[/latex]
• [latex]\overline{x} + \text{EBM} = 68 + 0.8225 = 68.8225[/latex]

The 90% confidence interval is (67.1775, 68.8225).

For this problem, we first need to find the point estimate, the sample mean [latex]\overline{x}[/latex], before we can find the confidence interval. Reminder from 2.9 Using Technology for Descriptive Statistics, that we can use our List and 1-Var Stats features to do this. We would find [latex]\overline{x}=1.024[/latex]. To find the [latex]\text{EBM}[/latex], because you are creating a 98% confidence interval, [latex]\text{CL} = 0.98[/latex], you need to find [latex]z_{0.01}[/latex] having the property that the area under the normal density curve to the right of [latex]z_{0.01}[/latex] is 0.01 and the area to the left is 0.99. Using your calculator to find the z-score, [latex]\text{invNorm}(0.99, 0, 1)[/latex], we would get [latex]z=2.326[/latex].

• [latex]EBM=(2.326)\frac{0.337}{\sqrt{30}}=0.1431[/latex]
• [latex]\overline{x} - \text{EBM} = 1.024 - 0.1431 = 0.8809[/latex]
• [latex]\overline{x} - \text{EBM} = 1.024 - 0.1431 = 1.1671[/latex]

Now, there is actually an even quicker way to use technology - and that's to have it construct the entire confidence interval.

Let's look at some examples to use the ZInterval.

A Confidence Interval with an Unknown Standard Deviation

Remember, in practice, we rarely know the population standard deviation. In the past, when the sample size was large, this did not present a problem to statisticians. They used the sample standard deviation s as an estimate for [latex]\sigma[/latex] and proceeded as before to calculate a confidence interval with close enough results. However, statisticians ran into problems when the sample size was small. A small sample size caused inaccuracies in the confidence interval.

In 7.2 A Single Population Mean using the Student t Distribution we saw how to find a confidence interval if the population standard deviation is not known and the population has an approximate normal distribution, by using the Student's t distribution.

Calculators and computers can easily calculate any Student's t-probabilities. The TI-83,83+, and 84+ have a [latex]\text{tcdf}[/latex] function to find the probability for given values of t. The syntax for the [latex]\text{tcdf}[/latex] command is [latex]\text{tcdf(lower bound, upper bound, degrees of freedom)}[/latex]. However, for confidence intervals, we need to use inverse probability to find the value of t when we know the probability.

For the TI-84+ you can use the [latex]\text{invT}[/latex] command on the [latex]\text{DISTRibution}[/latex] menu. The [latex]\text{invT}[/latex] command works similarly to the invnorm. The invT command requires two inputs: [latex]\text{invT(area to the left, degrees of freedom)}[/latex]. The output is the t-score that corresponds to the area we specified. Unfortunately, the TI-83 and 83+ do not have the [latex]\text{invT}[/latex] command.

But we now know that there is an even quicker way to use technology - and that's to have it construct the entire confidence interval.

Confidence Interval for a Proportion

Let's recall - how do you know you are dealing with a proportion problem?

1. First, the underlying distribution is a binomial distribution. (There is no mention of a mean or average.)
2. If [latex]X[/latex] is a binomial random variable, then [latex]X \sim B(n, p)[/latex] where [latex]n[/latex] is the number of trials and [latex]p[/latex] is the probability of a success.
3. To form a proportion, take [latex]X[/latex], the random variable for the number of successes and divide it by [latex]n[/latex], the number of trials (or the sample size).
4. The random variable [latex]P^{\prime}[/latex](read "P prime") is that proportion, [latex]{P}^{\prime }=\frac{X}{n}[/latex]

By-hand, the calculation for the confidence interval of a proportion in 7.3 A Population Proportion is the most difficulty. Remember, we needed to find [latex]{p}^{\prime}[/latex], [latex]{q}^{\prime}[/latex], and [latex]\text{EBP}[/latex]:

• [latex]{p}^{\prime }=\frac{x}{n}[/latex]
• [latex]{q}^{\prime}= 1 - {p}^{\prime}[/latex]
• [latex]\text{EBP}=\left({z}_{\frac{\alpha }{2}}\right)\sqrt{\frac{{p}^{\prime }{q}^{\prime }}{n}}[/latex]

The confidence interval for the true binomial population proportion is [latex](p^{\prime} - \text{EBP}, p^{\prime} + \text{EBP})[/latex]. But who really wants to calculate that [latex]\text{EBP}=\left({z}_{\frac{\alpha }{2}}\right)\sqrt{\frac{{p}^{\prime }{q}^{\prime }}{n}}[/latex]?

Well, [latex]P^{\prime}[/latex] follows a normal distribution for proportions. Which means, to find [latex]{z}_{\frac{\alpha }{2}}[/latex] in the [latex]\text{EBP}[/latex] formula, we could use the TI-83, 83+, or 84+ calculator command [latex]\text{invNorm}()[/latex]. So that would cut down on some of the work!

But we now know that there is an even quicker way to use technology - and that's to have it construct the entire confidence interval.

• Mean Confidence Intervals Using Student's t-Distribution (TI-84 Only)
• Calculate a Confidence Interval for a Population Proportion on a TI-84
• 1-Proportion Confidence Interval (TI-83 & TI-84)

Additional Technology Tools

In addition to the graphing calculator, there are some additional technology tools that can be used for the concepts covered on this page. Below are links to helpful videos for those tools:

• Microsoft Excel:
  • Confidence Interval Sigma Known
  • Confidence Interval Sigma Known From Raw Data
  • Confidence Interval Sigma Not Known TINV function
  • Confidence Interval For Advertising TINV function
  • Confidence Interval Data Analysis Add-in Descriptive Statistics
  • Confidence Intervals for Proportions #1
  • Confidence Intervals for Proportions #2
• Desmos: (Desmos Graphing Calculator Link)
  • Determine a Mean Confidence Interval Using Desmos
  • Determine a Sample Size for a Mean Confidence Interval Using Desmos
  • Determine a Mean Confidence Interval From Data Using Desmos (t-distribution)
  • Determine a Proportion Confidence Interval Using Desmos
  • Determine a Sample Size for a Proportion Confidence Interval Using Desmos

Section Practice

Among various ethnic groups, the standard deviation of heights is known to be approximately three inches. We wish to construct a 95% confidence interval for the mean height of male Swedes. Forty-eight male Swedes are surveyed. The sample mean is 71 inches. The sample standard deviation is 2.8 inches.

1. [latex]\overline{x} =\underline{\hspace{2cm}}[/latex]
2. [latex]\sigma = \underline{\hspace{2cm}}[/latex]
3. [latex]n =\underline{\hspace{2cm}}[/latex]
4. In words, define the random variables [latex]X[/latex] and [latex]\overline{X}[/latex].
5. Which distribution should you use for this problem? Explain your choice.
6. Construct a 95% confidence interval for the population mean height of male Swedes.
   • State the confidence interval.
   • Sketch the graph.
   • Calculate the error bound.
7. What will happen to the level of confidence obtained if 1,000 male Swedes are surveyed instead of 48? Why?

Solution

1. 71
2. 3
3. 48
4. [latex]X[/latex] is the height of a Swiss male and is the mean height from a sample of 48 Swiss males.
5. Normal. We know the standard deviation for the population, and the sample size is greater than 30.
6. Construct a 95% confidence interval for the population mean height of male Swedes.
   • [latex]\text{CI}: (70.151, 71.49)[/latex]
   • Graph below.
   • [latex]\text{EBM} = 0.849[/latex]
7. The confidence interval will decrease in size, because the sample size increased. Recall, when all factors remain unchanged, an increase in sample size decreases variability. Thus, we do not need as large an interval to capture the true population mean.

A camp director is interested in the mean number of letters each child sends during his or her camp session. The population standard deviation is known to be 2.5. A survey of 20 campers is taken. The mean from the sample is 7.9 with a sample standard deviation of 2.8.

1. Identify the following:
   • [latex]\overline{x} =\underline{\hspace{2cm}}[/latex]
   • [latex]\sigma =\underline{\hspace{2cm}}[/latex]
   • [latex]n =\underline{\hspace{2cm}}[/latex]
2. Define the random variables [latex]X[/latex] and [latex]\overline{X}[/latex] in words.
3. Which distribution should you use for this problem? Explain your choice.
4. Construct a 90% confidence interval for the population mean number of letters campers send home.
   • State the confidence interval.
   • Sketch the graph.
   • Calculate the error bound.
5. What will happen to the error bound and confidence interval if 500 campers are surveyed? Why?

Solution

1. Identify the following:
   • 7.9
   • 2.5
   • 20
2. [latex]X[/latex] is the number of letters a single camper will send home. [latex]\overline{X}[/latex] is the mean number of letters sent home from a sample of 20 campers.
3. N[latex]7.9\left(\frac{2.5}{\sqrt{20}}\right)[/latex]
4. Construct a 90% confidence interval for the population mean number of letters campers send home
   • [latex]\text{CI}: (6.98, 8.82)[/latex]
   • Graph below.
   • [latex]\text{EBM}: 0.92[/latex]
5. The error bound and confidence interval will decrease.

The Federal Election Commission collects information about campaign contributions and disbursements for candidates and political committees each election cycle. During the 2012 campaign season, there were 1,619 candidates for the House of Representatives across the United States who received contributions from individuals. Below are the total receipts from individuals for a random selection of 40 House candidates rounded to the nearest $100. The standard deviation for this data to the nearest hundred is [latex]\sigma = \$ 909,200[/latex].

$3,600; $1,243,900; $10,900; $385,200; $581,500; $7,400; $2,900; $400; $3,714,500; $632,500; $391,000; $467,400; $56,800; $5,800; $405,200; $733,200; $8,000; $468,700; $75,200; $41,000; $13,300; $9,500; $953,800; $1,113,500; $1,109,300; $353,900; $986,100; $88,600; $378,200; $13,200; $3,800; $745,100; $5,800; $3,072,100; $1,626,700; $512,900; $2,309,200; $6,600; $202,400; $15,800

1. Find the point estimate for the population mean.
2. Using 95% confidence, calculate the error bound.
3. Create a 95% confidence interval for the mean total individual contributions.
4. Interpret the confidence interval in the context of the problem.

Solution

1. [latex]\overline{x}[/latex] = $568,873
2. [latex]\text{CL} = 0.95[/latex]; [latex]\alpha = 1 – 0.95 = 0.05[/latex]; [latex]{z}_{\frac{\alpha }{2}} = 1.96[/latex]
   [latex]\text{EBM} = {z}_{0.025}\frac{\sigma }{\sqrt{n}}[/latex] = 1.96 [latex]\frac{909200}{\sqrt{40}} = \$ 281,764[/latex]
3. [latex]\overline{x} - \text{EBM} = 568,873 - 281,764 = 287,109[/latex]
   [latex]\overline{x} + \text{EBM} = 568,873+ 281,764 = 850,637[/latex]

   Alternate solution:

   1. Press STAT and arrow over to TESTS.
   2. Arrow down to 7:ZInterval.
   3. Press ENTER.
   4. Arrow to Stats and press ENTER.
   5. Arrow down and enter the following values:
      • [latex]\sigma: 909,200[/latex]
      • [latex]\overline{x}: 568,873[/latex]
      • [latex]n: 40[/latex]
      • [latex]\text{CL}: 0.95[/latex]
   6. Arrow down to Calculate and press ENTER.
   7. The confidence interval is ($287,114, $850,632).
   8. Notice the small difference between the two solutions—these differences are simply due to rounding errors in the hand calculations.
4. We estimate with 95% confidence that the mean amount of contributions received from all individuals by House candidates is between $287,109 and $850,637.

A pharmaceutical company makes tranquilizers. It is assumed that the distribution for the length of time they last is approximately normal. Researchers in a hospital used the drug on a random sample of nine patients. The effective period of the tranquilizer for each patient (in hours) was as follows: 2.7; 2.8; 3.0; 2.3; 2.3; 2.2; 2.8; 2.1; and 2.4.

1. Identify the following:
   • [latex]\overline{x} = \underline{\hspace{2cm}}[/latex]
   • [latex]{s}_{x} = \underline{\hspace{2cm}}[/latex]
   • [latex]n = \underline{\hspace{2cm}}[/latex]
   • [latex]n – 1 = \underline{\hspace{2cm}}[/latex]
2. Define the random variable [latex]X[/latex] in words.
3. Define the random variable [latex]\overline{X}[/latex] in words.
4. Which distribution should you use for this problem? Explain your choice.
5. Construct a 95% confidence interval for the population mean length of time.
   • State the confidence interval.
   • Sketch the graph.
   • Calculate the error bound.
6. What does it mean to be “95% confident” in this problem?

 

Solution

1. Identify the following:
   • [latex]\overline{x} = 2.51[/latex]
   • [latex]{s}_{x} = 0.318[/latex]
   • [latex]n = 9[/latex]
   • [latex]n - 1 = 8[/latex]
2. the effective length of time for a tranquilizer
3. the mean effective length of time of tranquilizers from a sample of nine patients
4. We need to use a Student’s-t distribution, because we do not know the population standard deviation.
5. Construct a 95% confidence interval
   • [latex]\text{CI}: (2.27, 2.76)[/latex]
   • Check student's solution.
   • [latex]\text{EBM}: 0.25[/latex]
6. If we were to sample many groups of nine patients, 95% of the samples would contain the true population mean length of time.

Suppose that 14 children, who were learning to ride two-wheel bikes, were surveyed to determine how long they had to use training wheels. It was revealed that they used them an average of six months with a sample standard deviation of three months. Assume that the underlying population distribution is normal.

1. Identify the following:
   • [latex]\overline{x} = \underline{\hspace{2cm}}[/latex]
   • [latex]{s}_{x} = \underline{\hspace{2cm}}[/latex]
   • [latex]n = \underline{\hspace{2cm}}[/latex]
   • [latex]n – 1 = \underline{\hspace{2cm}}[/latex]
2. Define the random variable [latex]X[/latex] in words.
3. Define the random variable[latex]\overline{X}[/latex] in words.
4. Which distribution should you use for this problem? Explain your choice.
5. Construct a 99% confidence interval for the population mean length of time using training wheels.
   • State the confidence interval.
   • Sketch the graph.
   • Calculate the error bound.
6. Why would the error bound change if the confidence level were lowered to 90%?

The Federal Election Commission (FEC) collects information about campaign contributions and disbursements for candidates and political committees each election cycle. A political action committee (PAC) is a committee formed to raise money for candidates and campaigns. A Leadership PAC is a PAC formed by a federal politician (senator or representative) to raise money to help other candidates’ campaigns.

The FEC has reported financial information for 556 Leadership PACs that operated during the 2011–2012 election cycle. The data below shows the total receipts during this cycle for a random selection of 20 Leadership PACs.

$46,500.00

$0

$40,966.50

$105,887.20

$5,175.00

$29,050.00

$19,500.00

$181,557.20

$31,500.00

$149,970.80

$2,555,363.20

$12,025.00

$409,000.00

$60,521.70

$18,000.00

$61,810.20

$76,530.80

$119,459.20

$0

$63,520.00

$6,500.00

$502,578.00

$705,061.10

$708,258.90

$135,810.00

$2,000.00

$2,000.00

$0

$1,287,933.80

$219,148.30

[latex]\overline{x}=$251,854.23[/latex]

[latex]s=\$521,130.41[/latex]

Use this sample data to construct a 96% confidence interval for the mean amount of money raised by all Leadership PACs during the 2011–2012 election cycle. Use the Student's t-distribution.

Solution

Enter the data as a list.

Press STAT and arrow over to TESTS.

Arrow down to 8:TInterval.

Press ENTER.

Arrow to Data and press ENTER.

Arrow down and enter the name of the list where the data is stored.

Enter Freq: 1

Enter C-Level: 0.96

Arrow down to Calculate and press Enter.

The 96% confidence interval is ($47,262, $456,447).

The difference between solutions arises from rounding differences.

Forbes magazine published data on the best small firms in 2012. These were firms that had been publicly traded for at least a year, have a stock price of at least $5 per share, and have reported annual revenue between $5 million and $1 billion. The data below shows the ages of the corporate CEOs for a random sample of these firms.

48

58

51

61

56

59

74

63

53

50

59

60

60

57

46

55

63

57

47

55

57

43

61

62

49

67

67

55

55

49

Use this sample data to construct a 90% confidence interval for the mean age of CEO’s for these top small firms. Use the Student's t-distribution.

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its mean number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard deviation is 4.1 seats.

1. Identify the following information:
   • [latex]\overline{x} = \underline{\hspace{2cm}}[/latex]
   • [latex]{s}_{x} = \underline{\hspace{2cm}}[/latex]
   • [latex]n = \underline{\hspace{2cm}}[/latex]
   • [latex]n – 1 = \underline{\hspace{2cm}}[/latex]
2. Define the random variables [latex]X[/latex] and [latex]\overline{X}[/latex] in words.
3. Which distribution should you use for this problem? Explain your choice.
4. Construct a 92% confidence interval for the population mean number of unoccupied seats per flight.
   • State the confidence interval.
   • Sketch the graph.
   • Calculate the error bound.

 

Solution

1. Identify the following information
   • [latex]\overline{x} = 11.6[/latex]
   • [latex]{s}_{x} = 4.1[/latex]
   • [latex]n = 225[/latex]
   • [latex]n - 1 = 224[/latex]
2. [latex]X[/latex] is the number of unoccupied seats on a single flight. [latex]\overline{X}[/latex] is the mean number of unoccupied seats from a sample of 225 flights.
3. We will use a Student’s-t distribution, because we do not know the population standard deviation.
4. Construct a 92% confidence interval
   • [latex]\text{CI}: (11.12 , 12.08)[/latex]
   • Check student's solution.
   • [latex]\text{EBM}: 0.48[/latex]

A national survey of 1,000 adults was conducted on May 13, 2013 by Rasmussen Reports. It concluded with 95% confidence that 49% to 55% of Americans believe that big-time college sports programs corrupt the process of higher education.

1. Find the point estimate and the error bound for this confidence interval.
2. Can we (with 95% confidence) conclude that more than half of all American adults believe this?
3. Calculate a 75% confidence interval for the proportion of American adults that believe that major college sports programs corrupt higher education.
4. Can we (with 75% confidence) conclude that at least half of all American adults believe this?

Solution

1. [latex]p^{\prime} = \frac{(0.55 + 0.49)}{2} = 0.52[/latex]; [latex]\text{EBP} = 0.55 - 0.52 = 0.03[/latex]
2. No, the confidence interval includes values less than or equal to 0.50. It is possible that less than half of the population believe this.
3. STAT TESTS A: 1-PropZinterval with [latex]x = (0.52)(1,000)[/latex], [latex]n = 1,000[/latex], [latex]\text{CL} = 0.75[/latex]. Answer is (0.502, 0.538)
4. Yes – this interval does not fall less than 0.50 so we can conclude that at least half of all American adults believe that major sports programs corrupt education – but we do so with only 75% confidence.

An article regarding interracial dating and marriage recently appeared in the Washington Post. Of the 1,709 randomly selected adults, 315 identified themselves as Latinos, 323 identified themselves as blacks, 254 identified themselves as Asians, and 779 identified themselves as whites. In this survey, 86% of blacks said that they would welcome a white person into their families. Among Asians, 77% would welcome a white person into their families, 71% would welcome a Latino, and 66% would welcome a black person.

1. We are interested in finding the 95% confidence interval for the percent of all black adults who would welcome a white person into their families. Define the random variables [latex]X[/latex] and [latex]P^{\prime}[/latex], in words.
2. Which distribution should you use for this problem? Explain your choice.
3. Construct a 95% confidence interval.
   • State the confidence interval.
   • Sketch the graph.
   • Calculate the error bound.

Use the information provided in the previous exercise to answer the following questions.

1. Construct three 95% confidence intervals:
   • percent of all Asians who would welcome a white person into their families.
   • percent of all Asians who would welcome a Latino into their families.
   • percent of all Asians who would welcome a black person into their families.
2. Even though the three-point estimates are different, do any of the confidence intervals overlap? Which?
3. For any intervals that do overlap, in words, what does this imply about the significance of the differences in the true proportions?
4. For any intervals that do not overlap, in words, what does this imply about the significance of the differences in the true proportions?

Solution

1. Construct three 95% confidence intervals
   • (0.72, 0.82)
   • (0.65, 0.76)
   • (0.60, 0.72)
2. Yes, the intervals (0.72, 0.82) and (0.65, 0.76) overlap, and the intervals (0.65, 0.76) and (0.60, 0.72) overlap.
3. We can say that there does not appear to be a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a Latino person into their families.
4. We can say that there is a significant difference between the proportion of Asian adults who say that their families would welcome a white person into their families and the proportion of Asian adults who say that their families would welcome a black person into their families.

Stanford University conducted a study of whether running is healthy for men and women over age 50. During the first eight years of the study, 1.5% of the 451 members of the 50-Plus Fitness Association died. We are interested in the proportion of people over 50 who ran and died in the same eight-year period.

1. Define the random variables [latex]X[/latex] and [latex]P^{\prime}[/latex] in words.
2. Which distribution should you use for this problem? Explain your choice.
3. Construct a 97% confidence interval for the population proportion of people over 50 who ran and died in the same eight–year period.
   • State the confidence interval.
   • Sketch the graph.
   • Calculate the error bound.
4. Explain what a “97% confidence interval” means for this study.

Refer to the previous example. Another question in the poll was “[How much are] you worried about the quality of education in our schools?” Sixty-three percent responded “a lot”. We are interested in the population proportion of adult Americans who are worried a lot about the quality of education in our schools.

1. Define the random variables [latex]X[/latex] and [latex]P^{\prime}[/latex] in words.
2. Which distribution should you use for this problem? Explain your choice.
3. Construct a 95% confidence interval for the population proportion of adult Americans who are worried a lot about the quality of education in our schools.
   • State the confidence interval.
   • Sketch the graph.
   • Calculate the error bound.
4. The sampling error given by Yankelovich Partners, Inc. (which conducted the poll) is [latex]\pm 3 \%[/latex]. In one to three complete sentences, explain what the [latex]\pm 3 \%[/latex] represents.