An FDA representative randomly selects 8 packages of ground chuck from a grocery store and measures the fat content (as a percent) of each package. Assume that the fat contents have an approximately normal distribution. The resulting measurements are given below.

Fat Contents (%)
19 18 11 17
17 19 12 15

Calculate the sample mean and the sample standard deviation of the fat contents. Round your answers to two decimal places, if necessary.
b. Construct a 95% confidence interval for the true mean fat content of all the packages of ground beef. Round the endpoints to two decimal places, if necessary.

A conservative investor would like to invest some money in a bond fund. The investor is concerned about the safety of her principal (the original money invested). Colonial Funds claims to have a bond fund which has maintained a consistent share price of $10. They claim that this share price has not varied by more than $0.25 on average since its inception. To test this claim, the investor randomly selects 24 days during the last year and determines the share price for the bond fund. The average share price of the sample is $11 with a standard deviation of $0.35. Assuming that the share prices of the bond fund have an approximately normal distribution, construct a 99% confidence interval for the standard deviation of the share price of the bond fund. Round any intermediate calculations to no less than six decimal places and round the endpoints of the interval to four decimal places.

Almost all smart devices (phones, tablets, and computers) are made with touch screens. A concern of many consumers is the shelf life of the “touch” component of the screens. A consumer advocacy group wanted to inform its members of a range that they can expect their touch screens to last. The group took a sample of 26 screens and measured the life of the “touch” function of the screens. That is, they used digital devices to simulate billions of touches to determine the life of the screens. Of the 26 screens sampled, the average “touch” life was 90 months with a standard deviation of seven months. Construct a 98% confidence interval for the standard deviation of the life of the touch screens. Assume that the life of the touch screens has an approximately normal distribution. Round any intermediate calculations to no less than six decimal places and round the endpoints of the interval to four decimal places.

A toy manufacturer wants to know how many new toys children buy each year. A sample of 686 children was taken to study their purchasing habits. Construct the 85% confidence interval for the mean number of toys purchased each year if the sample mean was found to be 6.8. Assume that the population standard deviation is 2.1. Round your answers to one decimal place.

Constructing a Confidence Interval for the Difference between Two Population Proportions
In order to determine if a new instructional technology improves students' scores, a professor wants to know if a larger percentage of students using the instructional technology passed the class than the percentage of students who did not use the new technology. Records show that 45 out of 50 randomly selected students who were in classes that used the instructional technology passed the class and 38 out of 51 randomly selected students who were in classes that did not use the instructional technology passed the class. Construct a 95%

confidence interval for the true difference between the proportion of students using the technology who passed and the proportion of students not using the technology who passed.

Solution

We are going to show how to construct the confidence interval first without a TI-83/84 Plus calculator and then with one.
Step 1: Find the point estimate.

First, we'll let Population 1 be those students who used the new technology and Population 2 be those students who did not. Next, we need to calculate the sample proportions. The sample proportion for Sample 1 (using instructional technology) is calculated as follows.

pˆ1=x1n1=4550=0.9

The sample proportion for Sample 2 (without the instructional technology) is found as follows.

pˆ2=x2n2=3851≈0.745098

Now that we have the sample proportions, we can calculate the point estimate.

pˆ1−pˆ2=0.9−0.745098=0.154902

Step 2: Find the margin of error.

Notice that the samples are indeed independent of one another. Because they are two separate groups of students, they are not connected in any way. We can assume that the other necessary conditions are met to allow us to use the standard normal distribution to calculate the margin of error. The level of confidence is c=0.95
, so the critical value is zα2/=z0.052/=z0.025=1.96

. Substituting the values into the formula gives us the following.

E=zα2/pˆ1(1−pˆ1)n1+pˆ2(1−pˆ2)n2⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√=1.960.9(1−0.9)50+0.745098(1−0.745098)51⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯√≈0.145675

Step 3: Subtract the margin of error from and add the margin of error to the point estimate.

Subtracting the margin of error from the point estimate and then adding the margin of error to the point estimate gives us the following endpoints of the confidence interval.

Lower endpoint: (pˆ1−pˆ2)−E=0.154902−0.145675≈0.009Upper endpoint: (pˆ1−pˆ2)+E=0.154902+0.145675≈0.301

Thus, the 95%
confidence interval for the difference between the two population proportions ranges from 0.009 to 0.301

. The confidence interval can be written mathematically using either inequality symbols or interval notation, as shown below.

0.009<p1−p2<0.301

or

(0.009,0.301)

Therefore, we are 95%
confident that the percentage of students who passed the class is between 0.9% and 30.1% higher for the population of students who used the new instructional technology (Population 1) than for the population of students who did not use the technology (Population 2). Thus, with 95%

confidence, the professor can conclude that the new instructional technology improves students' scores.

To calculate the confidence interval for the difference between two proportions on the calculator, we don't need to find the individual sample proportions; we just need to enter the number of successes and the sample size for each sample, as well as the level of confidence. Press STAT , scroll to TESTS, and then choose option B:2-PropZInt. x1 is the number of successes from the first sample and n1 is the first sample's size. Similarly, x2 is the number of successes from the second sample and n2 is the second sample's size. As usual, C-Level is the confidence level, which must be entered as a decimal. The data should be entered as shown in the first screenshot below. After you select Calculate and press ENTER , the results will be displayed on the screen as shown in the second screenshot below.
2-PropZInt data entry screen with x_1 equal to 45, n_1 equal to 50, x_2 equal to 38, n_2 equal to 51, and C-Level equal to .95. 2-PropZInt results screen shows ( .00923 , .30057 ), p hat_1 equal to .9 , p hat_2 equal to .7450980392, n_1 equal to 50, and n_2 equal to 51.

Notice that the calculator gives the same interval but with more decimal places. The interpretation of the confidence interval is still the same. The proportion of students passing the class was higher for the population of students who used the new instructional technology than for the population of students who did not use the technology.