Exam 7: Inferences Based on a Single Sample: Estimation With Confidence Intervals

How much money does the average professional football fan spend on food at a single football game? That question was posed to ten randomly selected football fans. The sampled results show That the sample mean and sample standard deviation were $70.00 and $17.50, respectively. Use This information to create a 95 percent confidence interval for the population mean.

Suppose it is desired to estimate the average time a customer spends in a particular store to within 5 minutes $( \text { e.g. } \pm 5 \text { minutes) }$ at 99% reliability. It is estimated that the standard deviation of the Times is 15 minutes. How large a sample should be taken to get the desired interval?

A marketing research company is estimating which of two soft drinks college students prefer. random sample of n college students produced the following 95% confidence interval for the Proportion of college students who prefer drink A: (.262, .622). Identify the point estimate for Estimating the true proportion of college students who prefer that drink.

An educator wanted to look at the study habits of university students. As part of the research, data was collected for three variables - the amount of time (in hours per week)spent studying, the Amount of time (in hours per week)spent playing video games and the GPA - for a sample of 20 Male university students. As part of the research, a 95% confidence interval for the average GPA of All male university students was calculated to be: (2.95, 3.10). What assumption is necessary for the Confidence interval analysis to work properly?

A random sample of 4000 U.S. citizens yielded 2250 who are in favor of gun control legislation. Estimate the true proportion of all Americans who are in favor of gun control legislation using a 90% confidence interval.

Suppose a large labor union wishes to estimate the mean number of hours per month a union member is absent from work. The union decides to sample 496 of its members at random and Monitor the working time of each of them for 1 month. At the end of the month, the total number Of hours absent from work is recorded for each employee. Which of the following should be used To estimate the parameter of interest for this problem?

The volumes (in ounces)of juice in eight randomly selected juice bottles are as follows: 15.0 15.9 15.8 15.0 15.7 15.4 15.9 15.4 15.7 15.4 15.9 15.4 Find a $99 \%$ confidence interval for the standard deviation, $\sigma$ , of the volumes of juice in all such bottles. Round to the nearest hundredth when necessary.

It is desired to estimate the proportion of college students who feel a sudden relief now that their statistics class is over. How many students must be sampled in order to estimate the true Proportion to within 2% at the 90% confidence level?

How much money does the average professional football fan spend on food at a single football game? That question was posed to 51 randomly selected football fans. The sample results provided a sample mean and standard deviation of $\$ 11.00$ and $\$ 3.10$ , respectively. Find and interpret a $99 \%$ confidence interval for $\mu$ .

A study was conducted to determine what proportion of all college students considered themselves as full-time students. A random sample of 300 college students was selected and 210 of The students responded that they considered themselves full-time students. A computer program Was used to generate the following 95% confidence interval for the population proportion: (0.64814, 0.75186). The sample size that was used in this problem is considered a large sample. What criteria should be used to determine if n is large?

One way of reducing the width of a confidence interval is to reduce the confidence level.

What is the confidence coefficient in a 95% confidence interval for $\mu ?$

Find $\mathrm { z } _ { \alpha / 2 }$ for the given value of $\alpha$ . $\alpha = 0.01$

A retired statistician was interested in determining the average cost of a $200,000.00 term life insurance policy for a 60-year-old male non-smoker. He randomly sampled 65 subjects (60-year-old male non-smokers)and constructed the following 95 percent confidence interval for The mean cost of the term life insurance: ($850.00, $1050.00). Explain what the phrase ʺ95 percent Confidentʺ means in this situation.

A confidence interval was used to estimate the proportion of statistics students who are female. A random sample of 72 statistics students generated the following 99% confidence interval: (.438, .642). State the level of reliability used to create the confidence interval.

Given the values of $\bar { x } , \mathrm {~s}$ , and $n$ , form a $99 \%$ confidence interval for $\sigma$ . $\overline { \mathrm { x } } = 19.7 , \mathrm {~s} = 4.8 , \mathrm { n } = 14$

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 Students and finds that 118 of them are receiving financial aid. The 95% confidence interval for p is 59 ± .07. Interpret this interval.

The mean systolic blood pressure for a random sample of 28 women aged 18 -24 is 115.2 mm Hg and the standard deviation is 13.1 mm Hg. Construct a 90% confidence interval for the standard Deviation σ, of the systolic blood pressures of all women aged 18-24. Round to the nearest Hundredth when necessary.

To help consumers assess the risks they are taking, the Food and Drug Administration (FDA) publishes the amount of nicotine found in all commercial brands of cigarettes. A new cigarette has Recently been marketed. The FDA tests on this cigarette yielded mean nicotine content of 26.5 Milligrams and standard deviation of 2.6 milligrams for a sample of n = 9 cigarettes. Construct a 90% confidence interval for the mean nicotine content of this brand of cigarette.

One way of reducing the width of a confidence interval is to reduce the size of the sample taken.