Question 1

85 students at a college were asked whether they had completed their required English 101 course, and 48 students said "yes". Find the best point estimate for the proportion of students at the college who have completed their required English 101 course. Round to four decimal places.

Question 2

A new fertilizer was applied to the soil of 165 bean plants. 45% showed increased growth.

Find the margin of error and 95% confidence interval for the percentage ofallbean plants which show increased growth after application of the fertilizer. Report answers to at least 2 decimal places.

Margin of Error (as a percentage): Confidence Interval: % to %

Question 3

103 students at a college were asked whether they had completed their required English 101 course, and 79 students said "yes". Construct the 95% confidence interval for the proportion of students at the college who have completed their required English 101 course. Enter your answers as decimals (not percents) accurate to three decimal places. The Confidence Interval is ( , )

Question 4

Assume that a sample is used to estimate a population proportionp. Find the 99% confidence interval for a sample of size 188 with 77 successes. Enter your answer as anopen-interval(i.e., parentheses) using decimals (not percents) accurate to three decimal places. C.I. =$$

Question 5

Out of 100 people sampled, 9 had kids. Based on this, construct a 95% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to four places < p <

Question 6

Assume that a sample is used to estimate a population proportionp. Find the 95% confidence interval for a sample of size 335 with 242 successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

<p<

Question 7

Out of 200 people sampled, 10 preferred Candidate A. Based on this, estimate what proportion of the voting population (p) prefers Candidate A.

Use a 90% confidence level, and give your answers as decimals, to three places.

Question 8

If n=350 and X = 280, construct a 99% confidence interval. Give your answers to three decimals < p <

Question 9

The result of an English test was collected, and the grades and gender are summarized below

	A	B	C	Total
Male	16	13	8	37
Female	18	2	9	29
Total	34	15	17	66

Letpprepresent the proportion of all female students who would receive a grade of A on this test. Use a 90% confidence interval to estimatepptothree decimal places. Enter your answer as a tri-linear inequality using decimals (not percents).

Question 10

Giving a test to a group of students, the grades and gender are summarized below

	A	B	C	Total
Male	10	5	13	28
Female	9	20	14	43
Total	19	25	27	71

Letpprepresent the percentage of all male students who would receive a grade of A on this test. Use a 99.5% confidence interval to estimate p to three decimal places. Enter your answer as a tri-linear inequality using decimals (not percents).

< p <

Question 11

Express the confidence interval49.8%3.1%in the form of a trilinear inequality. %< p< %

Question 12

Express the confidence interval89.8%6.5%in the form of a trilinear inequality. %< p< %

Question 13

In our Condor Cafe, the Oxnard College Culinary Program offers the most delicious and affordable breakfast and lunch choices among the 152 Community Colleges in California. This week, a random sample of 183 orders was recorded, and 117 orders were for the $5-meals. To find the 99% confidence interval for the true proportion of the orders for the $5-meals, you need to use which one of the following calculators?

• Confidence Interval for a Population Mean Given Statistics
• Chi-Square Test for Independence
• Confidence Interval for a Population Mean Given Data
• One-Way ANOVA
• Hypothesis Test for a Population Proportion
• Two Independent Sample Means Comparison Given Data
• Two Independent Proportions Comparison
• Confidence Interval for a Population Proportion
• Hypothesis Test for a Population Mean Given Data
• Two Independent Sample Means Comparison Given Statistics
• Chi-Square Test for Goodness of Fit
• Hypothesis Test for a Population Mean Given Statistics
• Two Dependent Sample Means Comparison Given Data

1. Enter your answer as anopen-interval(i.e., parentheses) using decimals (not percents) accurate to three decimal places.

Confidence interval =$$

1. Express the same answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

<p<

1. Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places.

p=

Question 14

We wish to estimate what percent of adult residents in a certain county are parents. Out of 200 adult residents sampled, 104 had kids. Based on this, construct a 99% confidence interval for the proportion p of adult residents who are parents in this county. Express your answer in tri-inequality form. Give your answers as decimals, to three places. < p < Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places. p =

Question 15

Assume that a sample is used to estimate a population proportionp. To find the 98% confidence interval for a sample of size 98 with 32 successes, you need to use which one the following calculator

• Confidence Interval for a Population Mean Given Statistics
• Two Dependent Sample Means Comparison Given Data
• Confidence Interval for a Population Proportion
• Chi-Square Test for Independence
• Hypothesis Test for a Population Mean Given Statistics
• Hypothesis Test for a Population Mean Given Data
• Two Independent Sample Means Comparison Given Data
• Confidence Interval for a Population Mean Given Data
• Two Independent Proportions Comparison
• Chi-Square Test for Goodness of Fit
• Hypothesis Test for a Population Proportion
• Two Independent Sample Means Comparison Given Statistics
• One-Way ANOVA

1. Enter your answer as anopen-interval(i.e., parentheses) using decimals (not percents) accurate to three decimal places.

Confidence interval =$$

1. Express the same answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

<p<

1. Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places.

p=

Question 16

If n=590 and X = 472, to construct a 95% confidence interval, you need to use which one of the following calculator

• Hypothesis Test for a Population Proportion
• Confidence Interval for a Population Mean Given Data
• Confidence Interval for a Population Proportion
• Chi-Square Test for Independence
• Chi-Square Test for Goodness of Fit
• Confidence Interval for a Population Mean Given Statistics
• Two Independent Sample Means Comparison Given Data
• Two Independent Proportions Comparison
• Hypothesis Test for a Population Mean Given Data
• Hypothesis Test for a Population Mean Given Statistics
• Two Dependent Sample Means Comparison Given Data
• Two Independent Sample Means Comparison Given Statistics
• One-Way ANOVA

1. Enter your answer as anopen-interval(i.e., parentheses) using decimals (not percents) accurate to three decimal places.

Confidence interval =$$

1. Express the same answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

<p<

1. Express the same answer using the point estimate and margin of error. Give your answers as decimals, to three places.

p=

Question 17

A newsgroup is interested in constructing a 90% confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 525 randomly selected Americans surveyed, 372 were in favor of the initiative. Round answers to 4 decimal places where possible.

a. With 90% confidence the proportion of all Americans who favor the new Green initiative is between and .

b. If many groups of 525 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about percent will not contain the true population proportion.

Hint:Hints Video[+]

Question 18

A newsgroup is interested in constructing a 90% confidence interval for the proportion of all Americans who are in favor of a new Green initiative. Of the 581 randomly selected Americans surveyed, 381 were in favor of the initiative.

a. With 90% confidence the proportion of all Americans who favor the new Green initiative is between and .

b. If many groups of 581 randomly selected Americans were surveyed, then a different confidence interval would be produced from each group. About .percent of these confidence intervals will contain the true population proportion of Americans who favor the Green initiative and about percent will not contain the true population proportion.

Question 19

A smart phone manufacturer is interested in constructing a 90% confidence interval for the proportion of smart phones that break before the warranty expires. 94 of the 1698 randomly selected smart phones broke before the warranty expired. Round answers to 4 decimal places where possible.

a. With 90% confidence the proportion of all smart phones that break before the warranty expires is between and .

b. If many groups of 1698 randomly selected smart phones are selected, then a different confidence interval would be produced for each group. About percent of these confidence intervals will contain the true population proportion of all smart phones that break before the warranty expires and about percent will not contain the true population proportion.

Question 20

A psychologist is interested in constructing a 95% confidence interval for the proportion of people who accept the theory that a person's spirit is no more than the complicated network of neurons in the brain. 56 of the 711 randomly selected people who were surveyed agreed with this theory.

a. With 95% confidence the proportion of all people who accept the theory that a person's spirit is no more than the complicated network of neurons in the brain is betweenand .

b. If many groups of 711 randomly selected people are surveyed, then a different confidence interval would be produced from each group. About percent of these confidence intervals will contain the true population proportion of all people who accept the theory that a person's spirit is no more than the complicated network of neurons in the brain and about percent will not contain the true population proportion.