Question 1

A sociologist develops a test to measure attitudes about public transportation,and 100 randomly selected subjects are given the test.Their mean score is 76.2 and their standard deviation is 21.4.Construct the 95% confidence interval for the mean score of all such subjects.

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Question 2

A company has surveyed a sample of its employees to find out how many might take advantage of a weight loss program.The table below shows the results.For which of the groups could we use our methods of inference to determine a 95% confidence interval for the proportion of employees who would be willing to participate in the program? $$\begin{array} { l | c | c | c } 
\text { Group } & \begin{array} { c } 
\text { Number of } \
\text { Employees }
\end{array} & \begin{array} { c } 
\text { Number } \
\text { Surveyed }
\end{array} & \begin{array} { c } 
\text { Percent to } \
\text { Participate }
\end{array} \
\hline \text { Building C } & 5200 & 250 & 8 \% \
\text { Building B } & 1190 & 150 & 6 \% \
\text { Building A } & 248 & 20 & 9 \%
\end{array}$$

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The answer of A company has surveyed a sample of...

Question 3

You want to estimate the average gas price in Canada for a litre of regular unleaded gasoline.You take a random sample of prices from 100 gas stations,recording an average cost of $1.37 and a standard deviation of $0.11.Create a 99% confidence interval for the mean price per litre of regular unleaded gasoline.

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Question 4

A pollster wishes to estimate the true proportion of voters in Ottawa,Ontario,who support the Conservative party.How many voters should be surveyed in order to be 95% confident that the true proportion is estimated to within 4%?

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Question 5

A laboratory tested 60 chicken eggs and found that the mean amount of cholesterol was 241 milligrams with s = 14.7 milligrams.Construct a 95% confidence interval for the true mean cholesterol content of all such eggs.

Accepted Answer

To construct the confidence interval, we can use the formula:

CI = x̄ ± t*(s/√n)

where x̄ is the sample mean (241), s is the sample standard deviation (14.7), n is the sample size (60), and t is the t-value corresponding to a 95% confidence level with 59 degrees of freedom (since n-1 = 59).

Using a t-table or calculator, we can find that the t-value is approximately 2.002. Substituting these values into the formula, we get:

CI = 241 ± 2.002*(14.7/√60)
CI = (237.3,244.7)

Therefore, the best choice is A) (237.3,244.7) as it is the correct confidence interval for the true mean cholesterol content of all such eggs with a 95% confidence level.

Question 6

A researcher wishes to estimate the proportion of fish in a certain lake that is inedible due to pollution of the lake.How large a sample should be tested in order to be 99% confident that the true proportion of inedible fish is estimated to within 6%?

Accepted Answer

We need to use the formula n = (z^2 * p * q)/E^2 where:
- n is the sample size we need to find
- z is the z-score associated with the desired level of confidence (99% in our case, so z = 2.58)
- p is the estimated proportion of inedible fish (unknown, we'll have to use 0.5 to get the maximum sample size)
- q is 1 - p
- E is the maximum error of estimate we allow, which is 6% or 0.06

Using these values, we get:
n = (2.58^2 * 0.5 * 0.5) / 0.06^2
n ≈ 460.89

Since we need a whole number for the sample size, we round up to the nearest integer:
n = 461

Therefore, we need to test 461 fish to be 99% confident that the true proportion of inedible fish is estimated to within 6%.

Question 7

A university's administrator proposes to do an analysis of the proportion of graduates who have not found employment in their major field one year after graduation.In previous years,the percentage averaged 13%.He wants the margin of error to be within 4% at a 99% confidence level.What sample size will suffice? Use 2.575 as the critical value for a 99% confidence interval.

Accepted Answer

The sample size can be calculated using the formula for margin of error: n = (Z^2 * p * (1-p)) / E^2, where Z is the critical value (2.575), p is the proportion (0.13), and E is the margin of error (0.04). Plugging in these values gives approximately 469.

Question 8

Among a sample of 100 students selected at random from one college,the mean number of siblings is 1.4 with a standard deviation of 1.3.Find a 95% confidence interval for the mean number of siblings for all students at this college.

Accepted Answer

The correct answer is D.The formula for a 95% confidence interval for the mean is:mean +/- 1.96 * (standard deviation / square root of sample size)Plugging in the values we have:1.4 +/- 1.96 * (1.3 / square root of 100)= 1.4 +/- 1.96 * (1.3 / 10)= 1.4 +/- 1.96 * 0.13= 1.4 +/- 0.25= (1.15, 1.65)

Question 9

A local bank needs information concerning the checking account balances of its local customers.A random sample of 144 accounts was checked and yielded a mean balance of $664.14 and a standard deviation of $297.29.Find a 98% confidence interval for the true mean checking account balance for local customers.

Accepted Answer

We can use the formula for a confidence interval:

CI = x̄ ± z* (σ/√n)

where x̄ is the sample mean, σ is the population standard deviation (or the sample standard deviation, which is used as an estimate when the population standard deviation is unknown), n is the sample size, and z* is the z-score for the desired confidence level.

For a 98% confidence level, the z-score is 2.33 (from a standard normal distribution table).

Plugging in the values we have:

CI = 664.14 ± 2.33 * (297.29/√144)

CI = 664.14 ± 2.33 * 24.77

CI = 664.14 ± 57.67

CI = (605.47, 722.81)

Rounding to two decimal places, we get (605.85, 722.43).

Therefore, the best choice is B.

Question 10

An advertising firm wants to confirm estimates that 18% of households will buy something on the internet next year.What sample size will produce a margin of error of 7% with 90% confidence level?

Accepted Answer

To find the sample size, use the formula for margin of error: ME = Z * sqrt[(p(1-p))/n]. For a 90% confidence level, Z is approximately 1.645. Plugging in the values: 0.07 = 1.645 * sqrt[(0.18(0.82))/n]. Solving for n gives approximately 82.

Question 11

A random sample of 64 statistics students at a small college were given a surprise quiz out of 40 marks.The average score was 27 with a standard deviation of 5.Find a 95% confidence interval for the mean quiz score for all statistics students at this college.

Accepted Answer

The 95% confidence interval is calculated using the formula: mean ± (Z * (standard deviation/√n)). Here, Z is 1.96 for 95% confidence, mean is 27, standard deviation is 5, and n is 64. This results in the interval (25.8, 28.2).

Question 12

In a survey of 1,000 television viewers,40% said they watch network news programs.For a 99% confidence level,the margin of error for this estimate is 3.99%.If we only want to be 90% confident,how will the margin of error change?

Accepted Answer

Lower confidence levels result in narrower confidence intervals, thus reducing the margin of error.

Question 13

A Statistics instructor wishes to estimate the proportion of first year statistics students that are over 30 years old.In his current class of 32,there are 2 students over 30 years old.Find a "Plus Four" 95% confidence interval for the true proportion.

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Question 14

A manufacturer wishes to estimate the proportion of washing machines leaving the factory that are defective.How large a sample should she check in order to be 90% confident that the true proportion is estimated to within 1.4%?

Accepted Answer

The answer of A manufacturer wishes to estimate the proportion...

Question 15

A company has surveyed a sample of its employees to find out how many might take advantage of a weight loss program.The table below shows the results.Give the 95% confidence interval for the group(s)for which we could use our methods of inference for the proportion of employees who would be willing to participate in the program. $$\begin{array} { l | c | c | c } 
\text { Group } & \begin{array} { c } 
\text { Number of } \
\text { Employees }
\end{array} & \begin{array} { c } 
\text { Number } \
\text { Surveyed }
\end{array} & \begin{array} { c } 
\text { Percent to } \
\text { Participate }
\end{array} \
\hline \text { Building A } & 5200 & 250 & 8 \% \
\text { Building B } & 1190 & 150 & 6 \% \
\text { Building C } & 248 & 20 & 9 \%
\end{array}$$

Accepted Answer

The answer of A company has surveyed a sample of...

Question 16

How much fat do reduced fat cookies typically have? You take a random sample of 120 reduced-fat cookies and test them in a lab,finding a mean fat content of 3.2 grams and a standard deviation of 1.9 grams of fat.Create a 99% confidence interval for the mean grams of fat.

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The answer of How much fat do reduced fat cookies...

Question 17

Eighty randomly selected students took the calculus final.If the sample mean was 76 and the standard deviation was 8.1,construct a 99% confidence interval for the mean score of all students.

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The answer of Eighty randomly selected students took the calculus...

Question 18

A survey of shoppers is planned to see what percentage use credit cards.Prior surveys suggest 56% of shoppers use credit cards.How many randomly selected shoppers must we survey in order to estimate the proportion of shoppers who use credit cards to within 1% with 95% confidence?

Accepted Answer

To determine the sample size needed for estimating a proportion with a specific margin of error and confidence level, we use the formula: n = (Z^2 * p * (1-p)) / E^2. For 95% confidence, Z is 1.96. Given p = 0.56 and E = 0.01, the calculation results in approximately 9466.

Question 19

A political party is interested in finding out what kind of popular support they might expect on an environmental initiative.Similar issues have gotten 87% support.The committee will set up a polling program to assure 95% confidence that the margin of error is less than 9%.What sample size will suffice?

Accepted Answer

We need to use the formula for sample size calculation for proportions:

n = (z^2 * p * q) / E^2

where:
z = z-score for the level of confidence (95% = 1.96)
p = estimated proportion (87% = 0.87)
q = 1 - p
E = maximum margin of error (9% = 0.09)

Plugging in the values:

n = (1.96^2 * 0.87 * 0.13) / 0.09^2
n = 53.64
n ≈ 54 (rounded up)

Therefore, the best choice is C with a sample size of 54.

Question 20

A company has surveyed a sample of its employees to find out how many might take advantage of a weight loss program.The table below shows the results.Give the 95% confidence interval for Building A for the proportion of employees who would be willing to participate in the program.Based on this interval,can we say that the proportion differs from 11%? At which level of significance are you making this decision? $$\begin{array} { l | c | c | c } 
\text { Group } & \begin{array} { c } 
\text { Number of } \
\text { Employees }
\end{array} & \begin{array} { c } 
\text { Number } \
\text { Surveyed }
\end{array} & \begin{array} { c } 
\text { Percent to } \
\text { Participate }
\end{array} \
\hline \text { Building A } & 5200 & 250 & 8 \% \
\text { Building B } & 1190 & 150 & 6 \% \
\text { Building C } & 248 & 20 & 9 \%
\end{array}$$

Accepted Answer

The answer of A company has surveyed a sample of...

Quiz 16: Confidence Intervals for Proportions

A sociologist develops a test to measure attitudes about public transportation,and 100 randomly selected subjects are given the test.Their mean score is 76.2 and their standard deviation is 21.4.Construct the 95% confidence interval for the mean score of all such subjects.

A pollster wishes to estimate the true proportion of voters in Ottawa,Ontario,who support the Conservative party.How many voters should be surveyed in order to be 95% confident that the true proportion is estimated to within 4%?

A laboratory tested 60 chicken eggs and found that the mean amount of cholesterol was 241 milligrams with s = 14.7 milligrams.Construct a 95% confidence interval for the true mean cholesterol content of all such eggs.

A researcher wishes to estimate the proportion of fish in a certain lake that is inedible due to pollution of the lake.How large a sample should be tested in order to be 99% confident that the true proportion of inedible fish is estimated to within 6%?

Among a sample of 100 students selected at random from one college,the mean number of siblings is 1.4 with a standard deviation of 1.3.Find a 95% confidence interval for the mean number of siblings for all students at this college.

An advertising firm wants to confirm estimates that 18% of households will buy something on the internet next year.What sample size will produce a margin of error of 7% with 90% confidence level?

A random sample of 64 statistics students at a small college were given a surprise quiz out of 40 marks.The average score was 27 with a standard deviation of 5.Find a 95% confidence interval for the mean quiz score for all statistics students at this college.

In a survey of 1,000 television viewers,40% said they watch network news programs.For a 99% confidence level,the margin of error for this estimate is 3.99%.If we only want to be 90% confident,how will the margin of error change?

A Statistics instructor wishes to estimate the proportion of first year statistics students that are over 30 years old.In his current class of 32,there are 2 students over 30 years old.Find a "Plus Four" 95% confidence interval for the true proportion.

A manufacturer wishes to estimate the proportion of washing machines leaving the factory that are defective.How large a sample should she check in order to be 90% confident that the true proportion is estimated to within 1.4%?

How much fat do reduced fat cookies typically have? You take a random sample of 120 reduced-fat cookies and test them in a lab,finding a mean fat content of 3.2 grams and a standard deviation of 1.9 grams of fat.Create a 99% confidence interval for the mean grams of fat.

Eighty randomly selected students took the calculus final.If the sample mean was 76 and the standard deviation was 8.1,construct a 99% confidence interval for the mean score of all students.

A survey of shoppers is planned to see what percentage use credit cards.Prior surveys suggest 56% of shoppers use credit cards.How many randomly selected shoppers must we survey in order to estimate the proportion of shoppers who use credit cards to within 1% with 95% confidence?