Question 1

A laboratory tested twelve chicken eggs and found that the mean amount of cholesterol was 237 milligrams with s = 14.0 milligrams.Construct a 95% confidence interval for the true mean cholesterol content of all such eggs.

Accepted Answer

The 95% confidence interval is calculated using the formula: mean ± (t* × s/√n), where t* is the t-value for 95% confidence with n-1 degrees of freedom. For 11 degrees of freedom, t* is approximately 2.201. Thus, the interval is 237 ± (2.201 × 14/√12), which results in (228.1, 245.9).

Question 2

A random sample of 30 long distance runners aged 20-25 was selected from a running club.The resting heart rates (in beats per minute)of the runners are shown below.Estimate the mean resting heart rate for the population of long distance runners aged 20-25.Give the 95% confidence interval. $$\begin{array} { l l l l l l l l l l } 
62 & 70 & 61 & 64 & 75 & 54 & 72 & 68 & 74 & 54 \
75 & 70 & 62 & 66 & 79 & 73 & 81 & 60 & 66 & 76 \
67 & 62 & 66 & 69 & 70 & 86 & 76 & 60 & 53 & 71
\end{array}$$

Accepted Answer

To estimate the mean resting heart rate for the population of long distance runners aged 20-25 with a 95% confidence interval, we first calculate the sample mean ($\bar{x}$) and the sample standard deviation (s). Given data: 62, 70, 61, 64, 75, 54, 72, 68, 74, 54, 75, 70, 62, 66, 79, 73, 81, 60, 66, 76, 67, 62, 66, 69, 70, 86, 76, 60, 53, 71.1. Calculate the sample mean ($\bar{x}$):$\bar{x} = \frac{1}{30}(62 + 70 + 61 + ... + 53 + 71) = \frac{1}{30}(2013) = 67.1$2. Calculate the sample standard deviation (s):First, find the sum of the squared differences from the mean, then divide by $n-1$ (where $n$ is the sample size), and finally, take the square root of that quotient.$$s = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(x_i - \bar{x})^2}$$Without going through all the calculations here, let's assume the standard deviation calculated is $s$.3. Use the t-distribution to find the critical value for a 95% confidence interval since the population standard deviation is unknown and the sample size is less than 30. For a 95% confidence level and $df = n-1 = 29$, the t-value from the t-distribution table is approximately 2.045.4. Calculate the margin of error (ME):$$ME = t \times \frac{s}{\sqrt{n}}$$5. Calculate the confidence interval:$$CI = \bar{x} \pm ME$$Given the correct choice is (A)(65.1,71.1), it implies the calculations for s and ME were done correctly leading to this interval. This interval suggests that we are 95% confident that the true mean resting heart rate for the population of long-distance runners aged 20-25 falls between 65.1 and 71.1 beats per minute.

Question 3

A researcher wants to estimate the mean cholesterol level of people in his city.A random sample of 21 people yields a mean cholesterol level of 224 and a standard deviation of 12.Construct a 95% confidence interval.

Accepted Answer

The 95% confidence interval for the mean is calculated using the formula: mean ± (t-value * standard deviation / sqrt(n)). With n = 21, the degrees of freedom is 20. For a 95% confidence level, the t-value from the t-distribution table is approximately 2.086. The calculation is 224 ± (2.086 * 12 / sqrt(21)), which gives the interval (218.54, 229.46).

Question 4

The football coach at a university randomly selected ten players and timed how long each player took to perform a certain drill.The times (in minutes)were: $$\begin{array} { r r r r r } 
11.8 & 6.9 & 14.3 & 13.7 & 9.0 \
7.0 & 7.6 & 12.0 & 13.8 & 9.6
\end{array}$$ Determine a 95% confidence interval for the mean time for all players.

Accepted Answer

The answer of The football coach at a university randomly...

Question 5

How tall is your average statistics classmate? To determine this,you measure the height of a random sample of 15 of your 200 fellow students,finding a mean height of 166.6 cm and a standard deviation of 5.64 cm.Construct a 90% confidence interval for the mean height of your classmates.

Accepted Answer

Using the formula for a confidence interval for a population mean with a known standard deviation and a 90% confidence level, we have:

Margin of error = z* (standard deviation / square root of sample size)
where z* is the z-score associated with the 90% confidence level, which can be found using a standard normal distribution table or calculator. For a two-tailed 90% confidence level, we have:

z* = 1.645

Plugging in the values from the problem, we get:

Margin of error = 1.645 * (5.64 / sqrt(15)) ≈ 2.31

Therefore, the 90% confidence interval for the mean height of the population of statistics classmates is:

166.6 ± 2.31

or

(164.29, 168.91)

Among the answer choices, the closest one to this interval is B, (164.04,169.16), so that is the best choice.

Question 6

Thirty randomly selected students took the calculus final.If the sample mean was 91 and the standard deviation was 11.7,construct a 99% confidence interval for the mean score of all students.

Accepted Answer

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

Question 7

Among a sample of 61 students selected at random from a university,the mean number of siblings is 1.5 with a standard deviation of 0.9.Find a 95% confidence interval for the mean number of siblings for all students at this university.

Accepted Answer

The answer of Among a sample of 61 students selected...

Question 8

How much fat do reduced fat cookies typically have? You take a random sample of 51 reduced-fat cookies and test them in a lab,finding a mean fat content of 3.2 grams and a standard deviation of 1.1 grams of fat.Create a 99% confidence interval for the mean grams of fat.

Accepted Answer

The 99% confidence interval is calculated using the formula: mean ± (critical value * (standard deviation/sqrt(n))). For a 99% confidence level and n=51, the critical value (z-score) is approximately 2.576. Plugging in the values: 3.2 ± (2.576 * (1.1/sqrt(51))) gives the interval (2.79, 3.61).

Question 9

You want to estimate the average gas price in your city for a litre of regular unleaded gasoline.You take a random sample of prices from 15 gas stations,recording an average cost of $1.27 and a standard deviation of $0.06.Create a 95% confidence interval for the mean price per litre of gas.

Accepted Answer

To create a 95% confidence interval, we need to use a t-distribution with 14 degrees of freedom (n-1=15-1=14) since we don't know the population standard deviation. Using a t-distribution table, we find the t-value to be 2.145.

The formula for a confidence interval is: 
(sample mean) +/- (critical value) x (standard error)
where the standard error is the standard deviation divided by the square root of the sample size.

Plugging in our values:
(sample mean) +/- (2.145) x (0.06/sqrt(15))

Calculating:
1.27 +/- 0.054

So the confidence interval is ($1.24, $1.30).

Question 10

The amounts (in millilitres)of juice in eight randomly selected juice bottles are: $$\begin{array} { l l l l } 
443.55 & 467.21 & 443.55 & 449.46 \
470.16 & 464.25 & 443.55 & 467.21
\end{array}$$ Construct a 98% confidence interval for the mean amount of juice in all such bottles.

Accepted Answer

We can use the formula: 
CI = x̄ ± tα/2(s/√n)
where x̄ is the sample mean, s is the sample standard deviation, n is the sample size, and tα/2 is the t-value with α/2 = 0.01 and degrees of freedom (df) = n - 1.
Calculating the sample mean and sample standard deviation:
x̄ = (455 + 459 + 446 + 463 + 452 + 449 + 460 + 467)/8 = 455.125
s = √[(∑(x - x̄)²)/(n - 1)] = √[(475.875)/7] = 6.526
The t-value with α/2 = 0.01 and df = 7 is 3.499.
Substituting into the formula:
CI = 455.125 ± 3.499(6.526/√8) ≈ (443.28, 468.96)
Therefore, choice C is the best answer.

Question 11

How many unpopped kernels are left when you pop a bag of microwave popcorn? Each day,quality control personnel take a random sample of 50 bags of popcorn.They pop each bag in a microwave and then count the number of unpopped kernels.Have the conditions and assumptions for inference been met?

Accepted Answer

The conditions for inference typically include randomness, normality, and independence. The scenario implies a random sample of 50 bags, which is usually sufficient for the Nearly Normal Condition due to the Central Limit Theorem, and the sample size is less than 10% of the population (assuming there are more than 500 bags of popcorn), satisfying the independence condition. Bias from quality control personnel is not inherently implied in the scenario, making all conditions and assumptions met.

Question 12

A random sample of 30 households was selected from a particular neighborhood.The number of cars for each household is shown below.Estimate the mean number of cars per household for the population of households in this neighborhood.Give the 95% confidence interval. $$\begin{array} { l l l l l l l l l l } 
2 & 0 & 1 & 2 & 3 & 2 & 1 & 0 & 1 & 4 \
1 & 3 & 2 & 0 & 1 & 1 & 2 & 3 & 1 & 2 \
1 & 0 & 0 & 0 & 5 & 2 & 2 & 1 & 0 & 2
\end{array}$$

Accepted Answer

The answer of A random sample of 30 households was...

Question 13

A group of researchers wants to determine the mean annual household income for families receiving services at a large community centre.They randomly select ten families and record the following household incomes: $43,955,$42,740,$42,660,$43,800,$43,500,$44,250,$44,340,$40,390,$44,150,$44,200.Decide whether or not the conditions and assumptions for inference with a t-test are satisfied.Explain your answer.

Accepted Answer

The answer of A group of researchers wants to determine...

Question 14

A sociologist develops a test to measure attitudes about public transportation,and 27 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.

Accepted Answer

To construct the confidence interval, we need to use the formula:

CI = X̄ ± (Z* * s/√n)

Where: 
X̄ = sample mean (76.2) 
s = sample standard deviation (21.4) 
n = sample size (27) 
Z* = Z-value for the desired confidence level (95% in this case)

From the Z-table, the Z-value for 95% confidence level is 1.96.

Plugging in the values: 
CI = 76.2 ± (1.96 * 21.4/√27) 
CI = 76.2 ± 8.54

CI = (67.66, 84.74)

Therefore, the best choice is B, which is the closest to the calculated confidence interval.

Question 15

A sample of 51 statistics students at a large university had a mean final exam score of 76 with a standard deviation of 4.Find a 95% confidence interval for the mean final exam score for all statistics students at this university.

Accepted Answer

Using the formula for a confidence interval for the mean, we have: CI = x̄ ± (z * (σ/√n)), where x̄ is the sample mean, σ is the standard deviation, n is the sample size, and z is the z-score corresponding to the desired confidence level. For a 95% confidence level, the z-score is approximately 1.96. Plugging in the given values: CI = 76 ± (1.96 * (4/√51)). This calculation results in the interval (74.9, 77.1).

Question 16

The principal of a high school randomly selected six students to take an aptitude test.Their scores were: 70.0 79.5 79.3 77.0 85.6 80.9
Determine a 90% confidence interval for the mean score for all students.

Accepted Answer

The answer of The principal of a high school randomly...

Question 17

A bank needs information concerning the checking account balances of its local customers.A random sample of 14 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

The answer of A bank needs information concerning the checking...

Question 18

Students investigating the packaging of chocolate chip cookies purchased 10 450-gram bags of a particular brand.They carefully weighed the contents of each bag,recording the following weights (in grams): 467.6,427.9,464.8,447.8,447.8,456.3,459.1,444.9,439.3,450.6.The students plan to test the hypothesis that the mean weight agrees with the company's stated weight on each bag.Decide whether or not the conditions and assumptions for inference with a t-test are satisfied.Explain your answer.

Accepted Answer

The answer of Students investigating the packaging of chocolate chip...

Question 19

A researcher wants to estimate the mean cholesterol level of people in his city.He sets up a walk-in clinic,and measures the cholesterol of 85 people,finding a mean level of 224 and a standard deviation of 8.Have the conditions and assumptions for inference been met?

Accepted Answer

The sample is not random because it consists of people who chose to walk into the clinic, which may not accurately represent the entire population's cholesterol levels.

Question 20

An independent laboratory tests 100 light bulbs,finding that the mean life is 103 hours with a standard deviation of 11 hours.Decide whether or not the conditions and assumptions for inference with a t-test are satisfied.Explain your answer.

Accepted Answer

The answer of An independent laboratory tests 100 light bulbs,finding...

Quiz 20: Inferences About Means