You have been hired as the quality control office of a pharmaceutical company that manufactures aspirin tablets. According to the quality assurance protocol, your job is to do the following. For each shipment of possibly thousands and thousands of aspirin tablets, randomly select and test 300 tablets, then accept the batch if there is no more than 13 tablets that don’t meet the test specification. If, at the pharmaceutical company, the probability that randomly selected tablet is defective is .04, what is the probability that a randomly selected shipment of 300 tablets will be accepted?
What is the probability that a randomly selected sample contains 20 that do not meet the requirements?
Is the value 20 unusual (from a above)
To calculate the probability that a randomly selected shipment of 300 tablets will be accepted, we need to determine the probability of having no more than 13 defective tablets in the sample.
First, let's find the probability of a tablet being defective: P(defective) = 0.04
Thus, the probability of a tablet passing the inspection is: P(passing) = 1 - P(defective) = 1 - 0.04 = 0.96
Using the binomial probability formula, the probability of having k defective tablets in a sample of size n is given by:
P(k defects) = C(n, k) * (P(passing))^k * (1 - P(passing))^(n - k)
In our case, n = 300, k <= 13, and P(passing) = 0.96.
Now, let's calculate the probability of having no more than 13 defective tablets:
P(acceptance) = P(0 defects) + P(1 defect) + ... + P(13 defects)
P(acceptance) = ∑ [ C(300, k) * (0.96)^k * (1 - 0.96)^(300 - k) ] for k = 0 to 13
Using a calculator or software to perform the calculations, we find that P(acceptance) ≈ 0.0776
Therefore, the probability that a randomly selected shipment of 300 tablets will be accepted is approximately 0.0776.
To find the probability that a randomly selected sample contains 20 tablets that do not meet the requirements, we use the same formula:
P(20 defects) = C(300, 20) * (0.04)^20 * (1 - 0.04)^(300 - 20)
Using a calculator or software to perform the calculations, we find that P(20 defects) ≈ 6.53 x 10^(-14)
The value of 20 is extremely small and unlikely to occur by chance, indicating that it is unusual from the above calculation.
1.Given that you were taking the following measurements, determine if they are discrete or continuous.
a. 10, 20, 20, 25, 30, 34, 37, 40
b. 0.5, 1, 2, 3, 4, 4.2, 4.4, 4.5, 8
2. Suppose that in a population of voters in a certain region 38% are in favor of particular bond issue. Nine hundred randomly selected voters are asked if they favor the bond issue. Find the probability that the sample proportion computed from a sample of size 900 will be within 5 percentage points of the true population proportion.
3. An online retailer claims that 90% of all orders are shipped within 12 hours of being received. A consumer group placed 121 orders of different sizes and at different times of day; 102 orders were shipped within 12 hours.
a. Compute the sample proportion of items shipped within 12 hours.
b. Assuming the retailer’s claim is true, find the probability that a sample of size 121 would produce a sample proportion so low as was observed in this sample.
c. Based on the answer to part (ii), draw a conclusion about the retailer’s claim.
4.Suppose you do a study of acupuncture to determine how effective it is in relieving pain. You measure sensory rates for 15 subjects with the results given. Use the sample data to construct a 95% confidence interval for the mean sensory rate for the population (assumed normal) from which you took the data. The collected data was as follows; 8.6, 9.4, 7.9, 6.8, 8.3, 7.3, 9.2, 9.6, 8.7, 11.4, 10.3, 5.4, 8.1, 5.5, 6.9
a. Find the z score for α = .04 for;
b. A left-tailed test on a standard normal distribution curve.
c. A right-tailed test on a standard normal distribution curve.
d. A two-tailed test on a standard normal distribution curve.
e. Find the t score for α = .05 for a sample size of 25 and degrees of freedom of 24. Assume an approximately normal distribution.
i. A left-tailed t test
ii. A right-tailed t test.
iii. A two-tailed t test.
5.The football coach randomly selected 10 players and timed how long each player took to perform a certain drill. The drill times in minutes were: 9, 5, 7,12,14,14,5, 11,6, 9
i. Calculate the mean of the sample?
ii. Calculate the variance of the sample?
iii. Calculate the standard deviation of the sample?
iv. Find the 95% confidence interval for the variance and standard deviation of the sample.
6. Suppose that a market research firm is hired to estimate the percent of adults living in a large city who have cell phones. Five hundred randomly selected adult residents in this city are surveyed to determine whether they have cell phones. Of the 500 people surveyed, 421 responded yes – they own cell phones.
i. Construct a 95% confidence interval for the proportion of adult residents of this city who have cell phones.
ii. Interpret the confidence interval found in part i.
7.You do a study of hypnotherapy to determine how effective it is in increasing the number of hours of sleep subjects get each night. You measure hours of sleep for 12 subjects with the following results. Construct a 95% confidence interval for the mean number of hours slept for the population (assumed normal) from which you took the data.
8.2; 9.1; 7.7; 8.6; 6.9; 11.2; 10.1; 9.9; 8.9; 9.2; 7.5; 10.5
1.
a. The measurements 10, 20, 20, 25, 30, 34, 37, 40 are discrete since they are whole numbers with a specific count.
b. The measurements 0.5, 1, 2, 3, 4, 4.2, 4.4, 4.5, 8 are continuous since they can have any value within a certain range.
2.
To find the probability that the sample proportion computed from a sample of size 900 will be within 5 percentage points of the true population proportion, we need to calculate the standard error.
Standard error = sqrt[(p(1-p))/n], where p is the true population proportion
Standard error = sqrt[(0.38(1-0.38))/900]
Using a calculator, we find that the standard error is approximately 0.0157.
To find the probability that the sample proportion is within 5 percentage points of the true population proportion, we need to find the area under the normal distribution curve within ±0.05 standard deviations of the mean.
Using a standard normal distribution table or calculator, we find that this probability is approximately 0.9904, or 99.04%.
3.
a. The sample proportion of items shipped within 12 hours is 102/121 ≈ 0.843.
b. Assuming the retailer's claim is true, we can use the binomial distribution to find the probability that a sample of size 121 would produce a sample proportion as low as observed. We calculate the probability of having 102 or fewer items shipped within 12 hours using the binomial distribution:
P(X ≤ 102) = ∑ [ C(121, x) * (0.90)^x * (1 - 0.90)^(121 - x) ] for x = 0 to 102
Using a calculator or software to perform the calculations, we find that P(X ≤ 102) ≈ 0.0069.
c. The probability of obtaining a sample proportion as low as observed under the assumption that the retailer's claim is true is very low (approximately 0.0069). This suggests that the retailer's claim may not be accurate or that the observed result is an outlier.
4.
a. To find the z score for α = 0.04, we need to find the z value that corresponds to an area of 0.04 in the left tail of the standard normal distribution. Using a standard normal distribution table or calculator, the z score is approximately -1.7507.
b. The critical region for a left-tailed test on a standard normal distribution with α = 0.04 is below -1.7507.
c. The critical region for a right-tailed test on a standard normal distribution with α = 0.04 is above 1.7507.
d. The critical region for a two-tailed test on a standard normal distribution with α = 0.04 is both below -1.96 and above 1.96.
e. To find the t score for α = 0.05 for a sample size of 25 and degrees of freedom of 24, we can use the t-distribution table or calculator. The t score is approximately 1.711.
i. The critical region for a left-tailed t-test is below -1.711.
ii. The critical region for a right-tailed t-test is above 1.711.
iii. The critical region for a two-tailed t-test is both below -1.711 and above 1.711.
5.
i. The mean of the sample is (9 + 5 + 7 + 12 + 14 + 14 + 5 + 11 + 6 + 9) / 10 = 8.8
ii. The variance of the sample can be calculated as:
[ (9 - 8.8)^2 + (5 - 8.8)^2 + (7 - 8.8)^2 + (12 - 8.8)^2 + (14 - 8.8)^2 + (14 - 8.8)^2 + (5 - 8.8)^2 + (11 - 8.8)^2 + (6 - 8.8)^2 + (9 - 8.8)^2 ] / 9
≈ 9.84
iii. The standard deviation of the sample is the square root of the variance, which is approximately √9.84 = 3.14
iv. To find the 95% confidence interval for the variance and standard deviation of the sample, we can use the chi-squared distribution. Assuming the sample is from a normally distributed population, the 95% confidence interval for the variance of the population is (n-1)s^2 / χ^2(α/2, n-1) to (n-1)s^2 / χ^2(1-α/2, n-1), where n is the sample size, s^2 is the sample variance, and χ^2 is the chi-squared distribution.
For the given sample, the 95% confidence interval for the variance is (9)(9.84) / χ^2(0.025, 9) to (9)(9.84) / χ^2(0.975, 9).
Using a chi-squared distribution table or calculator, we find that the 95% confidence interval for the variance is approximately 4.86 to 23.06.
The 95% confidence interval for the standard deviation is the square root of the interval for the variance, which is approximately √4.86 to √23.06.
Thus, the 95% confidence interval for the standard deviation is approximately 2.21 to 4.80.
6.
i. To construct a 95% confidence interval for the proportion of adult residents who have cell phones, we can use the formula:
p̂ ± Z * √( p̂(1-p̂) / n)
where p̂ is the sample proportion (421/500), Z is the Z-score for a 95% confidence level (approximately 1.96), and n is the sample size (500).
Calculating the confidence interval:
Confidence interval = p̂ ± 1.96 * √( p̂(1-p̂) / n)
Confidence interval = 0.842 ± 1.96 * √( 0.842(1-0.842) / 500)
Calculating the confidence interval using a calculator, the 95% confidence interval for the proportion of adult residents who have cell phones is approximately 0.809 to 0.875.
ii. The confidence interval found in part i means that we are 95% confident that the true proportion of adult residents in this city who have cell phones is between 0.809 and 0.875.
7.
To construct a 95% confidence interval for the mean number of hours slept for the population, we can use the formula:
x̄ ± t * (s / √n)
where x̄ is the sample mean, t is the t-score for a 95% confidence level with degrees of freedom n-1 (12-1=11), s is the sample standard deviation, and n is the sample size.
Calculating the confidence interval:
Confidence interval = x̄ ± t * (s / √n)
Given data: 8.2, 9.1, 7.7, 8.6, 6.9, 11.2, 10.1, 9.9, 8.9, 9.2, 7.5, 10.5
Calculating the sample mean: (8.2 + 9.1 + 7.7 + 8.6 + 6.9 + 11.2 + 10.1 + 9.9 + 8.9 + 9.2 + 7.5 + 10.5) / 12 ≈ 8.875
Calculating the sample standard deviation: √[ (8.2 - 8.875)^2 + (9.1 - 8.875)^2 + (7.7 - 8.875)^2 + (8.6 - 8.875)^2 + (6.9 - 8.875)^2 + (11.2 - 8.875)^2 + (10.1 - 8.875)^2 + (9.9 - 8.875)^2 + (8.9 - 8.875)^2 + (9.2 - 8.875)^2 + (7.5 - 8.875)^2 + (10.5 - 8.875)^2 ] / 11 ≈ 1.527
Calculating the confidence interval using a t-score table or calculator, we find that the 95% confidence interval for the mean number of hours slept for the population is approximately 8.875 ± t * (1.527 / √12).
Calculating the t-score for a 95% confidence level with 11 degrees of freedom, we find t ≈ 2.201.
The 95% confidence interval for the mean number of hours slept for the population is approximately 8.875 ± 2.201 * (1.527 / √12), which is approximately 8.214 to 9.536.
1. Describe the concept of statistics
2. Describe data
3. Describe variables
4. Analyze statistical problems using probabilities
5. Use sampling concepts in sampling procedures
6. Use statistical methods to test hypothesis
7. Estimate population parameters
8. Use statistical tests to interpret data
1. The concept of statistics involves the collection, analysis, interpretation, presentation, and organization of data. It is a branch of mathematics that deals with gathering information, summarizing it, and making inferences or predictions based on the data. Statistics helps us make sense of data by providing tools and techniques for analyzing patterns, relationships, and trends.
2. Data refers to the information or facts that are collected, observed, or measured from various sources. It can be in the form of numbers, words, measurements, or observations. Data can be classified into two main types: qualitative data, which describes qualities or characteristics, and quantitative data, which represents quantities or numerical values.
3. Variables are characteristics or attributes that can have different values and can vary among different individuals or objects. In statistics, variables are used to measure, record, or quantify specific characteristics or aspects of a population or sample. There are two main types of variables: independent variables, which are manipulated or controlled, and dependent variables, which are measured or observed and are expected to change in response to the independent variables.
4. Analyzing statistical problems using probabilities involves quantifying uncertainty and making informed predictions about future events or outcomes based on probability theory. Probability is a measure of the likelihood that a particular event or outcome will occur, and it is expressed as a number between 0 and 1. By using probabilities, statisticians can assess the likelihood of different outcomes, calculate expected values, and make optimal decisions based on available information.
5. Sampling concepts are used in sampling procedures to select a subset of individuals or objects from a larger population. The purpose of sampling is to gather representative data from a population without having to study or analyze every individual or object. Different sampling techniques, such as random sampling or stratified sampling, are used to ensure that the selected sample is unbiased and accurately represents the population of interest. Sampling allows us to make inferences about the population based on the analysis of the sample data.
6. Statistical methods are used to test hypotheses or propositions about a population based on sample data. Hypothesis testing involves making a claim or statement about a population parameter and then collecting and analyzing data to determine if there is enough evidence to support or reject the claim. Statistical tests, such as t-tests, chi-square tests, or ANOVA, are used to calculate the probability of obtaining the observed data under the null hypothesis and to make conclusions about the population based on the results.
7. Estimating population parameters involves using sample data to make inferences about unknown characteristics or properties of a population. By analyzing the sample data, statisticians can estimate population parameters, such as the mean, standard deviation, or proportion. Estimation techniques, such as point estimation or interval estimation, are used to provide estimates of the population parameters along with measures of uncertainty or
1.1 Definition of statistics
1.2 Statistic vs. statistics
1.3 Divisions of statistics
2.1 Types of data
2.2 Sources of data
2.3 Numerical presentation of data
2.4 Graphical and tabular presentation of data
2.5 Measurement scales
3.1 Types of variables
3.2 Measurement scales
.1 Definition of statistics
1.2 Statistic vs. statistics
1.3 Divisions of statistics
2.1 Types of data
2.2 Sources of data
2.3 Numerical presentation of data
2.4 Graphical and tabular presentation of data
2.5 Measurement scales
3.1 Types of variables
3.2 Measurement scales