Waiting in LineOne aspect of queuing theory is to consider waiting time in lines. A fast-food chain is trying to determine whether it should switch from having four cash registers with four separate lines to four cash registers with a single line. It has been determined that the mean wait-time in both lines is equal. However, the chain is uncertain about which line has less variability in wait time. From experience, the chain knows that the wait times in the four separate lines are normally distributed with σ = 1.2 minutes. In a study, the chain reconfigured five restaurants to have a single line and measured the wait times for 50 randomly selected customers. The sample standard deviation was determined to be s = 0.84 minute. Is the variability in wait time less for a single line than for multiple lines at the α = 0.05 level of significance?
Table of contents
- 1. Intro to Stats and Collecting Data1h 14m
- 2. Describing Data with Tables and Graphs1h 55m
- 3. Describing Data Numerically2h 5m
- 4. Probability2h 16m
- 5. Binomial Distribution & Discrete Random Variables3h 6m
- 6. Normal Distribution and Continuous Random Variables2h 11m
- 7. Sampling Distributions & Confidence Intervals: Mean3h 23m
- Sampling Distribution of the Sample Mean and Central Limit Theorem19m
- Distribution of Sample Mean - Excel23m
- Introduction to Confidence Intervals15m
- Confidence Intervals for Population Mean1h 18m
- Determining the Minimum Sample Size Required12m
- Finding Probabilities and T Critical Values - Excel28m
- Confidence Intervals for Population Means - Excel25m
- 8. Sampling Distributions & Confidence Intervals: Proportion1h 25m
- 9. Hypothesis Testing for One Sample3h 29m
- 10. Hypothesis Testing for Two Samples4h 50m
- Two Proportions1h 13m
- Two Proportions Hypothesis Test - Excel28m
- Two Means - Unknown, Unequal Variance1h 3m
- Two Means - Unknown Variances Hypothesis Test - Excel12m
- Two Means - Unknown, Equal Variance15m
- Two Means - Unknown, Equal Variances Hypothesis Test - Excel9m
- Two Means - Known Variance12m
- Two Means - Sigma Known Hypothesis Test - Excel21m
- Two Means - Matched Pairs (Dependent Samples)42m
- Matched Pairs Hypothesis Test - Excel12m
- 11. Correlation1h 24m
- 12. Regression1h 50m
- 13. Chi-Square Tests & Goodness of Fit2h 21m
- 14. ANOVA1h 57m
9. Hypothesis Testing for One Sample
Steps in Hypothesis Testing
Problem 8.T.5b
Textbook Question
According to the National Center for Health Statistics, 22.4% of adults are smokers. A random sample of 300 adults is obtained.
b. In a random sample of 300 adults, what is the probability that at least 50 are smokers?
Verified step by step guidance1
Identify the type of distribution to use. Since we are dealing with the number of smokers in a sample and each adult can be considered a Bernoulli trial (smoker or not), the number of smokers follows a Binomial distribution with parameters \(n = 300\) and \(p = 0.224\).
Define the random variable \(X\) as the number of smokers in the sample. Then, \(X \sim \text{Binomial}(n=300, p=0.224)\).
The problem asks for the probability that at least 50 adults are smokers, which can be written as \(P(X \geq 50)\).
Since calculating binomial probabilities directly for large \(n\) can be cumbersome, consider using a normal approximation to the binomial distribution. The mean and standard deviation of \(X\) are given by \(\mu = np\) and \(\sigma = \sqrt{np(1-p)}\) respectively.
Apply the continuity correction by calculating \(P(X \geq 50) \approx P\left(Y \geq 49.5\right)\) where \(Y\) is a normal random variable with mean \(\mu\) and standard deviation \(\sigma\). Then convert to the standard normal variable \(Z = \frac{Y - \mu}{\sigma}\) and find the corresponding probability using standard normal distribution tables or software.
Verified video answer for a similar problem:This video solution was recommended by our tutors as helpful for the problem above
Video duration:
2mPlay a video:
Was this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Binomial Distribution
The binomial distribution models the number of successes in a fixed number of independent trials, each with the same probability of success. Here, each adult either is a smoker (success) or not, with probability 0.224. The distribution helps calculate probabilities for different counts of smokers in the sample.
Recommended video:
Guided course
Mean & Standard Deviation of Binomial Distribution
Normal Approximation to the Binomial
When the sample size is large, the binomial distribution can be approximated by a normal distribution with mean np and variance np(1-p). This simplifies probability calculations, especially for values like 'at least 50 smokers' in a sample of 300, where exact binomial calculations are complex.
Recommended video:
Using the Normal Distribution to Approximate Binomial Probabilities
Continuity Correction
When using the normal approximation for a discrete binomial variable, a continuity correction adjusts for the difference between discrete and continuous distributions. For example, to find P(X ≥ 50), we use P(X ≥ 49.5) in the normal distribution to improve accuracy.
Recommended video:
Using the Normal Distribution to Approximate Binomial Probabilities
Watch next
Master Step 1: Write Hypotheses with a bite sized video explanation from Patrick
Start learningRelated Videos
Related Practice
Textbook Question
3
views
