A government agency reports that the mean amount of earnings for full-time workers ages 18 to 24 with a bachelor’s degree in a recent year is $52,133. In a random sample of 15 full-time workers ages 18 to 24 with a bachelor’s degree, the mean amount of earnings is $48,400 and the standard deviation is $6679. At α=0.05, is there enough evidence to reject the claim? Assume the population is normally distributed.
Table of contents
- 1. Intro to Stats and Collecting Data55m
- 2. Describing Data with Tables and Graphs1h 55m
- 3. Describing Data Numerically1h 45m
- 4. Probability2h 16m
- 5. Binomial Distribution & Discrete Random Variables2h 33m
- 6. Normal Distribution and Continuous Random Variables1h 38m
- 7. Sampling Distributions & Confidence Intervals: Mean1h 53m
- 8. Sampling Distributions & Confidence Intervals: Proportion1h 12m
- 9. Hypothesis Testing for One Sample2h 19m
- 10. Hypothesis Testing for Two Samples3h 22m
- 11. Correlation1h 6m
- 12. Regression1h 4m
- 13. Chi-Square Tests & Goodness of Fit1h 20m
- 14. ANOVA1h 0m
9. Hypothesis Testing for One Sample
Performing Hypothesis Tests: Means
Problem 7.T.3
Textbook Question
A travel analyst says that the mean price of a meal for a family of 4 in a resort restaurant is at most $100. A random sample of 33 meal prices for families of 4 has a mean of $110 and a standard deviation of $19. At α=0.01, is there enough evidence to reject the analyst’s claim?

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Step 1: Identify the null hypothesis (H₀) and the alternative hypothesis (H₁). The null hypothesis is H₀: μ ≤ 100 (the mean price is at most $100), and the alternative hypothesis is H₁: μ > 100 (the mean price is greater than $100). This is a one-tailed test.
Step 2: Determine the test statistic formula for a one-sample t-test. The formula is t = (x̄ - μ₀) / (s / √n), where x̄ is the sample mean, μ₀ is the hypothesized mean, s is the sample standard deviation, and n is the sample size.
Step 3: Substitute the given values into the formula. Here, x̄ = 110, μ₀ = 100, s = 19, and n = 33. Calculate the test statistic t using these values.
Step 4: Determine the critical t-value for a one-tailed test at α = 0.01 with degrees of freedom df = n - 1 = 33 - 1 = 32. Use a t-distribution table or statistical software to find the critical t-value.
Step 5: Compare the calculated t-value to the critical t-value. If the calculated t-value is greater than the critical t-value, reject the null hypothesis H₀. Otherwise, fail to reject H₀. Interpret the result in the context of the problem.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Hypothesis Testing
Hypothesis testing is a statistical method used to make decisions about a population based on sample data. It involves formulating a null hypothesis (H0) and an alternative hypothesis (H1). In this case, the null hypothesis states that the mean price of a meal is at most $100, while the alternative suggests it is greater. The goal is to determine if the sample data provides sufficient evidence to reject the null hypothesis.
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Step 1: Write Hypotheses
P-Value
The p-value is a measure that helps determine the significance of the results in hypothesis testing. It represents the probability of observing the sample data, or something more extreme, assuming the null hypothesis is true. A smaller p-value indicates stronger evidence against the null hypothesis. In this scenario, the p-value will be compared to the significance level (α = 0.01) to decide whether to reject the analyst's claim.
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Step 3: Get P-Value
Confidence Intervals
A confidence interval is a range of values, derived from sample statistics, that is likely to contain the population parameter with a specified level of confidence. It provides an estimate of the uncertainty around the sample mean. In this case, constructing a confidence interval for the mean meal price can help assess whether the true mean could be above $100, thus supporting or refuting the analyst's claim.
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