Ask Question
29 March, 19:40

A recent survey was conducted with a simple random sample of 35 shoppers at the food court in a local mall. The mall claims that the mean price for lunch is less than $4.65. The mean of the sample is $4.49, and the standard deviation of the population of lunch shoppers is $0.36. A P-value of 0.0043 is found using a 0.01 significance level to test the claim that the mean price for lunch is less than $4.65. State the conclusion about the null hypothesis.

+2
Answers (1)
  1. 29 March, 21:01
    0
    There is enough confidence to reject the claim that the mean price is less than $ 4.65

    Step-by-step explanation:

    Claim: The mean price for lunch is less than $4.65

    The Null hypothesis will be: The mean price for lunch is equal to or greater than $ 4.65

    Let u represents the mean price of lunch, the null and alternate hypothesis can be written as:

    Null Hypothesis: u ≥ 4.65

    Alternate Hypothesis: u < 4.65

    Calculated p-value = 0.0043

    Significance Level = α = 0.01

    Since the p value is less than the significance level, we can reject the claim that the mean price if less than $ 4.65

    Conclusion: There is enough confidence to reject the claim that the mean price is less than $ 4.65
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “A recent survey was conducted with a simple random sample of 35 shoppers at the food court in a local mall. The mall claims that the mean ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers