Arecent 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.

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

4+n=5

step-by-step explanation:

this is because the four is four hours and the n represents the 1 extra hour over the weekend.

b= i'm srry i don't know that one.

4+n=5 n=1 because if vernon plays 5 hours a week then 4+1=5 hours.

hope this : )

your answer is 228

step-by-step explanation:

add the top and bottom bases together, then divide them by 2, then multiply by the height.