A recent survey was conducted with a simple random sample of 35 shoppers at the food court tin 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 tot test the claim that the mean price for lunch is less than $4.65. State the conclusion about the null hypothesis. Could anyone show me how to do this?

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1 Answer

SAMPLE STATISTICS: mean price of lunch (x bar)=$4.49, size (n)=35 shoppers.

POPULATION STATISTIC: standard deviation (σ)=$0.36.

CLAIM: Mean price of lunch is less than $4.65.

COUNTERCLAIM: Mean price of lunch is greater than or equal to $4.65.

Null hypothesis, H₀: µ=$4.65 (from the counterclaim, because greater than or equal specifically includes the equality)

Alternative hypothesis, H₁: µ<$4.65 (the claim)

SIGNIFICANCE LEVEL: ɑ=0.01 (99% confidence level), left tail

TEST STATISTIC: Z=(x bar-µ)/(σ/√n)=(4.49-4.65)/(0.36/√35)=-2.63, which does indeed correspond to a P-value of 0.0043.

ɑ=0.01 (99% confidence level) and sample P-value 0.0043<0.01, which means that H₀ is rejected, and H₁ is accepted to be true. The conclusion is the claim that the mean price of lunch is less than $4.65.

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