Quiz-10

Quiz-10

Problem: Last years Christmas shoppers spent a mean of \$643, \$438, \$409, and \$762 per family on holiday gifts. Assume the standard deviation of the average amount is \$100. Test whether with a level of significance of 10%, the amount spent on average each year is more than \$525

Solution: The following solution will be broken up into 4 steps.

Step 1.) Define your Hypotheses and Known Variables
-Alternative Hypothesis (Ha): The amount spent on average each year exceeds \$525 (m > \$525)
-Null Hypothesis (Ho): The amount spent on average each year does not exceed \$525 (m < \$525)
- Significance Level (Alpha): 0.1
-Standard Deviation (Sigma): 100
- Sample Size (n): 4
-Sample Mean (X-bar): 563
-Boundary between Ha & Ho (Mu)

Step 2.) Calculate Z-Value by Plugging Known Variables into Equation
Z= (X-bar - Mu)/(Sigma/Sqrt(Sample Size))
Or in this case,
(563 - 525)/(150/Sqrt(4)) = Z = .76

Step 3.) Calculate the P-Value of Our Yielded Test Statistic (Z-Value)
*NOTE*: P-Value direction ALWAYS follows that of the ALTERNATIVE HYPOTHESIS (m > \$525)
So in this case,
P(Z > .76)
1 - .7764 = .2236

Step 4.) Compare Resultant P-Value to the Level of Significance (Alpha) and Make a Decision Based on the Data
*NOTE*: The SMALLER value is ALWAYS the one that is ACCEPTED
Significance Level: 0.1
P-Value: .2236
So in this case,
0.1 < .2236

Conclusion: One can definitively say that based on the data, the Alternative Hypothesis can be rejected in favor of the Null Hypothesis. The amount spent on average each year does not exceed \$525 (m < \$525)