Statistical Hypothesis Testing for Discretionary Spending
The Federal Reserve stated in its latest report that the average amount of discretionary spending for a family of four last year was $3,457. You would like to test The Federal Reserves claim to see if this changed at the .05 level of significance, so you randomly select 50 families and get a sample mean of $3,635 with a standard deviation of $925. Chris OByrne 2024
A. What will be the alternative hypothesis? (just need to state 1-tail -Lower, 1-tail-Upper or 2-tail - just put one of those 3 answers in the body of the discussion board).
B. The value of the test statistic and which test statistic you used (example: z = 1.23 or t = 1.234) and make sure to round correctly depending on which test statistic you will be using.
C. What is the probability of the test statistic for this specific problem? (take into consideration if this is 1-tail or 2-tail test) Chris OByrne 2024
D. What conclusion will you come to? (just need to state Accept Ho or Reject Ho and Accept Ha just put one of those 2 answers in the body of the discussion board).
E. Using the critical value approach, at what critical value or values will you start rejecting the null hypothesis? (make sure you are very careful and detailed with your answer, pay attention to signs and values)
Statistical Hypothesis Testing for Discretionary Spending
In this analysis, we will evaluate the Federal Reserve's claim regarding the average discretionary spending of a family of four using a hypothesis testing approach. The key components of hypothesis testing will be addressed step-by-step.
A. Alternative Hypothesis
Since we want to see if there's a change in the average amount of discretionary spending (either an increase or decrease), the appropriate alternative hypothesis is:
- 2-tail
B. Test Statistic Calculation
To calculate the test statistic, we will use the following formula for the t-test since the sample size is less than 30 and we have the sample standard deviation:
[
t = \frac{\bar{x} - \mu}{s / \sqrt{n}}
]
Where:
- (\bar{x}) = sample mean = $3,635
- (\mu) = population mean (Federal Reserve's claim) = $3,457
- (s) = sample standard deviation = $925
- (n) = sample size = 50
Plugging in the values:
[
t = \frac{3635 - 3457}{925 / \sqrt{50}}
]
[
t = \frac{178}{130.43} \approx 1.36
]
Thus, the value of the test statistic is:
- t ≈ 1.36
C. Probability of the Test Statistic
To find the probability associated with this t-value, we need to refer to the t-distribution table (or use statistical software) with:
- Degrees of freedom (df) = n - 1 = 50 - 1 = 49
Using a t-table or calculator, we find:
- For t = 1.36 and df = 49, the corresponding two-tailed p-value is approximately 0.18.
D. Conclusion
Given our significance level ((\alpha)) of 0.05, and since our p-value (0.18) is greater than 0.05, we conclude:
- Accept (H_0)
E. Critical Value Approach
For a two-tailed test at the 0.05 significance level, we need to find the critical t-values for df = 49.
Using a t-table:
- The critical values for a two-tailed test at (\alpha = 0.05) are approximately ±2.009.
Thus, we will reject the null hypothesis if our test statistic is less than -2.009 or greater than +2.009.
In summary:
- Critical values: -2.009 and +2.009
- We will not reject (H_0) since our calculated t-value (1.36) falls within the range of -2.009 and +2.009.
This thorough analysis leads us to conclude that there is insufficient evidence to suggest that the average discretionary spending for a family of four has significantly changed from the Federal Reserve's reported amount of $3,457.