Question 1197705
**a) Sample size of 15**

* **Given:**
    * Sample mean (x̄) = $350
    * Sample variance (s²) = 40 
    * Sample standard deviation (s) = √40 = $6.32
    * Sample size (n) = 15
    * Confidence level: 95% 
    * Degrees of freedom (df) = n - 1 = 15 - 1 = 14

* **Find the critical t-value:**
    * For a 95% confidence level and 14 degrees of freedom, the critical t-value (tα/2) from a t-distribution table is approximately 2.145.

* **Calculate the standard error:** 
    * Standard Error (SE) = s / √n = 6.32 / √15 ≈ 1.63

* **Calculate the margin of error:**
    * Margin of Error = tα/2 * SE = 2.145 * 1.63 ≈ 3.50

* **Calculate the confidence interval:**
    * Lower limit: x̄ - Margin of Error = 350 - 3.50 = $346.50
    * Upper limit: x̄ + Margin of Error = 350 + 3.50 = $353.50

* **95% Confidence Interval:** ($346.50, $353.50)

**b) Sample size of 22**

* **Prediction:** 
    * The confidence interval with a larger sample size (n = 22) will be narrower than the interval with a smaller sample size (n = 15). 
    * This is because a larger sample size generally provides a more precise estimate of the population parameter.

* **Calculations:**
    * Degrees of freedom (df) = 22 - 1 = 21
    * Critical t-value (tα/2) for 95% confidence and 21 degrees of freedom is approximately 2.080.
    * Standard Error (SE) = 6.32 / √22 ≈ 1.35
    * Margin of Error = 2.080 * 1.35 ≈ 2.81
    * Lower limit: 350 - 2.81 = $347.19
    * Upper limit: 350 + 2.81 = $352.81

* **95% Confidence Interval:** ($347.19, $352.81)

**Verification:**

* The confidence interval for the sample size of 22 is indeed narrower than the interval for the sample size of 15, as predicted.

**In Summary:**

* The 95% confidence interval for the mean monthly value of credit card purchases for a sample size of 15 is ($346.50, $353.50).
* The 95% confidence interval for the mean monthly value of credit card purchases for a sample size of 22 is ($347.19, $352.81).
* As expected, the larger sample size results in a narrower confidence interval, indicating a more precise estimate of the population mean.