Question 1197705: In a sample of credit card holders the mean monthly value of credit card purchases was $ 350 and the sample variance was 40 ($ squared). Assume that the population distribution is normal. Answer the following, rounding your answers to two decimal places where appropriate.
(a) Suppose the sample results were obtained from a random sample of 15 credit card holders. Find a 95% confidence interval for the mean monthly value of credit card purchases of all card holders.
(b) Suppose the sample results were obtained from a random sample of 22 credit card holders. Find a 95% confidence interval for the mean monthly value of credit card purchases of all card holders.
Before you answer (b) consider whether the confidence interval will be wider than or narrower than the confidence interval found for (a). Then check that your answer verifies this.
Answer by onyulee(41)   (Show Source):
You can put this solution on YOUR website! **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.
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