A teller at a branch of a savings bank located in a certain community has averaged 300 transactions daily over the past year. A random sample of 20 days during this year indicates a mean of 304.4 transactions with a variance of 479.61 transactions. Assuming that the daily transactions are normally distributed, test at a 5% level of significance whether the population mean daily transactions has increased. Calculate the test statistic.

Answer 1

To test whether the mean daily transactions have increased, we can perform a one-sample t-test. Here's how to set it up:

1. State the hypotheses:

   • Null hypothesis (\(H_0\)): The population mean daily transactions is 300 (i.e., \(\mu = 300\)).
   • Alternative hypothesis (\(H_a\)): The population mean daily transactions has increased (i.e., \(\mu > 300\)).

2. Identify the sample statistics:

   • Sample mean (\(\bar{x}\)) = 304.4
   • Sample variance (\(s^2\)) = 479.61
   • Sample size (\(n\)) = 20

3. Calculate the sample standard deviation: The sample standard deviation (\(s\)) is the square root of the variance: \[ s = \sqrt{479.61} \approx 21.9 \]

4. Calculate the test statistic: The t-test statistic is computed using the formula: \[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \] where \(\mu\) is the population mean (300), \(\bar{x}\) is the sample mean (304.4), \(s\) is the sample standard deviation (21.9), and \(n\) is the sample size (20).

   Plugging in the values: \[ t = \frac{304.4 - 300}{21.9 / \sqrt{20}} \]

   First, calculate the denominator: \[ s / \sqrt{n} = 21.9 / \sqrt{20} \approx 21.9 / 4.472 \approx 4.9 \]

   Now, calculate \(t\): \[ t = \frac{304.4 - 300}{4.9} \approx \frac{4.4}{4.9} \approx 0.898 \]

5. Conclusion: The computed t-statistic is approximately 0.898.

Next, we would compare this value to the critical t-value from t-distribution for \(n-1 = 20-1 = 19\) degrees of freedom at a 5% significance level for a one-tailed test. The critical t-value can be found in a t-distribution table or using statistical software.

If we find that our computed t-value is less than the critical t-value, we do not reject the null hypothesis. Conversely, if it exceeds the critical value, we would reject the null hypothesis, indicating that the population mean daily transactions has indeed increased.

For \(df = 19\) at a 5% significance level for a one-tailed test, the critical t-value is approximately 1.645. Since 0.898 < 1.645, we do not reject the null hypothesis.

Thus, there is not enough evidence to conclude that the population mean daily transactions have increased.