To determine whether the average checkout rate of the alternate system differs from that of the existing system, we will conduct a hypothesis test for the mean.
Step 1: Define the hypotheses
-
Null hypothesis (\(H_0\)): There is no difference in average checkout rates.
- \(H_0: \mu = 9.45\)
-
Alternative hypothesis (\(H_a\)): There is a difference in average checkout rates.
- \(H_a: \mu \neq 9.45\)
Step 2: Gather the necessary information
- Sample size (\(n\)) = 45
- Sample mean (\(\bar{x}\)) = 9.61
- Population mean (\(\mu_0\)) = 9.45
- Sample standard deviation (\(s\)) = 0.62
- Significance level (\(\alpha\)) = 0.01
Step 3: Calculate the test statistic
We will use a t-test since the sample size is relatively small, and we are estimating the population standard deviation from the sample.
The formula for the test statistic \(t\) is:
\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]
Substituting the values:
\[ t = \frac{9.61 - 9.45}{0.62 / \sqrt{45}} \]
Calculating the denominator:
\[ s / \sqrt{n} = 0.62 / \sqrt{45} \approx 0.62 / 6.708 = 0.0923 \]
Now, substituting back into the \(t\) formula:
\[ t = \frac{9.61 - 9.45}{0.0923} \approx \frac{0.16}{0.0923} \approx 1.73 \]
Step 4: Determine the critical value
For a two-tailed test at the 1% level of significance, we need to find the critical values of \(t\) from the t-distribution table. The degrees of freedom (df) is calculated as:
\[ df = n - 1 = 45 - 1 = 44 \]
Using the \(t\) distribution table, for \(df = 44\) and \(\alpha = 0.01\) (for a two-tailed test, we divide \(\alpha\) by 2):
\[ \alpha/2 = 0.005 \]
The critical t-values for \(df = 44\) at a 1% significance level (two-tailed) are approximately:
\(-t_{0.005, 44} \approx -2.688\) and \(t_{0.005, 44} \approx 2.688\).
Step 5: Decision
Now we compare the calculated t-value to the critical values:
- If \(t < -2.688\) or \(t > 2.688\), we reject \(H_0\).
- If \(-2.688 \leq t \leq 2.688\), we fail to reject \(H_0\).
In our case,
\[ t \approx 1.73 \]
Since \(1.73\) is within the range \(-2.688\) and \(2.688\), we fail to reject the null hypothesis.
Conclusion
At the 1% level of significance, there is not enough evidence to conclude that the average checkout rate of the alternate system differs from that of the scanning equipment used by the supermarket chain. The critical values are approximately \(-2.688\) and \(2.688\).