To conduct a hypothesis test, we first define our null and alternative hypotheses:
- Null Hypothesis (\(H_0\)): \(\mu = 17.10\) (the mean monthly billing for long-distance calls is R17.10)
- Alternative Hypothesis (\(H_a\)): \(\mu > 17.10\) (the mean monthly billing for long-distance calls is greater than R17.10)
We will perform a one-sample t-test since the sample size is small (n = 10) and the population variance is unknown.
Step 1: Calculate the test statistic
We use the t-test formula: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \] Where:
- \(\bar{x}\) = sample mean = R22.10
- \(\mu_0\) = population mean under the null hypothesis = R17.10
- \(s\) = sample standard deviation (the square root of the variance)
- \(n\) = sample size = 10
Given:
- Sample variance = R45
- Sample standard deviation \(s = \sqrt{45} \approx 6.708\)
Now, substitute the values into the t-test formula: \[ t = \frac{22.10 - 17.10}{6.708 / \sqrt{10}} \]
Calculating \(s / \sqrt{n}\): \[ s / \sqrt{n} = \frac{6.708}{\sqrt{10}} \approx \frac{6.708}{3.162} \approx 2.12 \]
Now compute \(t\): \[ t \approx \frac{5.00}{2.12} \approx 2.36 \]
Step 2: Determine the critical value
For a one-tailed t-test at a 5% significance level (α = 0.05) with \(n - 1 = 10 - 1 = 9\) degrees of freedom, we can look up the critical t-value from a t-distribution table.
The critical t-value for a one-tailed test at \(df = 9\) and \(\alpha = 0.05\) is approximately: \[ t_{critical} \approx 1.833 \]
Results
- Calculated value of the test statistic: \(t \approx 2.36\)
- Critical value: \(t_{critical} \approx 1.833\)
Thus, to summarize:
- Test Statistic (t): 2.36
- Critical Value (t): 1.833
Since \(2.36 > 1.833\), we would reject the null hypothesis that the mean long-distance call billing is R17.10, suggesting that it is indeed greater.
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