Question

A telephone company’s records indicate that private customers pay on average R17.10 per month for long distance calls. A random sample of 10 customers’ accounts during the given month produced a sample mean of R22.10 expended for long distance calls and a sample variance of R45. A 5% significance test is to be performed to determine if the mean level of billing for long distance calls per month is more than R17.10. The calculated value of the test statistic and the critical value respectively are:

Answer 1

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.