Question

In order to check whether iron ore is supplied to the specification of 62% Fe, a steel company has conducted a hypothesis test with the null hypothesis as \( H_0: \mu_{\text{Fe}} = 62% \) and alternative hypothesis \( H_a: \mu_{\text{Fe}} < 62% \). A random sample of 5 observations reveal the following grade values of the lot: 58%, 56%, 60%, 64%, 62%. The t-test statistic for the hypothesis is

• A. \( -3.000 \)
• B. \( 1.414 \)
• C. \( -1.414 \)
• D. \( 3.000 \)

Hint:

To calculate the t-statistic for a one-sample t-test, use the formula \( t = \frac{\overline{x} - \mu_0}{s/\sqrt{n}} \), where \( \overline{x} \) is the sample mean and \( s \) is the sample standard deviation.

Answer 1

The Correct Option is C

We are performing a one-sample t-test to check if the mean grade \( \mu_{\text{Fe}} \) is less than 62%. We have the following data: \[ \text{Sample: } 58%, 56%, 60%, 64%, 62% \] We are given the null hypothesis \( H_0: \mu = 62% \) and the alternative hypothesis \( H_a: \mu < 62% \). The formula for the t-statistic is: \[ t = \frac{\overline{x} - \mu_0}{\frac{s}{\sqrt{n}}} \] where:
- \( \overline{x} \) is the sample mean,
- \( \mu_0 \) is the population mean (62%),
- \( s \) is the sample standard deviation,
- \( n \) is the sample size.
Step 1: Calculate the sample mean \( \overline{x} \) The sample mean \( \overline{x} \) is the sum of the sample values divided by the number of observations: \[ \overline{x} = \frac{58 + 56 + 60 + 64 + 62}{5} = \frac{300}{5} = 60. \] Step 2: Calculate the sample standard deviation \( s \) The sample standard deviation is given by: \[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \overline{x})^2} \] Substitute the sample values and the mean: \[ s = \sqrt{\frac{1}{4} \left( (58-60)^2 + (56-60)^2 + (60-60)^2 + (64-60)^2 + (62-60)^2 \right)} \] \[ s = \sqrt{\frac{1}{4} \left( (-2)^2 + (-4)^2 + (0)^2 + (4)^2 + (2)^2 \right)} \] \[ s = \sqrt{\frac{1}{4} \left( 4 + 16 + 0 + 16 + 4 \right)} = \sqrt{\frac{40}{4}} = \sqrt{10} \approx 3.162. \] Step 3: Calculate the t-statistic Now that we have the sample mean and sample standard deviation, we can calculate the t-statistic: \[ t = \frac{\overline{x} - \mu_0}{\frac{s}{\sqrt{n}}} = \frac{60 - 62}{\frac{3.162}{\sqrt{5}}} = \frac{-2}{\frac{3.162}{2.236}} = \frac{-2}{1.414} \approx -1.414. \] Thus, the t-test statistic is \( -1.414 \), which corresponds to option (C).