A simple random sample of 50 adults is obtained, and each person’s red blood cell count (in cells per microliter) is measured. The sample mean is 5.23 and the sample standard deviation is 0.54. Find the test statistic used to test the claim that the sample is from a population with a mean less than 5.4.

Question

A) t = –2.226
B) t = 2.226
C) t = –1.636
D) t = 1.636

Answer 1

To calculate the test statistic for a sample mean, we use the formula for the t-statistic:

\[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \]

where:

• \(\bar{x}\) is the sample mean,
• \(\mu\) is the population mean under the null hypothesis,
• \(s\) is the sample standard deviation, and
• \(n\) is the sample size.

In this case:

• \(\bar{x} = 5.23\)
• \(\mu = 5.4\) (we want to test if the mean is less than this)
• \(s = 0.54\)
• \(n = 50\)

Now we can plug in the values:

1. Calculate the numerator: \[ \bar{x} - \mu = 5.23 - 5.4 = -0.17 \]

2. Calculate the denominator: \[ s / \sqrt{n} = 0.54 / \sqrt{50} \approx 0.54 / 7.071 = 0.0763 \]

3. Now calculate the t-statistic: \[ t = \frac{-0.17}{0.0763} \approx -2.227 \]

However, based on the options provided, this value appears to round to:

\[ t \approx -2.226 \]

So, the correct answer is:

A) t = –2.226