Question

question 3

8 pts

a researcher wants to determine if the average height of adult males in a city is different from the national average (\mu), which is known to be 175 cm. the researcher collects a random sample of 49 adult males from the city and measure their heights (in cm). the average height of the sample (\bar{x}) was found to be 171.8 cm with a sample standard deviation (s) of 4.6 cm.

(a) calculate the test statistic to test whether the average height of adult males in a city differs from the national average of 175 cm:
(round your answer to 2 decimal places)

(b) determine the degrees of freedom for the test:

(c) calculate the p-value for the test:

(d) state your decision at the 0.05 level of significance. enter r if you reject the null hypothesis, or n if you do not reject it:

Explanation:

⚡ Using what you learned: Hypothesis Testing

Step 1: Identify the given values

Extract the parameters from the problem description:

• Null hypothesis mean: \(\mu_0 = 175\)
• Sample size: \(n = 49\)
• Sample mean: \(\bar{X} = 171.8\)
• Sample standard deviation: \(s = 4.6\)

Step 2: Calculate the test statistic

Since the population standard deviation \(\sigma\) is unknown and we use the sample standard deviation \(s\), we use a one-sample \(t\)-test:

$$ t = \frac{\bar{X} - \mu_0}{\frac{s}{\sqrt{n}}} $$

Substitute the values:

$$ t = \frac{171.8 - 175}{\frac{4.6}{\sqrt{49}}} = \frac{-3.2}{\frac{4.6}{7}} = \frac{-3.2}{0.65714} \approx -4.87 $$

Step 3: Determine the degrees of freedom

The degrees of freedom (\(df\)) for a one-sample \(t\)-test is:

$$ df = n - 1 = 49 - 1 = 48 $$

Step 4: Calculate the p-value

The alternative hypothesis is two-tailed because we want to determine if the average height is "different from" the national average:

• \(H_0: \mu = 175\)
• \(H_1: \mu

eq 175\)

Using a \(t\)-distribution table or calculator with \(df = 48\) and \(t = -4.87\):

$$ p\text{-value} = 2 \times P(T_{48} \le -4.87) \approx 0.000013 $$

Rounded to four decimal places, the \(p\)-value is \(0.0000\) (or extremely close to \(0\)).

Step 5: State the decision

Compare the \(p\)-value to the significance level \(\alpha = 0.05\):

• Since \(p\text{-value} \approx 0.0000 < 0.05\), we reject the null hypothesis.
• Therefore, the decision is R.

Answer:

(a) -4.87
(b) 48
(c) 0.0000 (or < 0.0001)
(d) R