The physician-recommended dosage of a new medication is 14 mg. Actual administered doses vary slightly from dose to dose and are normally distributed with mean μ. A representative of a medical review board wishes to see if there is any evidence that the mean dosage is more than recommended and intends to test the hypotheses:

𝐻
𝑂
:
𝜇
=
14
𝑎
𝑛
𝑑
𝐻
𝐴
:
𝜇

14

To do this, he selects 17 doses at random and determines the weight of each. He finds the sample mean to be 13.75 mg, the sample standard deviation to be 0.24 mg. How is the condition of normality for the t-test on the mean met? Output for this data and t-test are below.

One Sample t-test
data: medicine$dosage
t = -0.18925, df = 16, p-value = 0.5739
alternative hypothesis: true mean is greater than 14
95 percent confidence interval:
13.74578 Inf
sample estimates:
mean of x
13.97514

Question

The physician-recommended dosage of a new medication is 14 mg. Actual administered doses vary slightly from dose to dose and are normally distributed with mean μ.  A representative of a medical review board wishes to see if there is any evidence that the mean dosage is more than recommended and intends to test the hypotheses:

𝐻
𝑂
:
𝜇
=
14
𝑎
𝑛
𝑑
𝐻
𝐴
:
𝜇
>
14

To do this, he selects 17 doses at random and determines the weight of each.  He finds the sample mean to be 13.75 mg, the sample standard deviation to be 0.24 mg. How is the condition of normality for the t-test on the mean met? Output for this data and t-test are below.

One Sample t-test
data:  medicine$dosage
t = -0.18925, df = 16, p-value = 0.5739
alternative hypothesis: true mean is greater than 14
95 percent confidence interval:
 13.74578      Inf
sample estimates:
mean of x 
 13.97514

BlackTom AI · Accepted Answer

The problem statement explicitly says the administered doses "are normally distributed with mean μ," so the population normality assumption required for a one-sample t-test is met directly.

1) Relevant knowledge points and concepts
- The one-sample t-test requires that the sampling distribution of the mean be approximately normal. This is satisfied in two ways: (a) if the underlying population is normal (any n), or (b) by the central limit theorem (CLT) when n is large (commonly n ≥ 30).  
- When the population is stated to be normally distributed, the small-sample t-test is valid even for modest sample sizes (here n = 17).

2) Step-by-step reasoning
- The stem: "Actual administered doses ... are normally distributed with mean μ." This is an explicit statement about the population distribution.  
- Sample size: n = 17. Because n < 30, we cannot rely on the CLT alone to justify approximate normality of the sample mean, so we need the population normality assumption.  
- Since the population normality is given, the t-test’s normality condition is satisfied.

3) Calculations / intermediate checks
- Sample size check: n = 17 (< 30), so CLT would not be the primary justification. No further numerical calculation is required to justify normality because it is stated in the problem.

4) Why the correct answer is correct
- The correct option, "We are told the data are normal," directly references the explicit population normality assumption in the question. That assumption is sufficient to meet the t-test requirement.

5) Why the other options are incorrect
- "The central limit theorem applies.": Incorrect here because n = 17 is small; the CLT justification is typically used for larger samples.  
- "Normality cannot be met.": Incorrect because the problem explicitly states the doses are normally distributed.  
- "The QQ plot indicates normality.": The QQ plot may look approximately linear but one or two deviations exist; more importantly, normality is already given in the statement, so the QQ plot is not the basis for meeting the condition.

6) Conclusion
- Therefore the t-test normality condition is met because the problem explicitly tells us the data are normally distributed. This confirms the provided answer: "We are told the data are normal."

Similar Questions

We conduct a test for a mean with a two-sided alternative hypothesis using a t-test. There were 30 observations in the sample and the calculated test statistic was 2.36. What is the p-value?

我们用t检验对平均值进行双侧备选假设检验。样本中有30项观测值，计算出的检验统计量为2.36。p值是什么？

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler