Hypothesis Tests Concerning the Mean

Hypothesis Testing for Mean: Z and T Distributions | CFA Level I Quantitative Methods

When it comes to hypothesis testing concerning a single mean, the critical value can be based either on the z distribution or the t distribution. In this lesson, we’ll refresh your memory on how to choose between these two test statistics and walk you through an example.

Criteria for Selecting the Test Statistic

Keep these three questions in mind when choosing between the z and t statistics:

  1. Is the population normal or can it be assumed normal?
  2. Is the population variance known?
  3. Is the sample size at least 30?

If the population is normal and the variance is known, use the z statistic. If the variance is unknown, it’s more conservative to use the t statistic.

If the population isn’t normal or you’re unsure, check if the sample size is at least 30. If it’s less than 30, the problem can’t be solved as the central limit theorem doesn’t apply. If it’s at least 30, you can use the z or t statistic based on whether the population variance is known.


A manufacturer claims their gadget has a mean battery life of 25 hours, with a standard deviation of 1 hour. They guarantee this at a 5% significance level. An acceptance test is performed on a batch of 10,000 gadgets by randomly selecting 30 sets. The mean battery life of the sample is 24.5 hours. Should the batch be accepted?

Let’s go through the steps to perform the hypothesis test for this example.

  1. State the hypothesis:
    • Define μ as the mean of the batch (population).
    • Define x̄ as the mean of the sample (30 randomly selected gadgets).
    • Null hypothesis (H₀): μ ≥ 25 (battery life is at least 25 hours).
    • Alternative hypothesis (H₁): μ < 25 (battery life is less than 25 hours).
  2. Select the test statistic: We can’t assume the population distribution is normal, but the sample size is at least 30 and the population variance is known. Therefore, we’ll use the z-distribution as the test statistic.
  3. Specify the significance level: Given as 5%.
  4. State the decision rule: This is a one-tailed test. At a 5% significance level, we want a 0.95 probability to the right of the critical value. From the z table, the closest z value is 1.65. For the left tail, the critical value is -1.65. The decision rule is: reject H₀ if the test statistic is less than -1.65.
  5. Calculate the test statistic: The value of x̄ is given as 24.5. Plugging in the numbers, we get a test statistic of -2.74.
  6. Make a statistical decision: Since -2.74 is less than the critical value of -1.65, it falls in the rejection region. There’s sufficient evidence to reject H₀.
  7. Make the economic decision: Based on the sample, the batch doesn’t meet the manufacturer’s specification of a 25-hour battery life. The decision should be to reject this batch of gadgets.

Key Takeaways

Understanding hypothesis testing for a single mean and the appropriate selection of test statistics is crucial for your CFA exam preparation. Here are some key takeaways:

Testing Differences Between Two Means

So far, what we have been looking at are tests of a single population mean, based on a random sample from the population. Let us now explore hypothesis testing for differences between means of two different populations. Keep these factors in mind when deciding what kind of test to apply:

  • Both population distributions must be normal.
  • Determine if the two populations are independent from each other.
  • Identify whether the population variances can be assumed equal.

Independent Populations

For independent populations, you’ll need to use different tests depending on whether their variances can be assumed equal:

  • Equal variances: Use a t-test with pooled variance estimation. The degrees of freedom for this t-distribution is n1 + n2 – 2.
  • Unequal variances: Use the t-distribution with modified degrees of freedom and non-pooled variance.

Dependent Populations

If the two populations are not independent, use the paired comparisons test. This method treats the differences between paired observations as a single random variable, allowing hypothesis testing as in the case of a single mean.


Kent Boyle suspects that stocks perform differently after a stock split than after a stock consolidation. He picks 31 stocks from each category and calculates their mean returns and standard deviations. Which approach to test the difference between the two means is most appropriate for this problem?

Answering the two key questions:

  • Are the returns of the two populations independent? Since the stocks are likely different and the events happened over different time periods, there is likely no correlation between the two populations. We can assume that the returns are independent.
  • Can we assume that the variances are the same? Since the two populations are outcomes from two unrelated events, there is no reason to assume that their variances are the same.

Since the two populations are independent and their variances are not the same, we use the t-distribution with modified degrees of freedom and non-pooled variance.


Continuing with the problem, Kent calculated the test statistic to be 1.672 and the degrees of freedom as 80. Perform a hypothesis test at 5% significance level to determine if Kent’s suspicion is substantiated.

Firstly, let’s state the hypotheses:

  • H₀: μ₁ – μ₂ = 0
  • H₁: μ₁ – μ₂ ≠ 0

Next, let’s state the decision rule. Looking at the t-table where degrees of freedom equals 80 and the two-tail cumulative probability equals 0.05, we get the critical values, ±1.99.

The decision rule can be stated as: reject H₀ if t is less than -1.99 or if t is greater than 1.99.

Given the test statistic of 1.672, it does not fall within the rejection regions. We therefore fail to reject H₀, and Kent Boyle’s suspicion cannot be substantiated at the 5% level of significance.

Paired Comparisons Test


Hanna wants to investigate an observation that stocks tend to be more volatile in the first hour of trading than the last hour. To make a fair comparison, Hanna chooses the same random sample of 30 stocks for the analysis of both the first and last hour volatilities. Which approach to testing the difference in mean volatilities between the first and the last hour is most appropriate?

As the same stocks are chosen for both samples, the volatilities of the two samples cannot be independent. Therefore, the most appropriate approach is the paired comparisons test.

Summary: Hypothesis Testing Concerning the Mean

In the case of a single mean:

  • If the population distribution is normal:
  • If the population distribution is unknown or not normal and the sample size is at least 30, approximate the sampling distribution as normal and proceed to use either the z or t distribution based on whether the variance is known.
  • If the sample size is less than 30, the problem cannot be solved.

In the case of testing the difference between 2 means:

  • If both population distributions are normal:
    • If the samples are independent:
      • If variances can be assumed equal, use the t-statistic with pooled variance approach. The degrees of freedom is the sum of the two sample sizes, minus 2.
      • If variances cannot be assumed equal, use the t-statistic with the degrees of freedom estimated from the population variances.
    • If the samples are dependent, use the paired comparisons test (essentially a standard t-test with n-1 degrees of freedom) by pairing the samples and calculating their differences.
  • If both population distributions are not normal, the solution is not within the scope of the CFA curriculum.

Practice on many problems to master hypothesis testing. Stay tuned for the next lesson where we’ll explore tests concerning the variance.

✨ Visual Learning Unleashed! ✨ [Premium]

Elevate your learning with our captivating animation video—exclusive to Premium members! Watch this lesson in much more detail with vivid visuals that enhance understanding and make lessons truly come alive. 🎬

Unlock the power of visual learning—upgrade to Premium and click the link NOW! 🌟