# Test of a Mean (Single sample, Population Variance Unknown)

| Home | | Advanced Mathematics |

## Chapter: Biostatistics for the Health Sciences: Tests of Hypotheses

We perform the following steps for a two-tailed test:

TEST OF A MEAN (SINGLE SAMPLE, POPULATION VARIANCE UNKNOWN)

In the case of a test of a mean (H0: μ = μ0) when the population variance is un-known, we estimate the population variance by using s2 and apply the t distribution to define rejection regions. We perform the following steps for a two-tailed test:

1. State the null hypothesis H0: μ = μ0 versus the alternative hypothesis H1: μ ≠ μ0. Note: This hypothesis set is exactly as stated in Section 9.3.

2. Choose a significance level α = α0 (often we take α0 = 0.05 or 0.01).

3. Determine the critical region for the appropriate t distribution, that is, the region of values of t in the upper and lower α/2 tails of the sampling distribution for Student’s t distribution with n – 1 degrees of freedom when μ= μ0 (i.e., the sampling distribution when the null hypothesis is true).

4. Compute the t statistic: t = ( – μ0)/(s/√n) for the given sample and sample size n where  is the sample mean and s is the sample standard deviation.

5. Reject the null hypothesis if the test statistic t computed in step 4 falls in the rejection region for this test; otherwise, do not reject the null hypothesis.

For example, reconsider the pig treatment data; take μ0 = 2200 ml (a plausible amount of blood to lose for a pig in the control group). In this case, because the sen-sible alternative would be one-sided, we could assume μ < 2200 for the alternative with the treatment group, as we expect the treatment to reduce blood loss and not to increase it. However, again assume we are totally naïve about the effectiveness of the drug; so we consider the two-sided alternative hypothesis, namely, H1: μ 0 2200.

In this section, we are illustrating the two-sided test, so we will look at the two-sided alternative hypothesis. We will use the sample data given in Section 8.9 but this time use the standard deviation s = 717.12. The sample mean is 1085.9 and the sample size n = 10. We now have enough information to run the test.

The five steps for hypothesis testing yield the following:

1. State the null hypothesis. The null hypothesis is H0: μ  = μ 0 = 2200 versus the alternative hypothesis H1: μ  μ0 = 2200.

2. Choose a significance level α = α0 = 0.05.

3. Determine the critical region, that is, the region of values of t in the upper and lower 0.025 tails of the sampling distribution for t (Student’s t distribution with 9 degrees of freedom) when μ = μ0 (i.e., the sampling distribution when the null hypothesis is true). For α0 = 0.05, the critical values are t = ±2.2622; the critical region includes all values of t > 2.2622 or t < –2.2622.

4. Compute the t statistic: t = ( – μ0)/(s/√n) for the given sample and sample size n = 10; since n = 10, the sample mean () is 1085.9, s = 717.12, and  μ0 = 2200. Then t = (1085.9 – 2200)/(717.12/10) = –1114.1/226.773 = –4.913.

5. Given that 4.913 (the absolute value of the t statistic) is clearly larger than 2.262, we reject H0 at the 5% level.

Later, in Section 9.6, we will see that a more meaningful quantity than the 5% level would be a specific p-value, which gives us more information as to the degree of significance of the test. In Section 9.6, we will calculate the p-value for a hypothesis test.

Related Topics