Test of a Mean (Single Sample, Population Variance Known)

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

The first and simplest case of hypothesis testing we will consider is the test of a mean (H₀: μ = μ₀). In this case, we will assume that the population variance is known; thus, we are able to use the Z statistic. We perform the following steps for a two-tailed test (in the next section we will look at both one-tailed and two-tailed tests):

1. State the null hypothesis H₀: μ = μ₀ versus the alternative hypothesis H₁ μ: ≠ μ ₀.

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

3. Determine the critical region, that is, the region of values of Z in the upper and lower α/2 tails of the sampling distribution for Z when μ = μ₀ (i.e., the sampling distribution when the null hypothesis is true).

4. Compute the Z statistic: Z = ( – ₀)/( σ/√n) for the given sample and sample size n.

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

As an example, consider the study that used blood loss data from pigs (refer to Table 8.1). Take μ₀ = 2200 ml (a plausible amount of blood to lose for a pig in the control group). In this case, the sensible alternative would be one-sided; we would assume μ < 2200 for the alternative with the treatment group, because we expect the treatment to reduce and not to increase blood loss.

However, if we are totally naïve about the effectiveness of the drug, we might consider the two-sided alternative, namely, H₁: μ₀ ≠ 2200. In this section we are il-lustrating the two-sided test, so we will look at the two-sided alternative. We will use the sample data given in Section 8.9 and assume that the standard deviation σ is known to be 720. The sample mean is 1085.9 and the sample size n = 10. We now have enough information to carry out the test. The five steps are as follows:

1. State the null hypothesis: The null hypothesis is H₀: μ = μ₀ = 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 Z in the upper and lower 0.025 tails of the sampling distribution for Z when μ = μ₀ (i.e., when the null hypothesis is true). For α₀ = 0.05, the critical values are Z = ±1.96 and the critical region includes all values of Z > 1.96 or Z < –1.96.

4. Compute the Z statistic: Z = ( – μ₀)/( σ/√n) for the given sample and sample size n = 10. We have the following data: n = 10; the sample mean () is 1085.9; σ = 720; and μ₀ = 2200. Z = (1085.9 – 2200)/(720/√10) = –1114.1/227.684 = –4.893.

5. Since 4.893 (the absolute value of the test statistic) is clearly larger than 1.96, we reject H₀ at the 5% level; i.e., –4.893 < –1.960. Therefore, we conclude that the treatment was effective in reducing blood loss, as the calculated Z is negative, implying that μ < μ₀.