
Test for Population Mean (small sample size) 
Test for Population Mean (small
sample size).
If the sample size is small () and the sample distribution is normal or approximately normal, then the Student's t distribution and associated statistics can be used to determine a test for whether the sample mean = population mean. That is, when the sample size is less than 30 we can use the student's t statistics, t to compare the sample mean against the population mean using value of the sample standard deviation to estimate the sample standard deviation if it is not known. There are three questions one may ask when comparing two means: Question 1: : Is ? H_{a} (Twotailed test) Question 2: : Is ? H_{a} (Righttailed test) Question 3: : Is ? H_{a} (Lefttailed test) 1. Know the statistics used to test for small sample size: The Student's t statistics. The test statistics is related to the student's t distribution, t. : The student's t statistics is similar to the zscore statistics, z The student's t distribution assumes that the sample distribution is
normal or approximately so.
As there are numerous Student's t distribution, so there are numerous tables of reference each approaches the question of the hypothesis differently. In this text we will use the Uppertailed test with alpha representing the area under the curve. See tstatistics reference table Using this table here are the decision rules for comparing means for
small sample size:
2. Know how to use appropriate statistics to test if a sample mean is equal to the population mean (small sample size). 3 Types of tests in comparing sample mean to population mean: When comparing the sample mean, to the population mean ( is known and is unknown) there are 3 questions to considered: Question 1: : Is ? H_{a} (Twotailed test) Question 2: : Is ? H_{a} (Righttailed test or Uppertailed) Question 3: : Is ? H_{a} (Lefttailed test or Lowertailed) Question 1:
Is
? H_{a} (Twotailed test)
By Examples: Problem 1. The output of a chemical process for manufacturing Snail Hail is monitored by taking a sample size of n = 25 vials to determine the level of impurities. The null hypothesis is that the mean level is exactly 0.05 gram per liter. If the mean level is too small, the process will be stopped and the tanks purged. Otherwise, the process will continue. A twotailed test applies, and the mean from the sample is 0.064 with a standard deviation of 0.017. At a significant level of alpha = 0.01, should the process be stopped and the tanks purged? Given , n = 25 (sp small sample size, use the student's t distribution), s = 0.017 and since, and Step 1  Hypothesis: The claim that or 0.064 = 0.05, the null hypothesis. The alternate hypothesis is that H_{0} : = 0.05 H_{a} :or Step 2. Select level of significance: This is given as (1%) So for twotailed test: Step 3. Test statistics and observed value.
Step 4. Determine the critical region (favors H_{a}) For
and df = 25  1 = (n1)= 24, t_{0.005}
= 2.7970
Step 5. Make decision. The observed z = 4.1176, and since 4.1176 > 2.797 and in the critical region, we reject H_{0} in favor of H_{a}. So the new mean of the process should be stopped and the tanks purged. Question 2:
Is
? H_{a} (Righttailed test)
By Examples: Problem 2. We need to test the hypothesis that a sample of n = 25, with a mean of 170 and standard deviation of 30at a significance level of alpha = 0.05 (a uppertailed test) whether: _{.} Given , n = 25 ( small, n< 30 so use tstatistics), s = 30 and since, Step 1  Hypothesis: The claim that or 170 = 160, the null hypothesis. Since xbar is smaller than , the alternate hypothesis is 170 > 160. H_{0} : = 160 H_{a} : _{.} Step 2. Select level of significance: This is given as (5%) Step 3. Test statistics and observed value.
Step 4. Determine the critical region (favors H_{a}) For alpha = 0.05 at the upper end of the acceptable region (Pr[1a]=0.95), t = 1.7109 From reference
table (search for t with df = 24 = 25 1, and alpha = 0.05.
Step 5. Make decision. The observed t = 1.6667, and since 1.67 < 1.71 and is not in the critical region, we have no reason to reject H_{0} in favor of H_{a}. So the new mean of 170 is not significantly greater than 160. Question 3:
Is
? H_{a} (Lefttailed test)
By Examples: Problem 3. A sample of 8 observations averages 5.8 and the standard deviation is 2.034. Is this mean less than the existing mean of 7.0? Test this hypothesis using a 2.5% level of significance. Given , n =8 (small so use t statistics), s = 2.034 and is given. Step 1  Hypothesis: The claim that or 5.8 = 7.0, the null hypothesis. Since xbar is smaller than , the alternate hypothesis is 5.8 < 7.0. H_{0} : = 5.8=7.0 H_{a} : _{.} Step 2. Select level of significance: This is given as (2.5%) Step 3. Test statistics and observed value.
Step 4. Determine the critical region (favors H_{a}) For alpha = 0.025 at the lower end of the acceptable region t = 2.3646 From reference
table (search for t with alpha = 0.025 and df = 7 =81)
and since lower tail t = 
Step 5. Make decision. The observed t = 1.67, and since 1.67 is not in the critical region (red ), we do not reject H_{0} in favor of H_{a}. So there is not enough evidence to justify the alternate hypothesis
that the new mean of 5.8 is not significantly lower than 7.0.
