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.
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