程序代写代做代考 9 Hypothesis testing (Part 2)

9 Hypothesis testing (Part 2)

Single sample procedures

9.1 Introduction

In this chapter we will discuss specific applications of hypothesis testing where

we have a single sample of data and wish to test hypotheses regarding the value

of a population mean parameter.

We focus our main discussion on the scenario in which the random sample is

from a N(µ, σ2) distribution with µ unknown and σ2 known. The ideas are then

extended to develop hypothesis tests for (i) the mean of a normal distribution

with unknown variance, (ii) the mean of a non-normal distribution, and (iii) a

population proportion p. In cases (ii) and (iii) it is not possible to calculate the

exact distribution of the test statistic under the null hypothesis, however we can

appeal to the central limit theorem to find an approximate normal distribution.

9.2 Inference about the mean of a normal distribution when the

variance is known

Let X1, . . . , Xn be a random sample from N(µ, σ
2), where the value of µ is

unknown but the value of σ2 is known. We would like to use the data to make

inferences about the value of µ and, in particular, we wish to test the following

hypotheses:

H0 : µ = µ0 vs H1 : µ > µ0 .

The null hypothesis H0 posits that the data are sampled from N(µ0, σ
2). In

contrast, the alternative hypothesis H1 posits that the data arise from N(µ1, σ
2),

where µ1 > µ0 is an unspecified value of µ. This is a one-sided test.

We know that the sample mean, X̄, is an unbiased estimator of µ. Hence, if

the true value of µ is µ0, then E[X̄ − µ0] = µ0 − µ0 = 0. In contrast, if H1 is
true, we would have that E[X̄ − µ0] = µ− µ0 > 0. This suggests that we should
reject H0 in favour of H1 if X̄ is ‘significantly’ larger than µ0, i.e. if X̄ > k, for

some k > µ0. The question is, how much greater than µ0 should x̄ be before we

reject H0? In other words, what value should we choose for k?

One way to decide this is to fix the probability of rejecting H0 if H0 is true,

i.e. the probability of making a Type I error; the critical value k can then be

determined on this basis. This is equivalent to fixing the significance level of the

1

test. Suppose that we do indeed use X̄ as the test statistic, with rejection region

C = {x̄ > k} ,

and suppose we wish to find k > µ0 to ensure that

P(type I error) = P(reject H0 | H0 true) = α .

Hence we have that

α = P(reject H0 | H0 true) = P(X̄ > k | H0 true)

= P

(
X̄ − µ0
σ/
√
n
>
k − µ0
σ/
√
n

)
= P

(
Z >

k − µ0
σ/
√
n

)
,

where Z =
X̄−µ0
σ/
√
n
∼ N(0, 1) under H0. Let zα denote the upper α point of

N(0, 1), i.e. P(Z > zα) = α. From this we see that zα =
k−µ0
σ/
√
n

and so

k = µ0 +
zα σ√
n
.

Thus, H0 is rejected in favour of H1 if the sample mean is greater than µ0 by zα

standard errors.

Equivalently, we reject H0 in favour of H1 at the 100α% significance level if

Z =
X̄ − µ0
σ/
√
n
> zα .

The standardized version of X̄ given by Z is the most frequently used form of

the test statistic in this scenario. The critical value zα can be obtained from

standard normal tables. In hypothesis testing it is common to use α = 0.05, and

in this case z0.05 = 1.645.

Suppose now that we wish to use our sample to test the hypotheses

H0 : µ = µ0 vs H1 : µ < µ0 . This is again a one-sided test. In this case we will reject H0 in favour of H1 if X̄ < k where k < µ0. Using analogous arguments to those used above, we will 2 reject H0 in favour of H1 at the 100α% significance level if X̄ < µ0 − zα σ√ n , or, equivalently, if Z = X̄ − µ0 σ/ √ n < −zα . For a test having a 5% significance level the critical value is −z0.05 = −1.645. If in fact our interest is in testing H0 : µ = µ0 vs H1 : µ 6= µ0 , then we now have a two-sided test. We will reject H0 in favour of H1 if X̄ is either significantly greater or significantly less than µ0, i.e. if X̄ < k1 or X̄ > k2 ,

The critical values k1 < µ0 and k2 > µ0 are chosen so that the significance level

is equal to α, i.e.

α = P(X̄ < k1 or X̄ > k2 | H0 true)

= P(X̄ < k1 |H0) + P(X̄ > k2 |H0) .

It seems natural to choose the values of k1 and k2 so that the probability of

rejecting H0 is split equally between the upper and lower parts of the rejection

region. In other words, we choose k1 and k2 such that

P(X̄ < k1 |H0) = P(X̄ > k2 |H0) = α/2 .

For illustration, see the figure overleaf which shows the p.d.f. of X̄, together with

the rejection region.

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µ0k1 k2

Reject H0Reject H0 Do not reject H0

α 2α 2 1 − α

Figure: illustration of a two-tailed test.

We now find appropriate values of k1 and k2 satisfying this property. We begin

with k2. Note that

α/2 = P(X̄ > k2 | H0 true ) = P
(
X̄ − µ0
σ/
√
n
>
k2 − µ0
σ/
√
n

)
= P

(
Z >

k2 − µ0
σ/
√
n

)
, with Z ∼ N(0, 1) .

However, we know that zα/2 satisfies P(Z > zα/2) = α/2. Hence,

k2 − µ0
σ/
√
n

= zα/2 ,

and so we have that

k2 = µ0 +
zα/2 σ√

n
.

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For k1, observe that

α/2 = P(X̄ < k1 | H0 true ) = P ( X̄ − µ0 σ/ √ n < k1 − µ0 σ/ √ n ) = P ( Z < k1 − µ0 σ/ √ n ) , with Z ∼ N(0, 1) . We know that P(Z < −zα/2) = α/2 and so k1−µ0 σ/ √ n = −zα/2. Hence k1 = µ0 − zα/2 σ√ n . To summarize the two-tailed test here, we reject H0 at significance level α if X̄ > µ0 +
zα/2 σ√

n
or if

X̄ < µ0 − zα/2 σ√ n . Equivalently, we reject H0 at significance level α if Z = X̄ − µ0 σ/ √ n > zα/2 or if

Z =
X̄ − µ0
σ/
√
n
< −zα/2 . 9.2.1 Connection between the two-tailed test and a confidence inter- val for the mean when the variance is known LetX1, . . . Xn be a random sample fromN(µ, σ 2) with µ unknown and σ2 known. Recall from Chapter 7 that a 100(1− α)% confidence interval for µ is given by[ X̄ − zα/2 σ√ n , X̄ + zα/2 σ√ n ] . From the preceding discussion, if we are testing the hypotheses H0 : µ = µ0 H1 : µ 6= µ0 , 5 then we will ‘accept’ H0 at the 100α% significance level if µ0 − zα/2 σ√ n ≤ X̄ ≤ µ0 + zα/2 σ√ n , or, equivalently, if X̄ − zα/2σ√ n ≤ µ0 ≤ X̄ + zα/2σ√ n . Thus, the values of µ in the confidence interval correspond to values of µ0 for which the corresponding null hypothesis H0 would not be rejected. In other words, informally, the 100(1 − α)% confidence interval is a set of values of µ which would ‘pass a hypothesis test at significance level α’. It is in this sense that we can regard the confidence interval as a set of plausible values of µ given the data. Example 9.1. (i) A random sample of n = 25 observations is taken from a normal distribution with unknown mean but known variance σ2 = 16. The sample mean is found to be x̄ = 18.2. Test H0 : µ = 20 vs H1 : µ < 20 at the 5% significance level. (ii) Find the probability that we reject H0 using this testing procedure when the true value of the mean µ is 19.0. 6 Example 9.2. Suppose now that we have a random sample of n = 50 obser- vations from a normal distribution with unknown mean and known variance σ2 = 36. It is found that x̄ = 30.8. (i) Test H0 : µ = 30 vs H1 : µ 6= 30 at the 5% significance level. 7 (ii) Find the probability that we reject H0 when the true value of the mean µ is 31.0. 8 9.3 Inference about the mean of a normal distribution when the variance is unknown Let X1, . . . , Xn be a random sample from the N(µ, σ 2) distribution, where the value of µ is unknown but that of σ2 is also unknown. We want to test the following hypotheses: H0 : µ = µ0 H1 : µ > µ0

at significance level α. Based on the discussion in the previous section, an

appropriate test statistic which measures the discrepancy between µ0 and the

sample estimator X̄ is given by

T =
X̄ − µ0
S/
√
n

where S is the sample standard deviation. This is an estimate of the standardized

difference between X̄ and µ0. As we have discussed previously, because the

statistic T involves the random quantities X̄ and S, its sampling distribution

is no longer N(0, 1). We have seen in Chapter 7 that T ∼ t(n − 1), under the
assumption that H0 is true, i.e. T has a Student t-distribution with n−1 degrees
of freedom.

Assuming that the significance level of the test is α, we use one of the fol-

lowing rejection regions, depending on the alternative hypothesis:

• For the one-sided alternative hypothesis H1 : µ > µ0,

reject H0 if T > tα ,

where tα is the upper α point of a t(n−1) distribution, i.e. P(T > tα) = α.

• For the one-sided alternative hypothesis H1 : µ < µ0, reject H0 if T < −tα . • For the two-sided alternative hypothesis H1 : µ 6= µ0, reject H0 if T < −tα/2 or T > tα/2 .

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Example 9.3. The drug 6-mP is used to treat leukaemia. A random sample of

21 patients using 6-mP were found to have an average remission time of x̄ = 17.1

weeks with a sample standard deviation of s = 10.00 weeks. A previously used

drug treatment had a known mean remission time of µ0 = 12.5 weeks. Assuming

that the remission times of patients taking 6-mP are normally distributed with

both the mean µ and variance σ2 being unknown, test at the 5% significance

level whether the mean remission time of patients taking 6-mP is greater than

µ0 = 12.5 weeks.

9.4 Using the central limit theorem

(i) Inference about the mean of a non-normal distribution.

Let X1, . . . , Xn be a random sample from a non-normal distribution, where

the value of the mean µ is unknown and that of the variance σ2 is also

unknown. We want to test the following hypotheses:

H0 : µ = µ0

H1 : µ > µ0

at significance level α. We can again use the test statistic

Y =
X̄ − µ0
S/
√
n

defined above which, by asymptotic (large n) results, has an approximate

N(0, 1) distribution when H0 is true (n ≥ 30). Aside from the choice of test
statistic, the rejection regions for the various versions of H1 are otherwise

10

identical to those defined in the case of normal data with a known variance.

(ii) Inference about the population proportion p.

Let X1, . . . , Xn be a random sample of Bi(1, p) random variables, where

the value of p is unknown. We want to test the following hypotheses:

H0 : p = p0

H1 : p > p0

at significance level α. As we have seen earlier in this module, an unbiased

sample estimator of the parameter p is given by

p̂ =
1

n

n∑
i=1

Xi = X̄n .

By the central limit theorem, p̂ ∼ N(p, p(1−p)/n) approximately for large
n. As a rule of thumb, n ≥ 9 max{p/(1 − p), (1 − p)/p} guarantees this
approximation has a good degree of accuracy. A suitable test statistic is

Y =
p̂− p0√

p0(1− p0)/n

Here we have estimated the standard error of p̂ by
√
p0(1− p0)/n which

uses the value of p specified under H0. If H0 is true then Y has an approx-

imate N(0, 1) distribution for large n. Thus, to achieve an approximate

significance level of α, we reject H0 in favour of the above H1 if Y > zα.

• For the one-sided alternative hypothesis H1 : p < p0, to achieve an approximate significance level of α, we reject H0 if Y < −zα. • For the two-sided alternative hypothesis H1 : p 6= p0, to achieve an approximate significance level of α, we reject H0 if Y < −zα/2 or Y > zα/2 .

Example 9.4. A team of eye surgeons has developed a new technique for

an eye operation to restore the sight of patients blinded by a particular

disease. It is known that 30% of patients who undergo an operation using

the old method recover their eyesight.

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A total of 225 operations are performed by surgeons in various hospitals

using the new method and it is found that 88 of them are successful in

that the patients recover their sight. Can we justify the claim that the

new method is better than the old one? (Use a 1% level of significance).

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Hypothesis testing (Part 2) Single sample procedures
Introduction
Inference about the mean of a normal distribution when the variance is known
Connection between the two-tailed test and a confidence interval for the mean when the variance is known

Inference about the mean of a normal distribution when the variance is unknown
Using the central limit theorem