Question

Is there a relationship between confidence intervals and
two-tailed hypothesis tests? Let *c* be the level of
confidence used to construct a confidence interval from sample
data. Let *α* be the level of significance for a two-tailed
hypothesis test. The following statement applies to hypothesis
tests of the mean.

For a two-tailed hypothesis test with level of significance
*α* and null hypothesis *H*_{0}: *μ* =
*k*, we *reject* *H*_{0} whenever
*k* falls *outside* the *c* = 1 – *α*
confidence interval for *μ* based on the sample data. When
*k* falls within the *c* = 1 – *α* confidence
interval, we do not reject *H*_{0}.

(A corresponding relationship between confidence intervals and
two-tailed hypothesis tests also is valid for other parameters,
such as *p*, *μ*_{1} −
*μ*_{2}, or *p*_{1} −
*p*_{2}, which we will study in later sections.)
Whenever the value of *k* given in the null hypothesis falls
*outside* the *c* = 1 – *α* confidence
interval for the parameter, we *reject H*_{0}. For
example, consider a two-tailed hypothesis test with *α* =
0.10 and

*H*_{0}: *μ* = 21

*H*_{1}: *μ* ≠ 21

A random sample of size 17 has a sample mean *x* = 20
from a population with standard deviation *σ* = 8.

(a) What is the value of *c* = 1 − *α*?

Using the methods of Chapter 7, construct a 1 − *α*
confidence interval for *μ* from the sample data. (Round
your answers to two decimal places.)

lower limit | |

upper limit |

What is the value of *μ* given in the null hypothesis (i.e.,
what is *k*)?

*k* =

Is this value in the confidence interval?

YesNo

Do we reject or fail to reject *H*_{0} based on this
information?

Fail to reject, since *μ* = 21 is not contained in this
interval.Fail to reject, since *μ* = 21 is contained in this
interval. Reject, since *μ* = 21 is
not contained in this interval.Reject, since *μ* = 21 is
contained in this interval.

(b) Using methods of Chapter 8, find the *P*-value for the
hypothesis test. (Round your answer to four decimal places.)

Do we reject or fail to reject *H*_{0}?

Reject the null hypothesis, there is sufficient evidence that
*μ* differs from 21.Fail to reject the null hypothesis,
there is insufficient evidence that *μ* differs from
21. Fail to reject the null hypothesis,
there is sufficient evidence that *μ* differs from 21.Reject
the null hypothesis, there is insufficient evidence that *μ*
differs from 21.

Compare your result to that of part (a).

These results are the same.We rejected the null hypothesis in part (b) but failed to reject the null hypothesis in part (a). We rejected the null hypothesis in part (a) but failed to reject the null hypothesis in part (b).

Answer #1

Is there a relationship between confidence intervals and
two-tailed hypothesis tests? Let c be the level of
confidence used to construct a confidence interval from sample
data. Let α be the level of significance for a two-tailed
hypothesis test. The following statement applies to hypothesis
tests of the mean.
For a two-tailed hypothesis test with level of significance
α and null hypothesis H0: μ =
k, we reject H0 whenever
k falls outside the c = 1 − α
confidence...

Is there a relationship between confidence intervals and
two-tailed hypothesis tests? Let c be the level of
confidence used to construct a confidence interval from sample
data. Let α be the level of significance for a two-tailed
hypothesis test. The following statement applies to hypothesis
tests of the mean.
For a two-tailed hypothesis test with level of significance
α and null hypothesis H0: μ =
k, we reject H0 whenever
k falls outside the c = 1
– α confidence interval...

The one sample t-test from a sample of n = 19 observations for
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H0: μ = 6
H1: μ ≠ 6
Has a t test statistic value = 1.93. You may assume that the
original population from which the sample was taken is symmetric
and fairly Normal.
Computer output for a t test:
One-Sample T: Test of mu = 6 vs not = 6
N Mean
StDev SE Mean 95%
CI
T P
19 6.200 ...

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(Reference: Merck Manual, a commonly used reference in
medical schools and nursing programs). A new drug for arthritis has
been developed. However, it is thought that this drug may change
blood pH. A random sample of 33 patients with arthritis took the
drug for 3 months. Blood tests showed that...

Confidence Intervals & Hypothesis Tests
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157 receipts, he used StatKey to construct a 95% bootstrap
confidence interval for r. The results were: [0.018, 0.292].
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Let x be a random variable representing dividend yield
of bank stocks. We may assume that x has a normal
distribution with σ = 3.3%. A random sample of 10 bank
stocks gave the following yields (in percents).
5.7
4.8
6.0
4.9
4.0
3.4
6.5
7.1
5.3
6.1
The sample mean is x = 5.38%. Suppose that for the
entire stock market, the mean dividend yield is μ = 4.8%.
Do these data indicate that the dividend yield of all...

Let x be a random variable representing dividend yield
of bank stocks. We may assume that x has a normal
distribution with σ = 2.3%. A random sample of 10 bank
stocks gave the following yields (in percents).
5.7
4.8
6.0
4.9
4.0
3.4
6.5
7.1
5.3
6.1
The sample mean is x = 5.38%. Suppose that for the
entire stock market, the mean dividend yield is μ = 4.9%.
Do these data indicate that the dividend yield of all...

Let x be a random variable representing dividend yield
of bank stocks. We may assume that x has a normal
distribution with σ = 2.0%. A random sample of 10 bank
stocks gave the following yields (in percents).
5.7
4.8
6.0
4.9
4.0
3.4
6.5
7.1
5.3
6.1
The sample mean is x = 5.38%. Suppose that for the
entire stock market, the mean dividend yield is μ = 4.6%.
Do these data indicate that the dividend yield of all...

Let x be a random variable representing dividend yield of bank
stocks. We may assume that x has a normal distribution with σ =
2.3%. A random sample of 10 bank stocks gave the following yields
(in percents).
5.7 4.8 6.0 4.9 4.0 3.4 6.5 7.1 5.3 6.1
The sample mean is x = 5.38%.
Suppose that for the entire stock market, the mean dividend
yield is μ = 4.4%.
Do these data indicate that the dividend yield of all...

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