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 H0: μ = k, we reject H0 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 H0. (A corresponding relationship between confidence intervals and two-tailed hypothesis tests also is valid for other parameters, such as p, μ1 − μ2, or p1 − p2, which we will study later.) Whenever the value of k given in the null hypothesis falls outside the c = 1 − α confidence interval for the parameter, we reject H0. For example, consider a two-tailed hypothesis test with α = 0.01 and H0: μ = 20 H1: μ ≠ 20 A random sample of size 30 has a sample mean x = 23 from a population with standard deviation σ = 6. (a) What is the value of c = 1 − α? Construct a 1 − α confidence interval for μ from the sample data. (Round your answers to two decimal places.) lower limit Using methods of this chapter, find the P-value for the hypothesis test. (Round your answer to four decimal places.)

Answer 1

SOLUTION:

A)

We are given that =23 and  σ = 6, n=30

Level of significance α = 0.01.

H₀: μ = 20

H₁: μ ≠ 20

c = 1 − α=1-0.01=0.99

=2.58(6/)

=2.82

The end points are given by

-E=23-2.82=20.18

+E=25.82

We are 99% certain that the interval from 20.18 to 25.82 is an interval that contains the population mean time .

The value of given in the null hypothesis is 20.This value is not in the confidence interval we reject H_0,, since the value fails outside of the specified range of the confidence interval .

B)

Level of significance α = 0.01

H₀: μ = 20

H₁: μ ≠ 20

Since,H₁: μ ≠ 20 this is a two tailed test.Since the x distribution is normal and σ is known,use the standard normal distribution with

The sample of 30 measurements has mean =23,Converting this measurement to Z we have

=3/10.954=2.738

P-value=2P(Z<2.738)=2(0.003)=0.006

Since P-value 0.006<0.01,we see that P-value<α.

We reject H₀.

The result are same in both the cases .In both the cases we reject the null hypothesis.