Question

You wish to test the following claim (
H
a
) at a significance level of
α
=
0.01
.


H
o
:
μ
=
61.5


H
a
:
μ
<
61.5


You believe the population is normally distributed and you know the standard deviation is
σ
=
17.6
. You obtain a sample mean of
M
=
59.1
for a sample of size
n
=
34
.

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =


What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =

Answer 1

Using the z-distribution, it is found that the desired measures are given as follows:

• The test statistic is z = -0.795.

• The p-value is of 0.2133.

What are the hypotheses tested?

At the null hypotheses, it is tested if the mean is of 61.5, that is:

$H_0: \mu = 61.5$

At the alternative hypotheses, it is tested if the mean is less than 61.5, that is:

$H_1: \mu < 61.5$

What is the test statistic?

The test statistic is given by:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

The parameters are:

• $\overline{x}$ is the sample mean.

• $\mu$ is the value tested at the null hypothesis.

• $\sigma$ is the standard deviation of the population.

• n is the sample size.

The values of the parameters are given as follows:

$\overline{x} = 59.1, \mu = 61.5, \sigma = 17.6, n = 34$

Hence, the test statistic is given by:

$z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{59.1 - 61.5}{\frac{17.6}{\sqrt{34}}}$

z = -0.795.

What is the p-value?

We have a left-tailed test, as we are testing if the mean is less than a value, with z = -0.795. Hence, using a z-distribution calculator, the p-value is of 0.2133.

More can be learned about the z-distribution at https://brainly.com/question/16162795

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