Question

You wish to test the following claim (
H
a
) at a significance level of
α
=
0.002
.


H
o
:
μ
=
63.5


H
a
:
μ
<
63.5


You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size
n
=
62
with mean
M
=
56.3
and a standard deviation of
S
D
=
13.2
.

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

• The test statistic is t = -4.29.

• The p-value is of 0.

What are the hypotheses tested?

At the null hypotheses, it is tested if the mean is of 63.5, that is:

$H_0: \mu = 63.5$

At the alternative hypotheses, it is tested if the mean is less than 63.5, that is:

$H_1: \mu < 63.5$

What is the test statistic?

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

• $\overline{x}$ is the sample mean.

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

• s is the standard deviation of the sample.

• n is the sample size.

The values of the parameters are given as follows:

$\overline{x} = 56.3, \mu = 63.5, s = 13.2, n = 62$

Hence, the test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{56.3 - 63.5}{\frac{13.2}{\sqrt{62}}}$

t = -4.29.

What is the p-value?

We have a left-tailed test, as we are testing if the mean is less than a value, with 62 - 1 = 61 df and t = -4.29. Hence, using a t-distribution calculator, the p-value is of 0.

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

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