### Chapter 7. Hypothesis Testing: Introduction to Hypothesis Testing (p-value Approach)

### Computing the p-value and Making a Decision

p-value

The #\boldsymbol{p}#**-value** of a test is the *probability*, when the null hypothesis #H_0# is assumed to be true, of observing a test statistic equal to or more extreme than what was actually observed.

For a *two-tailed *test, the #p#-value is a two-tailed probability, which means it is evenly split between both tails.

#H_a: \mu \neq \mu_0#

For a *one-tailed *test, the #p#-value is a one-tailed probability, which means it is entirely located in one of the tails.

#H_a: \mu \lt \mu_0#

#H_a: \mu \gt \mu_0#

Interpretation of a p-value

The #p#-value is not the probability that the null hypothesis #H_0# is false. The status of #H_0# is non-random.

The sample is random, so consequently, the *test statistic *is also random. Thus the #p#-value concerns the test statistic.

A small #p#-value implies that the value of the test statistic is not in the range of values that we would expect if #H_0# is true.

Calculating the p-value of a Z-test for a Population Mean with Statistical Software

The calculation of the #p#-value of a #Z#-test for #\mu# is dependent on the *direction *of the test and can be performed using either Excel or R.

To calculate the #p#-value of a #Z#-test for #\mu# in **Excel**, make use of one of the following commands:

\[\begin{array}{lllll}

\phantom{0}\text{Direction}&\phantom{0000}H_0&\phantom{0000}H_a&\phantom{000}p\text{-value}&\phantom{0000000000}\text{Excel Command}\\

\hline

\text{Two-tailed}&H_0:\mu = \mu_0&H_a:\mu \neq \mu_0&2\cdot \mathbb{P}(Z\geq |z|)&=2 \text{ * }(1 \text{ - }\text{NORM.DIST}(\text{ABS}(z),0,1,1))\\

\text{Left-tailed}&H_0:\mu \geq \mu_0&H_a:\mu \lt \mu_0&\mathbb{P}(Z\leq z)&=\text{NORM.DIST}(z,0,1,1)\\

\text{Right-tailed}&H_0:\mu \leq \mu_0&H_a:\mu \gt \mu_0&\mathbb{P}(Z\geq z)&=1 \text{ - }\text{NORM.DIST}(z,0,1,1)\\

\end{array}\]

To calculate the #p#-value of a #Z#-test for #\mu# in **R**, make use of one of the following commands:

#\begin{array}{lllll}

\phantom{0}\text{Direction}&\phantom{0000}H_0&\phantom{0000}H_a&\phantom{000}p\text{-value}&\phantom{0000000}\text{R Command}\\

\hline

\text{Two-tailed}&H_0:\mu = \mu_0&H_a:\mu \neq \mu_0&2\cdot \mathbb{P}(Z\geq |z|)&2 \text{ * }\text{pnorm}(\text{abs}(z),0,1, \text{FALSE})\\

\text{Left-tailed}&H_0:\mu \geq \mu_0&H_a:\mu \lt \mu_0&\mathbb{P}(Z\leq z)&\text{pnorm}(z,0,1, \text{TRUE})\\

\text{Right-tailed}&H_0:\mu \leq \mu_0&H_a:\mu \gt \mu_0&\mathbb{P}(Z\geq z)&\text{pnorm}(z,0,1, \text{FALSE})\\

\end{array}#

Note: #|z|# denotes the *absolute value *of #z#, i.e. the distance of #z# from #0#, regardless of whether #z < 0# or #z > 0#.

#\phantom{0}#

Once the #p#-value of the test has been computed, the time has come to make a decision regarding the null hypothesis #H_0#.

#\phantom{0}#

Making a Decision

If the computed #p#-value is *smaller* than or *equal *to the significance level (i.e. #p \leq \alpha#), this means that the observed data is highly unlikely to occur, given that #H_0# is true. We then say that we reject #H_0# in favor of #H_a#.

If the computed #p#-value is *larger *than the significance level (i.e. #p \gt \alpha#), this means that the observed data falls in the range of values that we would expect to observe if #H_0# is true. We then say we do not reject #H_0#.

A researcher collects one-liter water specimens from #105# randomly-selected locations along a river and measures the amount of dissolved oxygen in each specimen.

The researcher plans on using a #Z#-test to determine if the mean oxygen content of the river significantly

*differs*from #6# mg per liter, at the #\alpha = 0.08# level of significance.

The sample mean #\bar{X}# turns out to be #5.98# mg per liter.

Calculate the #p#-value of the test and make a decision regarding #H_0#. Round your answer to #4# decimal places.

On the basis of this #p#-value, #H_0# should not be rejected, because #\,p# #\gt# #\alpha#.

There are a number of different ways we can calculate the #p#-value of the test. Click on one of the panels to toggle a specific solution.

Because the sample is drawn from a normally-distributed population, we know that the test statistic

\[Z=\cfrac{\bar{X} - \mu_0}{\sigma/\sqrt{n}}\]

has the #N(0,1)# distribution, under the assumption that #H_0# is true.

Calculate the value of *test statistic* #z#:

\[z = \cfrac{\bar{X}-\mu_0}{\sigma/\sqrt{n}} =\cfrac{5.98 - 6}{0.53/\sqrt{105}} = -0.38668\]

For a *two-tailed *#Z#-test, the #p#-value is defined as #2\cdot \mathbb{P}(Z \geq |z|)#. To calculate this value in Excel, make use of the following function:

NORM.DIST(x, mean, standard_dev, cumulative)

x: The value at which you wish to evaluate the distribution function.mean: The mean of the distribution.standard_dev: The standard deviation of the distribution.cumulative: A logical value that determines the form of the function.

- TRUE - uses the cumulative distribution function, #\mathbb{P}(X \leq x)#
- FALSE - uses the probability density function

Thus, to calculate #p = 2\cdot \mathbb{P}(Z \geq |z|)#, run the following command:

\[

=2 \text{ * }(1 \text{ - } \text{NORM.DIST}(\text{ABS}(z),0,1,1))\\

\downarrow\\

=2 \text{ * }(1 \text{ - } \text{NORM.DIST}(\text{ABS}(\text{-}0.38668),0,1,1))

\]

This gives:

\[p = 0.6990\]

Since #\,p# #\gt# #\alpha#, #H_0: \mu = 6# should not be rejected.

Because the sample is drawn from a normally-distributed population, we know that the test statistic

\[Z=\cfrac{\bar{X} - \mu_0}{\sigma/\sqrt{n}}\]

has the #N(0,1)# distribution, under the assumption that #H_0# is true.

Calculate the value of *test statistic* #z#:

\[z = \cfrac{\bar{X}-\mu_0}{\sigma/\sqrt{n}} =\cfrac{5.98 - 6}{0.53/\sqrt{105}} = -0.38668\]

For a *two-tailed *#Z#-test, the #p#-value is defined as #2\cdot \mathbb{P}(Z \geq |z|)#. To calculate this value in R, make use of the following function:

pnorm(q, mean, sd, lower.tail)

q: The value at which you wish to evaluate the distribution function.mean: The mean of the distribution.sd: The standard deviation of the distribution.lower.tail: If TRUE (default), probabilities are #\mathbb{P}(X \leq x)#, otherwise, #\mathbb{P}(X \gt x)#.

Thus, to calculate #p = 2\cdot \mathbb{P}(Z \geq |z|)#, run the following command:

\[

2 \text{ * } \text{pnorm}(q = \text{abs}(z), mean = 0, sd = 1,lower.tail = \text{FALSE})\\

\downarrow\\

2\text{ * } \text{pnorm}(q = \text{abs}(\text{-}0.38668), mean = 0, sd = 1,lower.tail = \text{FALSE})

\]

This gives:

\[p = 0.6990\]

Since #\,p# #\gt# #\alpha#, #H_0: \mu = 6# should not be rejected.

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