P-Value [One-Sample Z-Procedure]

The Problem

Example #176: Let us test the null hypothesis that the mean height of a population is 47 cm. In symbols, this is:

H₀ : μ = 47 cm

To test this hypothesis, we collect data. These data are the heights of a sample from this population. The sample values are:

52, 62, 48, 16, 38, 44, 57, 38, 18, 27, 34, 34, 48, 30, 31, 35, 52, 61, 52, 44, 71, 30, 49, 23, 52, 67, 58, 40, 54, 49, 74, 57, 49, 33, 49, 34, 44, 47, 52, 24, 30, 77, 51, 60, 57, 46, 44, 32, 31, 36, 40, 34, 81, 65, 24, 63, 21, 59, 38, 47, 53, 27, 45, 63, 59, 41, 55, 25, 28, 64, 48, 61, 44, 58, 53

In addition to these data, we also know that the data are generated from a Normal process (they come from a Normally distributed population) and that the standard deviation of this population is σ = 16.

With this information, calculate the test statistic corresponding to the null hypothesis.

Information given:

To summarize the above, the values of import are:

Summary statistics from the problem
$\mu_0$	=	47
$\bar{x}$	=	45.8267
$\sigma$	=	16
$n$	=	75
$z$	=	-0.6351

Calculate them yourself, then hover your mouse over the grey space to see if you calculated them correctly.

Your Answer

In the box below, please enter the p-value for the null hypothesis and the data given above. Once you have done so, click on the “Check your answer!” button. Please round your answer to the ten-thousandths place.

Your guess:

Another Set of Data

Would you like to continue working on this topic? If so, click here for another data set.

Assistance

Show Formula

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Here is the formula you can use to calculate the z test statistic.

$\text{p-value} = 2\ \Phi \left( -|z| \right)$

In this formula, there are a few symbols to know:

$z$

the observed test statistic (see The One-Sample Z-Procedure Test Statistic for practice)
In this example, $z = -0.6351$ .

$\Phi(z)$

the cumulative distribution function (CDF) function for the standard Normal distribution evaluated at the value of the test statistic, $P[Z \le z]$
To determine this, use a Z-Table to find the probability (area) corresponding to the calculated z-score.

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$\begin{align} \text{p-value} &= 2\ \Phi \left( {-|z|} \right) \\[1em] &= 2\ \Phi \Big(\, -\big|(-0.6351)\big|\ \Big) \\[1em] &= 2\ \Phi \Big(\, -0.6351\ \Big) \\[1em] &= 2\times 0.2627 \\[1em] &= 0.5254 \\ \end{align}$

Thus, the p-value for these data and this hypothesis is 0.5254. As this p-value is greater than the usual α = 0.05, we fail to reject the null hypothesis. What we observed is reasonably common if the null hypothesis is true.

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The z-procedures are sensitive to knowing the population variance. Logic dictates that if we do not know the population mean, then we will not know the population variance. As such, the z-procedures are rarely used now. As such, there is no z-test in the base Excel program.

Copy and paste the following code into your Excel spreadsheet window, making sure the value x ends up in A1 after pasting.

How to calculate the expected value in Excel.
sample
52	mu0:	47
62	sigma:	16
48	ts:	=(AVERAGE(A:A)-C2)/C3*SQRT(COUNT(A:A))
16	p.value:	=2*NORM.DIST(-ABS(C4),0,1,TRUE)
38
44
57
38
18
27
34
34
48
30
31
35
52
61
52
44
71
30
49
23
52
67
58
40
54
49
74
57
49
33
49
34
44
47
52
24
30
77
51
60
57
46
44
32
31
36
40
34
81
65
24
63
21
59
38
47
53
27
45
63
59
41
55
25
28
64
48
61
44
58
53

The z test statistic is the number calculated in cell C4.

Again, when you paste this code into Excel, make sure that you start the pasting in call A1. To help with that, you may want to also copy this notice. It seems to help.

© Ole J. Forsberg, Ph.D. 2025. All rights reserved.		.

Calculating the P-Value

The Problem

Information given:

Your Answer

Another Set of Data

Assistance