convert z to p value

[Reginal]

Hi,

I'm familiar with the Sig.Chisq and Sig.F statistical functions in SPSS which
allow for the statistical significance testing of chi-square and F values,
respectively. Does anyone know how to execute the same function for z values?
I realise I could square z values to transform them into chi-square values,
but I'd like to stick with z values for didactic reasons.

I'm also familiar with the CDF.Normal function which nearly does what I am
requesting, but not quite. For example, CDF.Normal(-1.96, 0, 1) yields p =
.025, however, I would prefer p = .05 (i.e., two-tailed), which is what you
get when you use SIG.CHISQ(3.8416,1). Also, with positive Z values,
CDF.Normal(1.96, 0, 1) yields p .975, when I would rather it yield p = .050.

I know, it seems picky, but I'm showing students how to test specific z values
for statistical significance using SPSS. Any ideas would be much appreciated.

Thanks,

Reg

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[Art Kendall]

One approach would be an opportunity to show the difference between one-tail and two-tail probabilities.
compute lower_one_tail_p = CDF.Normal(-1.96, 0, 1).
compute upper_one_tail_p = CDF.Normal( 1.96, 0, 1).
compute two_tail_p = lower_one_tail_p + upper_one_tail_p.
On 5/31/2012 5:08 AM, Reginal wrote:

Hi,

I'm familiar with the Sig.Chisq and Sig.F statistical functions in SPSS which
allow for the statistical significance testing of chi-square and F values,
respectively. Does anyone know how to execute the same function for z values?
I realise I could square z values to transform them into chi-square values,
but I'd like to stick with z values for didactic reasons.

I'm also familiar with the CDF.Normal function which nearly does what I am
requesting, but not quite. For example, CDF.Normal(-1.96, 0, 1) yields p =
.025, however, I would prefer p = .05 (i.e., two-tailed), which is what you
get when you use SIG.CHISQ(3.8416,1). Also, with positive Z values,
CDF.Normal(1.96, 0, 1) yields p .975, when I would rather it yield p = .050.

I know, it seems picky, but I'm showing students how to test specific z values
for statistical significance using SPSS. Any ideas would be much appreciated.

Thanks,

Reg

=====================
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[Art Kendall]

If you use http://psych.colorado.edu/~mcclella/java/normal/normz.html
you can demo z scores by plugging +/-1.96 as the y  and the z-score.

On 5/31/2012 10:14 AM, Art Kendall wrote:

One approach would be an opportunity to show the difference between one-tail and two-tail probabilities.
compute lower_one_tail_p = CDF.Normal(-1.96, 0, 1).
compute upper_one_tail_p = CDF.Normal( 1.96, 0, 1).
compute two_tail_p = lower_one_tail_p + upper_one_tail_p.
On 5/31/2012 5:08 AM, Reginal wrote:

Hi,

I'm familiar with the Sig.Chisq and Sig.F statistical functions in SPSS which
allow for the statistical significance testing of chi-square and F values,
respectively. Does anyone know how to execute the same function for z values?
I realise I could square z values to transform them into chi-square values,
but I'd like to stick with z values for didactic reasons.

I'm also familiar with the CDF.Normal function which nearly does what I am
requesting, but not quite. For example, CDF.Normal(-1.96, 0, 1) yields p =
.025, however, I would prefer p = .05 (i.e., two-tailed), which is what you
get when you use SIG.CHISQ(3.8416,1). Also, with positive Z values,
CDF.Normal(1.96, 0, 1) yields p .975, when I would rather it yield p = .050.

I know, it seems picky, but I'm showing students how to test specific z values
for statistical significance using SPSS. Any ideas would be much appreciated.

Thanks,

Reg

=====================
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[hidden email] (not to SPSSX-L), with no body text except the
command. To leave the list, send the command
SIGNOFF SPSSX-L
For a list of commands to manage subscriptions, send the command
INFO REFCARD

===================== To manage your subscription to SPSSX-L, send a message to [hidden email] (not to SPSSX-L), with no body text except the command. To leave the list, send the command SIGNOFF SPSSX-L For a list of commands to manage subscriptions, send the command INFO REFCARD

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[Bruce Weaver]

And if you're always working with the standard normal, you can use CDFNORM in place of CDF.Normal.  The former takes only the z value as a parameter.  E.g.,

data list free / z (f5.3).
begin data
-2.57 -2.33 -1.96 -1.645 -1.28 0 1.28 1.645 1.96 2.33 2.57
end data.

* Computing p-values for z-values is easier if you always use
* a negative z-value.

compute #zneg = 0-abs(z). /* z-value in lower tail .
compute p1tailed = cdfnorm(#zneg).
compute p2tailed = cdfnorm(#zneg)*2.
compute pfromchisq = sig.chisq(z**2,1).
formats p1tailed to pfromchisq (f5.3).
list.

OUTPUT:

     z p1tailed p2tailed pfromchisq
-2.570    .005     .010      .010
-2.330    .010     .020      .020
-1.960    .025     .050      .050
-1.645    .050     .100      .100
-1.280    .100     .201      .201
  .000    .500    1.000     1.000
 1.280    .100     .201      .201
 1.645    .050     .100      .100
 1.960    .025     .050      .050
 2.330    .010     .020      .020
 2.570    .005     .010      .010

Number of cases read:  11    Number of cases listed:  11

HTH.

Art Kendall wrote

One approach would be an opportunity to show the
      difference between one-tail and two-tail probabilities.
      compute lower_one_tail_p = CDF.Normal(-1.96, 0, 1).
    compute upper_one_tail_p = CDF.Normal( 1.96,
      0, 1).
    compute two_tail_p = lower_one_tail_p + upper_one_tail_p.
    Art Kendall
Social Research Consultants
   
    On 5/31/2012 5:08 AM, Reginal wrote:
   
      Hi,

I'm familiar with the Sig.Chisq and Sig.F statistical functions in SPSS which
allow for the statistical significance testing of chi-square and F values,
respectively. Does anyone know how to execute the same function for z values?
I realise I could square z values to transform them into chi-square values,
but I'd like to stick with z values for didactic reasons.

I'm also familiar with the CDF.Normal function which nearly does what I am
requesting, but not quite. For example, CDF.Normal(-1.96, 0, 1) yields p =
.025, however, I would prefer p = .05 (i.e., two-tailed), which is what you
get when you use SIG.CHISQ(3.8416,1). Also, with positive Z values,
CDF.Normal(1.96, 0, 1) yields p .975, when I would rather it yield p = .050.

I know, it seems picky, but I'm showing students how to test specific z values
for statistical significance using SPSS. Any ideas would be much appreciated.

Thanks,

Reg

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