basic stat question
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Hello,
Apologies for a non-SPSS question.
I would like to compare two proportions but I only have data for one group. In other words I need to compare the proportion in my data to a fixed value. What is the best way? I would like to use the one sample means t-test, but am not sure if it is valid. A colleague recommends a 2x2 chi-sq test which is equal to a z-test of proportions, I believe. I cannot find a link between the z-test and one sample t-test.
Is it valid to use the one sample t-test to compare a proportion to a standard?
Thank you,
John |
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No, don't use a t-test. The usual test would be a z-test for a single proportion. You can see an example here (mind the line-wrap): http://www.cliffsnotes.com/study_guide/Test-for-a-Single-Population-Proportion.topicArticleId-25951,articleId-25941.html But SPSS does not compute this test directly, AFAIK. However, a chi-square goodness of fit test is equivalent to the z-test. Remember than z-squared = chi-square with df = 1. Here's some code to analyze the example given on that web-page. new file. dataset close all. data list list / category count (2f5.0). begin data 1 32 2 38 end data. weight by count. NPAR TESTS /CHISQUARE=category /EXPECTED=.4 .6 /MISSING ANALYSIS. * Or, if you have the "exact" statistics module, you can do it this way. NPAR TESTS /CHISQUARE=category /EXPECTED=.4 .6 /MISSING ANALYSIS /METHOD=EXACT TIMER(5). IIRC, the "exact" p-value from this second analysis is computed using the binomial distribution. HTH.
--
Bruce Weaver bweaver@lakeheadu.ca http://sites.google.com/a/lakeheadu.ca/bweaver/ "When all else fails, RTFM." PLEASE NOTE THE FOLLOWING: 1. My Hotmail account is not monitored regularly. To send me an e-mail, please use the address shown above. 2. The SPSSX Discussion forum on Nabble is no longer linked to the SPSSX-L listserv administered by UGA (https://listserv.uga.edu/). |
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In reply to this post by J P-6
You need to run a one sample z test. Please note that the hypothetical standard deviation in the population, and the distributionv of sample proportions (given a specific sample size) are determined by the hypothetical population proportion (the population SD is not a parameter that you need to estimate from sample statistics).
HTH, Steve Brand www.StatisticsDoc.com From: J P <[hidden email]>
Date: Wed, 3 Feb 2010 13:25:11 -0800 To: <[hidden email]> Subject: basic stat question Hello,
Apologies for a non-SPSS question.
I would like to compare two proportions but I only have data for one group. In other words I need to compare the proportion in my data to a fixed value. What is the best way? I would like to use the one sample means t-test, but am not sure if it is valid. A colleague recommends a 2x2 chi-sq test which is equal to a z-test of proportions, I believe. I cannot find a link between the z-test and one sample t-test.
Is it valid to use the one sample t-test to compare a proportion to a standard?
Thank you,
John |
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In reply to this post by Bruce Weaver
Hi Bruce,
Although I've never tried, it should also be possible to use a generalized linear model (intercept only), assuming the expected proportions are .50. GENLIN {Variable Name} /MODEL INTERCEPT=YES DISTRIBUTION=BINOMIAL LINK=LOGIT /PRINT SOLUTION. Best, Ryan
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Thanks to everyone for your assistance. The NPARS does the trick.
From: rblack <[hidden email]> To: [hidden email] Sent: Thu, February 4, 2010 8:50:39 AM Subject: Re: basic stat question Hi Bruce, Although I've never tried, it should also be possible to use a generalized linear model (intercept only), assuming the expected proportions are .50. GENLIN {Variable Name} /MODEL INTERCEPT=YES DISTRIBUTION=BINOMIAL LINK=LOGIT /PRINT SOLUTION. Best, Ryan Bruce Weaver wrote: > > > J P-6 wrote: >> >> Hello, >> >> Apologies for a non-SPSS question. >> >> I would like to compare two proportions but I only have data for one >> group. In other words I need to compare the proportion in my data to a >> fixed value.  What is the best way? I would like to use the one sample >> means t-test, but am not sure if it is valid. A colleague recommends a >> 2x2 chi-sq test which is equal to a z-test of proportions, I believe. I >> cannot find a link between the z-test and one sample t-test. >> >> Is it valid to use the one sample t-test to compare a proportion to a >> standard? >> >> Thank you, >> John >> >> > > No, don't use a t-test. The usual test would be a z-test for a single > proportion. You can see an example here (mind the line-wrap): > > http://www.cliffsnotes.com/study_guide/Test-for-a-Single-Population-Proportion.topicArticleId-25951,articleId-25941.html > > But SPSS does not compute this test directly, AFAIK. However, a > chi-square goodness of fit test is equivalent to the z-test. Remember > than z-squared = chi-square with df = 1. Here's some code to analyze the > example given on that web-page. > > new file. > dataset close all. > > data list list / category count (2f5.0). > begin data > 1 32 > 2 38 > end data. > > weight by count. > > NPAR TESTS > /CHISQUARE=category > /EXPECTED=.4 .6 > /MISSING ANALYSIS. > > * Or, if you have the "exact" statistics module, you can do it this way. > > NPAR TESTS > /CHISQUARE=category > /EXPECTED=.4 .6 > /MISSING ANALYSIS > /METHOD=EXACT TIMER(5). > > IIRC, the "exact" p-value from this second analysis is computed using the > binomial distribution. > > HTH. > > > -- View this message in context: http://old.nabble.com/basic-stat-question-tp27443741p27452194.html Sent from the SPSSX Discussion mailing list archive at Nabble.com. ===================== 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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