ANSWERS CHAPTER 9. TIO 9.2: If the values are the same, the difference is 0, therefore the null hypothesis cannot be rejected.

Transcription

1 ANSWERS CHAPTER 9 THINK IT OVER thnk t over TIO 9.: χ 2 k = ( f e ) = 0 e Breakng the equaton down: the test statstc for the ch-squared dstrbuton s equal to the sum over all categores of the expected frequency for category subtracted from the observed frequency for category, all dvded by the expected frequency for category. TIO 9.2: If the values are the same, the dfference s 0, therefore the null hypothess cannot be rejected. TIO 9.3: Fred can t st down and run a marathon at the same tme, he has to choose one or the other, therefore the events are mutually exclusve. Hs ftness level wll have a huge mpact on hs performance, therefore hs result s dependent on hs ftness level. TIO 9.4: 5 4 x+ y y= x= Wrtng ths out long hand : set y to have the value : ( + ) + (2 + ) + (3 + ) + (4 + ) = 4. Next set x =: ( + ) + ( + 2) + ( + 3) + ( + 4) + ( + 5) = Therefore, x+ y = = 34 y= x= y= x= TIO 9.5: Each dstrbuton has certan assumptons assocated wth t Bnomal: dscrete random varable, events are ndependent wth the possblty of two outcomes, e.g. con tossng. Normal: the random varable s contnuous, e.g. test scores. Posson: dscrete random varable, and the probablty of events s the same for any two ntervals of equal length, events are ndependent, e.g. number of arrvals at a café. TIO 9.6: It would mean we would accept the alternatve H : the number of cars enterng the drve-n lne durng 0-mnute ntervals does not have a Posson dstrbuton. Whch just tells us to thnk agan about the arrval patterns! TIO 9.7: Thnk back to a standardsed normal dstrbuton. We know that a z-score of corresponds to standard devaton from the mean. If you look up a z-score of n a table of areas under the normal dstrbuton you wll see t corresponds to an area of 0.343, or approxmately 34%. In our case we want an area of 0% so we look for a value close to 0. n the table, then read off the correspondng z-score. In ths case t s Remember we are workng wth one half of the dstrbuton and the z-value we read off s close to the mean not n the tal. We do the same for 20%, whch s 0.53, 30% s 0.84, 40% s.28. TIO 9.8: The Central Lmt Theorem states that: the dstrbuton of a mean tends to be normal, even when the dstrbuton from whch the mean s computed s decdedly non-normal. You would be testng the mean of the mean.

2 TIO 9.9: SPSS assumes the data s from the populaton and therefore only subtracts n the degrees of freedom calculaton. In the example ths means the number of categores mnus, whereas the hand calculaton used the number of categores mnus the two estmated parameters, gvng a value of 7 for the degrees of freedom. It could potentally cause a problem snce f you look at the ch-squared table and compare the values for 7 and 9 degrees of freedom at a sgnfcance of 0.05, you wll see dfferent values. Wth 7 degrees of freedom the value s 2.7 and for 9 t s 3.33 so f the result from the test fell between these values, how would a decson be reached? Ths s where you must use your judgement, because f the value of the test statstc s ths close to the value gven n the table, you wll need to look at other measures. Ths s the major advantage of usng SPSS snce you can ask t to dsplay the mean and standard devaton and nstgate other descrptve tests. You should remember for data to be normally dstrbuted, the standard devaton should be small compared to the mean. You could also check the kurtoss and skewness and use these values to help you reach a decson. After all that, n the example just done the test statstc of 8.4 was suffcently large to enable us to reach a reasonably robust decson. EXERCISES. If the confdence level s set to 0.05, any value smaller than ths would be n the tal of the dstrbuton, meanng there s suffcent evdence to reject the null hypothess. 2. It means f one event occurs another cannot. For example, rollng a de: f you roll a 6 you cannot roll a at the same tme x = set j = x = = 8 j j= = 4 x = = j = 2 = = 9 Contnue to set j up to 5. The answer s then = ,.e. rubbsh! 5. The sample data s a random samplng from a fxed dstrbuton or populaton where every collecton of members of the populaton of the gven sample sze has an equal probablty of selecton. A sample wth a suffcently large sze s assumed. Adequate expected cell counts. Some requre 5 or more, and others requre 0 or more. A common rule s 5 or more n all cells of a 2 2 table, and 5 or more n 80% of cells n larger tables, but no cells wth zero expected count. The observatons are always assumed to be ndependent of each other. 6.

3 result Observed N Expected N Resdual red blue purple Total 333 Test Statstcs result Ch-Square.468 a df 2 Asymp. Sg..480 a. 0 cells (0.0%) have expected frequences less than 5. The mnmum expected cell frequency s A ch-squared value of.468 s less than the 95% level (5.99), the null hypothess cannot be rejected and therefore the results are consstent wth the theory. Also, SPSS reports a sgnfcance value of 0.48 whch s far greater than the 0.05 value of Goodness of ft refers to how close the expected frequency s to the observed frequency. In other words how well your model fts wth realty! 8. Null hypothess: students show no preference. Alternatve hypothess: students do show a preference. Case Processng Summary Cases Vald Mssng Total N Percent N Percent N Percent Answer * Bar % 0 0.0% % Answer * Bar Crosstabulaton Bar Plate of spuds Bucket of Frogs Total Answer Yes Count Expected Count No Count Expected Count Total Count Expected Count

4 Ch-Square Tests Value df Asymp. Sg. (2- Exact Sg. (2- Exact Sg. (- Pearson Ch-Square a.003 Contnuty Correcton b Lkelhood Rato Fsher's Exact Test Lnear-by-Lnear Assocaton N of Vald Cases 489 a. 0 cells (0.0%) have expected count less than 5. The mnmum expected count s b. Computed only for a 2x2 table 2 χ = 9.053, f p < 0.05 then we do not have suffcent grounds to reject the null hypothess,.e. there s lttle evdence to suggest students show a preference. 9. Degrees of freedom = (n )(m ) = 2 where n s the number of rows and m the number of columns. 0. Reason * bar Crosstabulaton bar Spuds Frogs Olves Total Reason drnk cost Count Expected Count locaton Count Expected Count state Count Expected Count other Count Expected Count Total Count Expected Count Ch-Square Tests Value df Asymp. Sg. (2- Pearson Ch-Square a Lkelhood Rato Lnear-by-Lnear Assocaton N of Vald Cases 87 a. 0 cells (0.0%) have expected count less than 5. The mnmum expected count s 5.2.

5 Dependng on the wordng of the null hypothess, the test statstc ndcates that f the null hypothess was there s no relatonshp between the prmary reason for not usng a bar and the bar, then t cannot be rejected. In Englsh, ths means that students went to the dfferent bars even though they complaned about them!. No answer requred. 2. No answer requred.