Unit 3 Homework

Note: All work that can be done in SPSS should be submitted that way, complete with tables, graphs with titles, and stat and English statements, as demonstrated in the work text.

1. Does lecithin enhance memory? To test this idea, nine pairs of people matched in terms of age, sex, IQ and neurological status are given capsules containing either lecithin or a placebo at meals for a week. All are scored on a recent memory talk. A higher score indicates better task performance.

1. Conduct and interpret the appropriate analysis to determine if lecithin significantly improved memory. (5 pts)
2. Create the appropriate graph to show the difference between the two treatment approaches. (2.5 pts)
3. People scoring below the 35 ^th percentile in the Placebo group are going to be ‘tagged’ for further assessment. What score would alert researchers to the need for such assessment?
4. After review of the scoring and assessment tool, researchers decided that equal intervals could not be assumed, and that the data should be treated as a rating. Conduct and interpret the appropriate analysis to determine if the lecithin group has significantly better memory. (5 pts)

Solution: (a) We need to run a right-tailed paired-samples t-test. Using SPSS we obtain:

The t-statistics is t = 2.927, and the corresponding right-tailed p-value is p = 0.019/2 = 0.095, which is less than the significance level 0.05. This means that we have enough evidence to support the claim that lecithin significantly improved memory

(b) We have the following plot:

We can also use the plot for the difference:

(c) Assuming normality, we conclude that the 35 ^th percentile for the placebo group is computed as

(d) In this case, only the ordinal level of measurement can be assumed, so we need a parametric test (Wilcoxon test). Using SPSS we get

The z-value is z = -2.178, and the corresponding p-value is p = 0.029, which means that we reject the null hypothesis of equal medians, reaching the same conclusion as the one obtained with the parametric test.

2. An instructor asks students in his class to indicate whether or not they agree with the opinions of three different speakers.

1. Conduct and interpret the appropriate analysis to determine if there is a connection between the speaker and the opinion of the students (20 pts).

Solution: Using SPSS the following results are obtained:

The Chi-Square statistic is \({{\chi }^{2}}=8.043\), and the corresponding p-value is p = 0.018, which means that we reject the null hypothesis of independence. This means that we have enough evidence to claim that the variables are related.

3. Suppose that a sociologist determines from the census report that the distribution of people in various occupations in a specific geographic area are as follows:

He samples individuals who are members of a particular political party to see if the distribution of occupations are the same.

1. Conduct and interpret the appropriate analysis to determine if the sample differs from the census report or is like the expected distribution(20 points)
   Solution: We need to test the following null hypothesis:
   \[{{H}_{0}}:\] \[{{p}_{\text{1}}}=\text{0}\text{.2}\] , \[{{p}_{\text{2}}}=\text{0}\text{.3}\] , \[{{p}_{\text{3}}}=\text{0}\text{.3}\] , \[{{p}_{\text{4}}}=\text{0}\text{.15}\] , \[{{p}_{\text{5}}}=\text{0}\text{.05}\]
   We need to construct the table with observed and expected values (under the null hypothesis). Based on the data provided, we find:

   	Agricultural	Laborers	Local, state or federal government	Professional and/or self-employed	Industrial Managers	Total
   Observed (fo)	145	310	305	78	26	864
   Expected (fe)	172.8	259.2	259.2	129.6	43.2	864
   (fo - fe)²/fe	4.472	9.956	8.093	20.544	6.848	49.914

   
   This means the Chi-Square statistics is computed as
   \[{{\chi }^{2}}=\sum\limits_{i=1}^{n}{\frac{{{\left( {{O}_{i}}-{{E}_{i}} \right)}^{2}}}{{{E}_{i}}}}=\] \[\frac{{{\left( \text{145}-\text{172}\text{.8} \right)}^{2}}}{\text{172}\text{.8}}\] + \[\frac{{{\left( \text{310}-\text{259}\text{.2} \right)}^{2}}}{\text{259}\text{.2}}\] + \[\frac{{{\left( \text{305}-\text{259}\text{.2} \right)}^{2}}}{\text{259}\text{.2}}\] + \[\frac{{{\left( \text{78}-\text{129}\text{.6} \right)}^{2}}}{\text{129}\text{.6}}\] + \[\frac{{{\left( \text{26}-\text{43}\text{.2} \right)}^{2}}}{\text{43}\text{.2}}\] \[=\text{49}\text{.914}\]
   The critical value for \(\alpha =\text{0}\text{.05}\) and \(d.f.=\text{4}\) is given by
   \[\chi _{C}^{2}=\text{9}\text{.488}\]
   and the corresponding p-value is
   \[p=\Pr \left( {{\chi }^{2}}>\text{49}\text{.914} \right)=\text{0}\text{.000}\]
   Since the p-value is less than the significance level \(\alpha =\text{0}\text{.05}\), then we reject \({{H}_{0}}\). This means that we have enough evidence to reject the null hypothesis of the given proportions, at the 0.05 significance level.
   4. A researcher was interested in the attitudes towards equal rights among stay at home moms and women with jobs outside the home. She creates a survey that measures attitudes toward equal rights, administers it to a sample of 20 women. However, since the survey is not standardizes, she realizes that the scores may not reflect equal intervals.
   
   

   Stay at Home	Jobs Outside the Home
   19 22 28 32 34 37 40 42 43 46	16 18 21 26 27 29 31 33 38 39

   
   a. Conduct and interpret the appropriate analysis (15 pts).
2. Construct the appropriate graph to illustrate the differences between the two groups. (5 pts.)
   Solution: (a) Since we don’t know whether or not the scores reflect equal intervals, we’ll use a non parametric test (Mann-Whitney). Using SPSS we obtain:

   Ranks
   	type	N	Mean Rank	Sum of Ranks
   score	Stay at Home	10	12.80	128.00
   	Job outside the home	10	8.20	82.00
   	Total	20

   Test Statistics b
   	score
   Mann-Whitney U	27.000
   Wilcoxon W	82.000
   Z	-1.739
   Asymp. Sig. (2-tailed)	.082
   Exact Sig. [2*(1-tailed Sig.)]	.089 a
   Not corrected for ties.
   b. Grouping Variable: type

   
   The p-value for the Mann-Whitney test is p = 0.082, which means that we fail to reject the null hypothesis of equal means
   (b) We have:
   
   5. Suppose you are interested in whether or not fluency with the English language is affected by the teaching method. You are working with a school system that has a large population of children for whom English is their second language. This school system is open to trying new teaching methods, so you set up a study where Classroom 1 follows the traditional rote-learning method of teaching, Classroom 2 utilizes word games to facilitate learning, and Classroom 3 uses some online video games designed around the vocabulary words. Experienced classroom teachers are asked to rate the fluency of the non-English speaking children while watching them in a free-play situation. Higher ratings indicate greater ability to communicate in English.

   Classroom 1	Classroom 2	Classroom 3
   9 6 4 3 1 7 8 9 8 6	12 14 8 10 6 15 8 11 12 10	18 16 15 12 20 10 4 8 3 10

   1. Conduct and interpret the appropriate analysis to determine the effect of the different teaching methods on ability to communicate (15 pts.)
      b. Construct the appropriate graph to illustrate the differences in the teaching methods (5 pts.)

Solution: (a) Since we are dealing with rating, an appropriate method would be applying Kruskal-Wallis test. Using SPSS we get:

The Chi-Square statistic is 9.325, and the corresponding p-value is p = 0.009, which means that we have enough evidence to reject the null hypothesis that the samples come from the same distribution.

(b) The effect of the teaching methods is depicted in the following graphs:

Price: $9.49

Solution: The downloadable solution consists of 4 pages, 549 words.
Deliverable: Word Document