EEB225H1 Lecture Notes - Lecture 3: Type I And Type Ii Errors, Null Hypothesis, Explained Variation

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blackgrasshopper510

28 Apr 2016

School

UTSG

Department

Ecology & Evolutionary Biology

Course

EEB225H1

Professor

Benjamin Gilbert

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Document Summary

Comparing the means of more than two groups: parametric test using an analysis of variance (anova). Type 1 error: difference between two things falsely rejecting null hypothesis. Saying there is a difference when there isn"t a difference. Using a=0. 5, probability of getting a type 1 error is. This probability increases when the number of tests comparing increases. So, we use a different test to compare more than two groups: anova. Anova with 2 groups is the same as two-tailed 2-sample t-test. Null hypothesis for simple anova is there is no variance among groups. Alternative hypothesis is at least one group varies from the others. Compare variability within (error mean square) and among (group mean square) groups. If group mean square (msgroup)>>error mean square (mserror), there are differences among groups. Group mean square: calculate the grand mean: calculate the mean normally if all group n are the same (take the average) or.

Related Questions

1	Which of the following statements about Type I and Type II errors is correct
a	Type I: Reject a true alternative hypothesis. Type II: Do not reject a false alternative.
b	Type I: Reject a true null hypothesis. Type II: Do not reject a false null hypothesis.
c	Type I: Reject a false null hypothesis. Type II: Reject a true null hypothesis.
d	Type I: Do not reject a false null hypothesis. Type II: Reject a true null hypothesis.

2	You are reading a report that contains a hypothesis test you are interested in. The writer of the report writes that the p-value for the test you are interested in is 0.0749, but does not tell you the value of the test statistic. From this information you can:


a	Not reject the hypothesis at a Probability of Type I error = .05, but reject the hypothesis at a Probability of Type I error = 0.10

b	Reject the hypothesis at a Probability of Type I error = .05, and reject at a Probability of Type I error = 0.10

c	Not reject the hypothesis at a Probability of Type I error = 0.05, and not reject at a Probability of Type I error = 0.10

d	Reject the hypothesis at a Probability of Type I error = .05, but not reject at a Probability of Type I error = 0.10


3	The random sample below is obtained to test the following hypothesis about the population mean.
	H?: ? ?	1500
	H?: ? >	1500
	620	1711	366	2528	2678	1661	442	725	1938
	409	330	2480	542	369	2124	549	2074	1665
	1873	873	2143	2061	1177	2509	1264	2397	1523
	1837	1958	1041	1639	2199	2232	387	2270	2136
	1111	1883	2612	2230	1597	1726	694	1990	1354
	2090	909	2128	1608	747	1121	2220	2390	2347
	1041	316	655	632	2064	1901	532	552	846
	2704	1410	2165	1065	937	1452	2539	410	656
	1169	527	809	2364	2350	2210	1459	2391	856
	2711	1985	2382	2289	1927	518	2177	437	1151
	2018	1580	607	2715	2188	1691	1394	2610	1186
	695	2428	2246	858	2036	1681	2449	1578	1971
	1846	1729	2389	1737	1913	1863	2072	2593	2287
	2220	2230	551	458	2626	2731	488	2551	1736
	1373	307	1803	2647	2679	1508	1468	1443	516
	1002	2116	2616	817	2522	460	1879	1999	1837

	The level of significance of the test is ? = 0.05. Compute the relevant test statistic.
	This is a(n) _______ (two-tail, upper-tail, lower-tail) test. The test statistic is TS = _______.
a	Two-tail test		TS =	1.81
	Do not reject H?: ? ? 1500. Conclude that the population mean is not greater than 1500.
b	Upper tail test.		TS =	1.52
	Do not reject H?: ? ? 1500. Conclude that the population mean is not greater than 1500.
c	Upper tail test.		TS =	1.81
	Reject H?: ? ? 1500. Conclude that the population mean is greater than 1500.
d	Lower tail test.		TS =	1.98
	Do not reject H?: ? ? 1500. Conclude that the population mean is no greater than 1500.

4	Consider the following hypothesis test.
	H?: ? ?	30
	H?: ? >	30
	A random sample of n = 15 yielded the following observations
	51	38	26	16	28
	57	20	33	35	23
	21	47	56	54	36
	Use ? =	0.05
	TS = ______	CV = ______	State the decision rule.
a	1.68	1.761	Do not reject H?. Conclude the mean is not greater than 30.
b	1.68	1.64	Reject H?. Conclude the mean is greater than 30.
c	1.847	2.145	Do not reject H?. Conclude the mean is not less than 30.
d	1.847	1.761	Reject H?. Conclude the mean is less than 30.

5	In a recent study, a major fast food restaurant had a mean service time of 165 seconds. The company embarks on a quality improvement effort to reduce the service time and has developed improvements to the service process. The new process will be tested in a sample of stores. The new process will be adopted in all of its stores, if it reduced mean service time by more than 45 seconds compared to the current mean service time. To perform the hypothesis test, the sample of 48 stores yields the following data (seconds).





	90	96	133	108	136	110	119	138
	129	98	101	92	135	124	115	90
	132	125	110	124	126	138	94	130
	108	96	140	135	102	114	109	137
	138	104	108	134	92	107	96	119
	105	111	96	136	126	116	98	131

	Use ? =	0.05
	\|TS\| = ______	\|CV\| = ______
a	1.548	1.678	Do not reject H?. The mean service time is not reduced by more than 45 seconds. Do not adopt the new process.

b	1.871	1.678	Reject H?. The mean service time is reduced by more than 45 seconds. Adopt the new process.

c	1.871	1.640	Do not reject H?. The mean service time is not reduced by more than 45 seconds. Do not adopt the new process.

d	1.548	1.640	Reject H?. The mean service time is reduced by more than 45 seconds. Adopt the new process.

Introduction: A Chi-square test is used to compare observed data with expected data according to a hypothesis. For instance, if you were crossbreeding 2 heterozygous pea plants, you would expect to see a 3:1 phenotypic ratio in the offspring. In this case, if you were to breed 400 pea plants, you would expect to see 300 plants showing the dominant trait and 100 showing the recessive trait. But what happens if you observe only 260 plants with the dominant trait and 140 plants with the recessive trait? Does this mean something is wrong with Mendelian genetics or is this difference in expected results just due to chance (random sampling error)? These are the questions that can be answered using Chi-square statistics. The results of this statistical test is used to either reject or accept (fail to reject) the null hypothesis. The null hypothesis states there is no significant difference between the observed results and the expected results. This means that if the null hypothesis is accepted, the difference in observed and expected results was just a matter of chance and so the observed results basically "fit" with what was expected. Degrees of freedom (df) = number of independent outcomes (Y) being compared less 1 df = Y-1 At the 95% confidence interval we are 95% confident that there is a significant difference between the observed and expected results, therefore rejecting the null hypothesis. Probability Value - Is the decimal value determined from the X2 table and is the probability of accepting the null hypothesis. A 0.05 probability value equates to a 95% confidence interval.

The Chi-squared test formula is: Example: If we cross two pea plants that are heterozygous yellow pods, we would expect a 3:1 phenotypic ratio. So let's say we actually did the cross and got 280 plants with green pods and 120 plants with yellow pods. Question: Is this a 3:1 phenotypic ratio? This is the value of Chi-squared Test. We have a total of 400 plants and we expect a 300 green:100 yellow phenotypic ratio If the calculated Chi-squared value is less than the critical value listed in the Chi-squared table, then we accept the null hypothesis. This means that there is no significant difference between the observed and the expected values. Our degrees of freedom (df) = 2 outcomes - 1, or df = 1. Now we go the X2 table below and using the df = 1 and probability value of 0.05, our critical value is 3.84. Since our calculated X2 value is 5.33, and is larger than the critical value, we reject the null hypothesis and can say (at 95% confidence) that there is a significant difference between our observed and expected values.

The parent generation is yellowed podded and green podded pea plants. You cross a yellow podded pea plant with a green podded pea plant and you get 100% yellow podded plants in the F1 Generation (Phenotypic ratio 4 : 0, yellow to green). What will be the expected phenotypic ratio when you allow the F1 generation to reproduce?

Fill out the Punnett square.

If we actually did the cross and got 1150 yellow and 350 green. Would this be a consistent with what was expected?

Learning Outcomes Questions

1. Why would you run a Chi-squared test?

To determine if our data is consistent with expected results.
		a To determine if our data is consistent with expected results. b To determine if our data exactly matches the expected results.
		c To determine the expected results.
	d	To compare the phenotypic ratios to the genotypic ratios.

2. Determine the degrees of Freedom of the phenotypic ratio for this genetic cross.

a. 1

b. 2

c. 3

d. 4

e. 5

3. Using the data given, what is the result of your Chi-squared analysis? x2= ___.

	a.	2.22
	b	2.71
	c	4.36
	d	187.78
	e	448.27

4. Using the results of your Chi-squared analysis, do we fail to reject or reject the null hypothesis?

a.		Fail to reject the null
b.		Reject the null
c.		It cannot be determined from the data given

silversalmon27

Several studies suggest that breast-fed babies become more intelligent children than formula-fed babies. One such study (Mortensen, Michaelsen, Sanders & Reinisch, 2002) involved a sample of over 3,000 women and men born in Copenhagen, Denmark, between October 1959 and December 1961. The samples were divided into five categories based on duration of breast-feeding, as assessed by physician interview with mothers at a one-year examination. The childâs intelligence was assessed using the Wechsler Adult Intelligence Scale (WAIS) at a mean age of 27.2 years. The results showed that duration of breast-feeding was associated with significantly higher scores on the verbal, performance, and full scale WAIS IAs. This difference was observed even after the researchers adjusted for differences in the social class and maternal education of the two groups. (This adjustment allowed the researchers to rule out any preexisting differences in the groups that might have independently contributed to IQ differences in their children. The authors acknowledged that other differences between the groups, such as the childrenâs genetic potential or their parentsâ caregiving skills or motivation to nurture, could explain the results. However, they believe that human milk contains various hormones and other factors that enhance brain growth and maturation.

1. State the research hypothesis in your own words. Identify the independent and dependent variables, and define all important concepts and terms as they are used in this study.

2. What evidence do the researchers offer as a test of their hypothesis? Is this evidence empirical (observable)? Is it valid?

3. What explanation do the researchers offer for their findings? Does this explanation make sense based on the evidence?

4. Given the results of this study, why canât the researchers draw a causal connection between type of food and later intelligence? What might be an alternative explanation for the results of this study? What could the researchers do in order to make a causal connection between the dependent and independent variables?

5. Are there any practical implications for this research?

EEB225H1 Lecture Notes - Lecture 3: Type I And Type Ii Errors, Null Hypothesis, Explained Variation

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