100013 Lecture 5: Hypothesis testing

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crimsoncoyote209

25 Feb 2019

School

UWS

Department

University of Western Sydney

Course

100013

Professor

Lee Shepherd

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Document Summary

Topic 5: the null hypothesis h0 is that there is no significant difference between specified population, the experimental hypothesis ha states that there is a statistically different relationship between variables. In psychology, we rarely examine just one persons score in relation to a population, because if the original population is skewed, the sampling distribution of the mean is normal. If we have access to the parameters of a population sampling distribution we can collect one sample mean and compare it to these estimates. Instead of comparing one observation to a population distribution of observations we compare a mean to an estimated distribution of means. One- tailed test: tests whether our sample mean is significantly larger than population mean, or tests whether score is significantly smaller than the mean. Two tailed test: tests whether score is significantly larger or smaller than mean (p<0. 025 each end, the sample mean could be higher or lower than the population mean.

Related Questions

Use the following scenario for questions 6 – 8. In 2016-17, the average SAT score of the UVA first year class was 1420 out of 1600. An advisor in the statistics department is interested in determining whether statistics majors from that class have higher SAT scores than the average. He tests H,:µ = 1420 against Hạ: µ > 1420, where u is the mean SAT score for statistics majors from the 2016-17 first year class. In a simple random sample of 30 statistics majors, he finds a sample mean SAT score of 1440 with sample standard deviation of 40.

6. What is the value of the test statistic for this significance test?

a. 0.09

b. 0.5

c. 2.73

d. 20

7. From what distribution is this test statistic?

a. N(0, 1)

b. N(1420, 7.30)

c. t distribution with 29 degrees of freedom

d. t distribution with 28 degrees of freedom

8.) The p-value for this test is 0.005. Which of the following is a proper interpretation of this p- value?

a.There is a 0.5% chance that the null hypothesis is true.

b. If the population mean SAT score really is 1420, there's a 0.5% chance that the sample a. mean SAT score from 30 statistics majors would equal 1440.

c.If the population mean SAT score really is 1420, there's a 0.5% chance that the sample mean SAT score from 30 statistics majors would be 1440 or lower.

d. If the population mean SAT score really is 1420, there's a 0.5% chance that the sample mean SAT score from 30 statistics majors would be 1440 or higher.

beigeserval271

Hypothesis testing with ANOVA opinions about whether caffeine enhances test performance differ. You design a study to test the impact of drinks with different caffeine contents on students' test-taking abilities. You choose 21 students at random from your introductory psychology course to participate in your study. You randomly assign each student to one of three drinks, each with a different caffeine concentration, such that there are seven students assigned to each drink. You then give each of them a plain capsule containing the precise quantity of caffeine that would be consumed in their designated drink and have them take an arithmetic test 15 minutes later.

The students receive the following arithmetic test scores:

	Cola	Black Tea	Coffee
Caffeine Content (mg/oz)	2.9	5.9	13.4
	85	85	92	∑X² = 147,641
	86	89	87	G = 1,755
	82	82	80	N = 21
	75	75	89	k = 3
	66	88	96
	78	76	83
	87	82	92
	T₁ = 559	T₂ = 559	T₃ = 559
	SS₁ = 338.86	SS₂ = 338.86	SS₃ = 338.86
	n₁ = 7	n₂ = 7	n₃ = 7
	M₁ = 79.8571	M₂ = 79.8571	M₃ = 79.8571

1.) You plan to use an ANOVA to test the impact of drinks with different caffeine contents on students' test-taking abilities. What is the null hypothesis?

(a) The population mean test score for the cola population is different from the population mean test score for the black tea population.

(b) The population mean test scores for all three treatments are equal.

(d) The population mean test scores for all three treatments are not all equal.

2.) Calculate the degrees of freedom and the variances for the following ANOVA table:

Source	SS	df	MS
Between	-	-	-
Within	702.28	-	-
Total	973.14	-

The formula for the F-ratio is:

3.) Using words, the formula of the F-ratio can be written as:

4.) Using the data from the ANOVA table given, the F-ratio can be written as:

5.) Calculate for F-ratio:

6.) At the level of significance, what is your conclusion?

(a)You can reject the null hypothesis; you do not have enough evidence to say that caffeine affects test performance.

(b)You cannot reject the null hypothesis; caffeine does appear to affect test performance.

(c)You cannot reject the null hypothesis; you do not have enough evidence to say that caffeine affects test performance.

(d)You can reject the null hypothesis; caffeine does appear to affect test performance.

ceruleanfly137

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

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100013 Lecture 5: Hypothesis testing

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