PBHL 5401 Chapter Notes - Chapter 7: Sampling Distribution, Statistical Hypothesis Testing, Type I And Type Ii Errors

24 views3 pages

mauvehamster904

11 Mar 2016

School

Department

Course

Professor

For unlimited access to Textbook Notes, a Class+ subscription is required.

Document Summary

This is the formula for t testing. x s. N this is the standard error of the mean or the standard deviation of the sampling distribution. Degrees of freedom: this data from the chart will correct the sampling error. You will calculate the t obtained value and the critical value comes from the chart. Note the which is the type 1 error rate. The alpha number will be given as a percent so it must be turned into a decimal. The book will have the figures in the back for the critical values. The significance level of error happening # % of the time. Degrees of freedom formula= df=(n which is sample size)-(1) Use the df number to look at the chart. This value will be on the left side of the chart going in order from 1 . = tells us the value we set as the cutoff.

Related Questions

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

2. (12 marks) In a study of 85 individuals, food questionnaires were used to measure the change in polyunsaturated fatty acids (PUFA) intake from the beginning to the end of the study. In addition researchers measured a genetic marker in the TNF-αgene; for this analysis, we will take the two values of this TNF-αvariable to be A- or A+. The researchers also collected information on gender, age and BMI. The primary question of interest is whether TNF-αmodiﬁes the eﬀect of PUFA on HDL.

(a) (1 mark) Let TNFa be a variable with value 1 for A+ and 0 for A- TNF-αstatus and

let gender be 1 for males and 0 for females. State a linear model for mean change in HDL that includes the predictors PUFA, TNFa, PUFA×TNFa interaction, and eﬀects for gender, age and BMI.

(b) (1 mark) A summary of the ﬁtted model is:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 0.037531 0.196980 0.191 0.8494

pufa 0.019143 0.009245 2.071 0.0417

TNFa 0.074433 0.034128 2.181 0.0322

gender -0.008988 0.031793 -0.283 0.7781

age -0.003895 0.002156 -1.806 0.0747

bmi 0.006777 0.004332 1.565 0.1217

pufa:TNFa -0.024575 0.017844 -1.377 0.1724

Write a short sentence to describe why we would even consider removing gender from the model (i.e., why we would not just be happy with the above model from

the outset).

(c) (2 marks) Below is a summary of the model ﬁt without gender. Using our 10% rule-of-thumb, is there any evidence that gender confounds the PUFA×TNFa eﬀect?

If so why, if not why not?

Estimate Std. Error t value Pr(>|t|)

(Intercept) 0.024467 0.190364 0.129 0.8981

pufa 0.018841 0.009130 2.064 0.0423

TNFa 0.074850 0.033897 2.208 0.0301

age -0.003783 0.002107 -1.795 0.0765

bmi 0.006835 0.004301 1.589 0.1161

pufa:TNFa -0.024496 0.017737 -1.381 0.1712

(d) (2 marks) The researchers decided to use the model with PUFA, TNFa, PUFA×TNFa, age and BMI for inference of PUFA×TNFa interaction. What are the degrees of

freedom for the t-distribution whose critical value we need to construct conﬁdence intervals for the interaction? What upper tail probability should you use for a 95%

conﬁdence interval?

(e) (1 mark) The appropriate critical value is 1.96. Construct a 95% conﬁdence interval for the interaction.

(f) (2 marks) State the null hypothesis and a two-sided alternative hypothesis for testing whether TNFa modiﬁes the eﬀect of PUFA on HDL.

(g) (1 mark) From the computer output, what is the p-value for this test?

(h) (2 marks) For ﬁxed age and BMI, what is the estimated eﬀect of a one-unit change in PUFA for a subject who is A- at TNF-α? For ﬁxed age and BMI, what is the estimated eﬀect of a one-unit change in PUFA for a subject who is A+ at TNF-α.

crimsonyellowjacket200

Refer to the example of the Cobb-Douglas production function discussed in class, where the GDP (Y), Total Employment (X2 ), and Stock of Fixed Capital (X3 ), are measured for the output, labor input and capital input, respectively. The following model is fitted and the R output is below. The data are yearly data from 1955 to 1974.

lnY = B1 +B2lnX2 + B3lnX3 + u

a. Interpret the coefficient estimate of B3 specific to this problem.

b. Find the rate of change of Y with respect to the change of X2 , at Y=218506.3 and X2=11152.6, and X3 = 340072.45.

c. Find the estimate of the return to scale parameter. (3 pts.)

d. Is there strong evidence that the GDP is inelastic relative to the stocks of fixed capital? Justify your answer at a=0.05.

e. Test if there is a positive (first order) autocorrelation in the error term using a=0.05 . Make sure to state the null and alternative hypotheses, to provide the values of the test statistic and the critical value(s), and to draw a conclusion. (6 pts.)

Regression output for Problem 3.

> fit<-lm(log(y) ~ log(x2) + log(x3))

> summary(fit)

Call:

lm(formula = log(y) ~ log(x2) + log(x3))

Residuals:

Min 1Q Median 3Q Max

-0.057142 -0.016250 -0.005793 0.023512 0.038739

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) -1.65233 0.60619 -2.726 0.0144 *

log(x2) 0.33972 0.18569 1.830 0.0849 .

log(x3) 0.84600 0.09335 9.063 6.42e-08 ***

Signif. codes: 0 0.001 0.01 0.05 0.1 1

Residual standard error: 0.02829 on 17 degrees of freedom

Multiple R-squared: 0.9951, Adjusted R-squared: 0.9945

F-statistic: 1719 on 2 and 17 DF, p-value: < 2.2e-16

Durbin-Watson Statistic = 0.425618

sangriakangaroo204

PBHL 5401 Chapter Notes - Chapter 7: Sampling Distribution, Statistical Hypothesis Testing, Type I And Type Ii Errors

Document Summary

Get access

Related Documents

PBHL 5401 Lecture Notes - Lecture 24: Confidence Interval, Point Estimation, Standard Score

PBHL 5401 Lecture Notes - Lecture 21: F-Distribution, Null Hypothesis, Confidence Interval

STAT 0200 Study Guide - Final Guide: Type I And Type Ii Errors, Null Hypothesis, Confidence Interval

Related Questions