HLT1RAE Lecture Notes - Lecture 10: Null Hypothesis, Confidence Interval, Alternative Hypothesis
Document Summary
Get access
Related Documents
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 |
1. You are given only three quarterly seasonal indices and quarterly seasonally adjusted data for the entire year. What is the raw data value for Q4? Raw data is not adjusted for seasonality.
Quarter Seasonal Index Seasonally Adjusted Data
Q1 .80 295
Q2 .85 299
Q3 1.15 270
Q4 --- 271
2. One model of exponential smoothing will provide almost the same forecast as a liner trend method. What are linear trend intercept and slope counterparts for exponential smoothing?
A. Alpha and Delta
B. Delta and Gamma
C. Alpha and Gamma
D. Standard Deviation and Mean
3. When performing correlation analysis what is the null hypothesis? What measure in Minitab is used to test it and to be 95% confident in the significance of correlation coefficient.
A. Ho: r = .05 p < .5
B. Ho: r = 0 p >.05
C. Ho: r ? 0 p?.05
D. Ho: r = 0 p?.05
| In decomposition what does the cycle factor (CF) of .80 represent for a monthly forecast estimate of a Y variable? |
A. The estimated value is 80% of the average monthly seasonal estimate.
B. The estimate is .80 of the forecasted Y trend value.
C. The estimated value is .80 of the historical average CMA values.
D. The estimated value has 20% more variation than the average historical Y data values.
| 5. A Wendy's franchise owner notes that the sales per store has fallen below the stated national Wendy's outlet average of $1,368,000. He asserts a change has occurred that reduced the fast food eating habits of Americans. What is his hypothesis (H1) and what type of test for significance must be applied? |
A. H1: u ? $1,368,000 A one-tailed t-test to the left.
B. H1: u = $1,368,000 A two-tailed t-test.
C. H1: u < $1,368,000 A one-tailed t-test to the left.
D. H1: p < $1,368,000 A one-tailed test to the right
A. The rejection region and the t-table value generally gets smaller for sample size below 31. |
A. Yes. The data are significantly correlated through the 12th lag. C. No. Only the 12 lag period is not correlated. D. You cannot tell since the number of sample observations is not provided. E. The p-value is above .05 so the data is correlated. |
A. Type 2 error |
A. Yes. They move in the same direction as statistical significance. |
A. The weight cannot be calculated since the data observation is not given. |
A. Yes. The correlation coefficient is .873 that is greater than .05. |
A. Yes, since the residuals randomly vary in magnitude. |
A. -101.0 |
|