Just the ones circled in red (C and D) **Analyzing Egg Laying Patterns in Modified Fruit Flies**A biologist is interested in determining whether fruit flies with a specific gene modification lay fewer eggs into rotting fruit compared to the average fruit fly. The standard average number of eggs for fruit flies is known to be 393. The following data represents the number of eggs laid by individual fruit flies with the gene modification: 360, 367, 373, 409, 407, 380, 382, 404, 382, 375, 405, 389, 396, 374, 401.### Statistical AnalysisTo investigate whether the modification results in a lower average, the study utilizes a statistical test with a significance level of $ \alpha = 0.01 $.**a. Choosing a Statistical Test**For this study, a *t-test for a population mean* is appropriate due to the sample size and the assumption of a normal population distribution.**b. Hypotheses Formulation**- **Null Hypothesis ($H_0$)**: The mean number of eggs $(\mu)$ = 393- **Alternative Hypothesis ($H_1$)**: The mean number of eggs $(\mu)$ < 393**c. Test Statistic Calculation**The test statistic is calculated using the sample data provided. It should be shown to three decimal places.**d. P-value Calculation**The p-value associated with the test statistic is calculated to determine if the results are statistically significant. Show the p-value to four decimal places.### ConclusionThe conclusion about whether the gene modification results in a statistically significant decrease in the number of eggs laid can be determined by comparing the p-value to the significance level. If the p-value is less than $ \alpha = 0.01 $, we reject the null hypothesis in favor of the alternative.

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The average fruit fly will lay 393 eggs into rotting fruit. A biologist wants to see if the average will be fewer for flies that have a certain gene modified. The data below shows the number of eggs that were laid into rotting fruit by several fruit flies that had this gene modified. Assume that the distribution of the population is normal. 360, 367, 373, 409, 407, 380, 382, 404, 382, 375, 405, 389, 396, 374, 401 What can be concluded at the the a = 0.01 level of significance level of significance? a. For this study, we should use t-test for a population mean b. The null and alternative hypotheses would be: Но: 393 H1: 393 C. The test statistic |t (please show your answer to 3 decimal places.) d. The p-value = (Please show your answer to 4 decimal places.)

The average fruit fly will lay 393 eggs into rotting fruit. A biologist wants to see if the average will be fewer for flies that have a certain gene modified. The data below shows the number of eggs that were laid into rotting fruit by several fruit flies that had this gene modified. Assume that the distribution of the population is normal. 360, 367, 373, 409, 407, 380, 382, 404, 382, 375, 405, 389, 396, 374, 401 What can be concluded at the the a = 0.01 level of significance level of significance? a. For this study, we should use t-test for a population mean b. The null and alternative hypotheses would be: Но: 393 H1: 393 C. The test statistic |t (please show your answer to 3 decimal places.) d. The p-value = (Please show your answer to 4 decimal places.)

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

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Just the ones circled in red (C and D)

**Analyzing Egg Laying Patterns in Modified Fruit Flies**

A biologist is interested in determining whether fruit flies with a specific gene modification lay fewer eggs into rotting fruit compared to the average fruit fly. The standard average number of eggs for fruit flies is known to be 393.

The following data represents the number of eggs laid by individual fruit flies with the gene modification:
360, 367, 373, 409, 407, 380, 382, 404, 382, 375, 405, 389, 396, 374, 401.

### Statistical Analysis

To investigate whether the modification results in a lower average, the study utilizes a statistical test with a significance level of $ \alpha = 0.01 $.

**a. Choosing a Statistical Test**
For this study, a *t-test for a population mean* is appropriate due to the sample size and the assumption of a normal population distribution.

**b. Hypotheses Formulation**
- **Null Hypothesis ($H_0$)**: The mean number of eggs $(\mu)$ = 393
- **Alternative Hypothesis ($H_1$)**: The mean number of eggs $(\mu)$ < 393

**c. Test Statistic Calculation**
The test statistic is calculated using the sample data provided. It should be shown to three decimal places.

**d. P-value Calculation**
The p-value associated with the test statistic is calculated to determine if the results are statistically significant. Show the p-value to four decimal places.

### Conclusion
The conclusion about whether the gene modification results in a statistically significant decrease in the number of eggs laid can be determined by comparing the p-value to the significance level. If the p-value is less than $ \alpha = 0.01 $, we reject the null hypothesis in favor of the alternative.

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