The average fruit fly will lay 415 eggs into rotting fruit. A biologist wants to see if the average will change 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. 435, 412, 442, 414, 412, 446, 441, 421, 441, 403, 439, 428 What can be concluded at the the αα = 0.05 level of significance level of significance? For this study, we should use Select an answer t-test for a population mean z-test for a population proportion The null and alternative hypotheses would be: H0:H0: ? p μ Select an answer < = > ≠ H1:H1: ? p μ Select an answer < > = ≠ The test statistic ? z t = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is ? ≤ > αα Based on this, we should Select an answer fail to reject accept reject the null hypothesis. Thus, the final conclusion is that ... The data suggest the population mean is not significantly different from 415 at αα = 0.05, so there is sufficient evidence to conclude that the population mean number of eggs that fruit flies with this gene modified will lay in rotting fruit is equal to 415. The data suggest the populaton mean is significantly different from 415 at αα = 0.05, so there is sufficient evidence to conclude that the population mean number of eggs that fruit flies with this gene modified will lay in rotting fruit is different from 415. The data suggest that the population mean number of eggs that fruit flies with this gene modified will lay in rotting fruit is not significantly different from 415 at αα = 0.05, so there is insufficient evidence to conclude that the population mean number of eggs that fruit flies with this gene modified will lay in rotting fruit is different from 415.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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The average fruit fly will lay 415 eggs into rotting fruit. A biologist wants to see if the average will change 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.

435, 412, 442, 414, 412, 446, 441, 421, 441, 403, 439, 428

What can be concluded at the the αα = 0.05 level of significance level of significance?

For this study, we should use Select an answer t-test for a population mean z-test for a population proportion
The null and alternative hypotheses would be:

H0:H0: ? p μ Select an answer < = > ≠

H1:H1: ? p μ Select an answer < > = ≠

The test statistic ? z t = (please show your answer to 3 decimal places.)
The p-value = (Please show your answer to 4 decimal places.)
The p-value is ? ≤ > αα
Based on this, we should Select an answer fail to reject accept reject the null hypothesis.
Thus, the final conclusion is that ...
- The data suggest the population mean is not significantly different from 415 at αα = 0.05, so there is sufficient evidence to conclude that the population mean number of eggs that fruit flies with this gene modified will lay in rotting fruit is equal to 415.
- The data suggest the populaton mean is significantly different from 415 at αα = 0.05, so there is sufficient evidence to conclude that the population mean number of eggs that fruit flies with this gene modified will lay in rotting fruit is different from 415.
- The data suggest that the population mean number of eggs that fruit flies with this gene modified will lay in rotting fruit is not significantly different from 415 at αα = 0.05, so there is insufficient evidence to conclude that the population mean number of eggs that fruit flies with this gene modified will lay in rotting fruit is different from 415.

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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