Questions: The average fruit fly will lay 405 eggs into rotting fruit. A biologist wants to see if the average will be higher 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. 422,404,409,418,402,392,418,421,388,405,415,425,389,420,421 What can be concluded at the 0.10 level of significance level of significance?

Solution Steps

The sample mean \( \bar{x} \) is calculated as follows: \[ \bar{x} = \frac{\sum_{i=1}^N x_i}{N} = \frac{6149}{15} = 409.933 \]

The sample standard deviation \( s \) is computed using the formula: \[ s = \sqrt{\frac{\sum_{i=1}^N (x_i - \bar{x})^2}{N-1}} \approx 12.646 \]

The standard error \( SE \) is given by: \[ SE = \frac{s}{\sqrt{n}} = \frac{12.646}{\sqrt{15}} \approx 3.265 \]

The test statistic \( t \) for a right-tailed test is calculated as: \[ t = \frac{\bar{x} - \mu_0}{SE} = \frac{409.933 - 405}{3.265} \approx 1.511 \]

For a right-tailed test, the P-value is determined as: \[ P = 1 - T(t) \approx 0.077 \]

Since the P-value \( 0.077 \) is less than the significance level \( \alpha = 0.10 \), we reject the null hypothesis.

Final Answer