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Actinomycin D A biologist hypothesizes that high concentrations of actinomycin D inhibit RNA synthesis in cells and hence the production of proteins as well. An experiment conducted to test this theory compared the RNA synthesis in cells treated with two concentrations of actinomycin D: .6 and .7 microgram per milliliter. Cells treated with the lower concentration (.6) of actinomycin D showed that 55 out of 70 developed normally, whereas only 23 out of 70 appeared to develop normally for the higher concentration (.7). Do these data provide sufficient evidence to indicate a difference between the rates of normal RNA synthesis for cells exposed to the two different concentrations of actinomycin D? a. Find the \(p\) -value for the test. b. If you plan to conduct your test using \(\alpha=.05\), what will be your test conclusions?

Step by step solution

01

Define the hypotheses

We will test the following hypotheses: \(H_0: p_1 - p_2 = 0\) (There is no difference in the rates of normal RNA synthesis for cells exposed to the two different concentrations of actinomycin D) \(H_1: p_1 - p_2 ≠ 0\) (There is a difference in the rates of normal RNA synthesis for cells exposed to the two different concentrations) where \(p_1\) is the proportion of normal RNA synthesis in cells treated with the lower concentration (0.6), and \(p_2\) is the proportion of normal RNA synthesis in cells treated with the higher concentration (0.7).

02

Calculate the sample proportions and their difference

For the lower concentration (0.6): \(n_1 = 70\) (number of trials) \(x_1 = 55\) (number of successes, i.e., cells that developed normally) \(\hat{p_1} = x_1/n_1 = 55/70 = 0.7857\) For the higher concentration (0.7): \(n_2 = 70\) (number of trials) \(x_2 = 23\) (number of successes, i.e., cells that developed normally) \(\hat{p_2} = x_2/n_2 = 23/70 = 0.3286\) Now, calculate the difference in sample proportions: \(\hat{p_1} - \hat{p_2} = 0.7857 - 0.3286 = 0.4571\)

03

Calculate the standard error and test statistic

The standard error for the difference in proportions is given by: \(SE = \sqrt{\frac{\hat{p_1}(1-\hat{p_1})}{n_1} + \frac{\hat{p_2}(1-\hat{p_2})}{n_2}} = \sqrt{\frac{0.7857(1-0.7857)}{70} + \frac{0.3286(1-0.3286)}{70}} = 0.0996\) The test statistic (\(z\)) is calculated as: \(z = \frac{(\hat{p_1} - \hat{p_2}) - 0}{SE} = \frac{0.4571 - 0}{0.0996} = 4.59\)

04

Calculate the p-value

The \(p\)-value for a two-tailed test is: \(p\)-value \(= 2 * P(Z > |z|)\), where \(Z\) follows a standard normal distribution. Using a standard normal table or calculator, we find that the \(p\)-value is close to 0 (\(<0.0001\)).

05

Compare the p-value to the significance level and make conclusions

a. The \(p\)-value we found is close to 0 (\(<0.0001\)). b. To make conclusions based on the provided significance level of \(\alpha = 0.05\), we compare our \(p\)-value with \(\alpha\): Since our \(p\)-value (\(<0.0001\)) is less than \(\alpha = 0.05\), we reject the null hypothesis (\(H_0\)). This means there is sufficient evidence to support the alternative hypothesis (\(H_1\)) that there is a difference in the rates of normal RNA synthesis for cells exposed to the two different concentrations of actinomycin D.

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