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Chapter 9: Problem 45

Treatment versus Control An experiment was conducted to test the effect of a new drug on a viral infection. After the infection was induced in 100 mice, the mice were randomly split into two groups of \(50 .\) The first group, the control group, received no treatment for the infection, and the second group received the drug. After a 30 -day period, the proportions of survivors, \(\hat{p}_{1}\) and \(\hat{p}_{2}\), in the two groups were found to be .36 and \(.60,\) respectively. a. Is there sufficient evidence to indicate that the drug is effective in treating the viral infection? Use \(\alpha=05\) b. Use a \(95 \%\) confidence interval to estimate the actual difference in the survival rates for the treated versus the control groups.

Short Answer

Expert verified
Answer: Yes, there is evidence to support the claim that the new drug is effective in treating the viral infection, as the hypothesis test resulted in rejecting the null hypothesis, and the 95% confidence interval for the difference in survival rates does not contain zero, suggesting a positive difference between the survival rates of the treated and control groups.

Step by step solution

01

State the null and alternative hypothesis

The null hypothesis \(H_{0}\) states that there is no difference between the survival proportions of the control group and the treated group, i.e \(d=0\). Alternative hypothesis \(H_{1}\) says the survival proportions is greater for the treated group than the control group, i.e \(d>0\). Mathematically: \(H_{0}: d=0\) \(H_{1}: d>0\)
02

Calculate the test statistic

The test statistic is given by: \(Z=\frac{(\hat{p}_{2}-\hat{p}_{1})-0}{\sqrt{\frac{\hat{p}_{C}(1-\hat{p}_{C})}{n_{1}}+\frac{\hat{p}_{C}(1-\hat{p}_{C})}{n_{2}}}}\) where \(\hat{p}_{C}\) is the pooled proportion calculated as \(\hat{p}_{C}=\frac{n_{1}\hat{p}_{1}+n_{2}\hat{p}_{2}}{n_{1}+n_{2}}\) We know \(n_{1}=n_{2}=50\). Now, calculate the pooled proportion: \(\hat{p}_{C}=\frac{50(0.36)+50(0.60)}{50+50}=0.48\) Next, calculate the test statistic Z: \(Z=\frac{(0.60-0.36)}{\sqrt{\frac{0.48(1-0.48)}{50}+\frac{0.48(1-0.48)}{50}}}=4.15\)
03

Find the critical value and make a conclusion

Since the significance level \(\alpha=0.05\), and it's a one-tailed test, we find the critical value \(Z_{\alpha}=1.645\). The test statistic \(Z=4.15\) is greater than the critical value \(Z_{\alpha}=1.645\). Therefore, we reject the null hypothesis \(H_{0}\) and conclude that there is sufficient evidence to support the claim that the new drug is effective in treating the viral infection. b. Confidence Interval
04

Calculate the confidence interval

We will now compute the \(95\%\) confidence interval for the difference in survival rates. The confidence interval is given by: \((\hat{p}_{2}-\hat{p}_{1})\pm Z_{\alpha/2}\sqrt{\frac{\hat{p}_{1}(1-\hat{p}_{1})}{n_{1}}+\frac{\hat{p}_{2}(1-\hat{p}_{2})}{n_{2}}}\) Since we are calculating a \(95\%\) confidence interval, the critical value \(Z_{\alpha/2}=1.96\). Plug in the values and calculate the confidence interval: \((0.60-0.36)\pm 1.96\sqrt{\frac{0.36(1-0.36)}{50}+\frac{0.60(1-0.60)}{50}}\) \((0.24)\pm 1.96\sqrt{\frac{0.2304}{50}+\frac{0.24}{50}}\) \((0.24)\pm 1.96\sqrt{0.009408}\) \((0.24)\pm 1.96 \times 0.09701\) \((0.24)\pm 0.1901\) So, the \(95\%\) confidence interval for the difference in survival rates is \((0.0499, 0.4301)\). This interval does not contain zero, suggesting a positive difference between the survival rates of the treated and control groups, further supporting the drug's effectiveness.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Interval
In statistics, a confidence interval gives us a range of values that likely includes an unknown population parameter. In the context of our drug efficacy study, it assesses the difference in survival rates between treated and untreated mice. The confidence interval provides a visual sense of variability, showing how much the observed effect can differ due to random sampling.

To calculate it, we look at the difference between the proportions, \((\hat{p}_{2}-\hat{p}_{1})\), and account for sampling variability. With a 95% confidence interval, we use a critical value of 1.96. This reflects where 95% of data points would lie in a normal distribution.

The computed confidence interval for our study \((0.0499, 0.4301)\) suggests the possible range of difference in survival rates. Importantly, this range does not include zero, indicating there's a significant difference in survival rates between groups. Therefore, it's strong evidence supporting the drug's efficacy.
Statistical Significance
Statistical significance helps determine if the observed effect in an experiment is genuine, or due to chance. In our study, the key question is whether the new drug truly impacts survival rates. We use a hypothesis test to figure this out.

First, we define our hypotheses:
  • Null hypothesis (\(H_0\)): No difference in survival rates (\(d=0\))
  • Alternative hypothesis (\(H_1\)): Survival rate is higher in the treated group (\(d>0\))
We then calculate the test statistic, which indicates how far our sample observation is from the null hypothesis. In this case, a Z-value of 4.15 was computed.

We compare this against a critical value of 1.645 for a significance level (\(\alpha\)) of 0.05. Since 4.15 exceeds 1.645, the results are statistically significant. It means there is enough evidence to reject the null hypothesis, suggesting the drug indeed improves survival rates.
Control and Treatment Groups
In experiments, we often use control and treatment groups to measure the effect of an intervention. Here, the control group consisted of untreated mice, while the treatment group received the drug. This setup helps isolate the drug's effect from other variables.

The primary purpose of having these groups is to make a valid comparison. Without a control group, it would be challenging to assess whether changes in the outcome are due to the treatment or were merely coincidental.

Random assignment of mice to groups ensures any differences are due to chance rather than biased grouping. By comparing outcomes, like survival rates here, we can confidently attribute significant differences to the drug, assuming potential confounding variables are minimal. This method is common in clinical trials and scientific studies to assess new treatments or products.

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Most popular questions from this chapter

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?

Bass Fishing The pH factor is a measure of the acidity or alkalinity of water. A reading of 7.0 is neutral; values in excess of 7.0 indicate alkalinity; those below 7.0 imply acidity. Loren Hill states that the best chance of catching bass occurs when the \(\mathrm{pH}\) of the water is in the range 7.5 to \(7.9 .{ }^{18}\) Suppose you suspect that acid rain is lowering the \(\mathrm{pH}\) of your favorite fishing spot and you wish to determine whether the \(\mathrm{pH}\) is less than 7.5 . a. State the alternative and null hypotheses that you would choose for a statistical test. b. Does the alternative hypothesis in part a imply a one- or a two-tailed test? Explain. c. Suppose that a random sample of 30 water specimens gave \(\mathrm{pH}\) readings with \(\bar{x}=7.3\) and \(s=.2\). Just glancing at the data, do you think that the difference \(\bar{x}-7.5=-.2\) is large enough to indicate that the mean \(\mathrm{pH}\) of the water samples is less than \(7.5 ?\) (Do not conduct the test.) d. Now conduct a statistical test of the hypotheses in part a and state your conclusions. Test using \(\alpha=.05 .\) Compare your statistically based decision with your intuitive decision in part \(\mathrm{c} .\)

A random sample of \(n=35\) observations from a quantitative population produced a mean \(\bar{x}=2.4\) and a standard deviation \(s=.29 .\) Suppose your research objective is to show that the population mean \(\mu\) exceeds 2.3 a. Give the null and alternative hypotheses for the tes b. Locate the rejection region for the test using a \(5 \%\) significance level. c. Find the standard error of the mean. d. Before you conduct the test, use your intuition to decide whether the sample mean \(\bar{x}=2.4\) is likely or unlikely, assuming that \(\mu=2.3 .\) Now conduct the test. Do the data provide sufficient evidence to indicate that \(\mu>2.3 ?\)

Find the \(p\) -value for the following large-sample \(z\) tests: a. A right-tailed test with observed \(z=1.15\) b. A two-tailed test with observed \(z=-2.78\) c. A left-tailed test with observed \(z=-1.81\)

Flextime A company was contemplating the installation of a flextime schedule in which a worker schedules his or her work hours or compresses work weeks. The company estimates that it needs a minimum mean of 7 hours per day per assembly worker in order to operate effectively. Each of a random sample of 80 of the company's assemblers was asked to submit a tentative flextime schedule. If the mean number of hours per day for Monday was 6.7 hours and the standard deviation was 2.7 hours, do the data provide sufficient evidence to indicate that the mean number of hours worked per day on Mondays, for all of the company's assemblers, will be less than 7 hours? Test using \(\alpha=.05\).
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