Question

1/ In clinical trials of 3066 patients taking treatment​ A, 790 reported nausea as a side effect. In clinical trials of 1013 patients taking treatment​ B, 216 reported nausea as a side effect. Researchers want to know whether the difference in sample proportions is statistically significant. Use the​ "Randomization test for two​ proportions" applet in StatCrunch with a seed of 30102. Obtain exactly 3000 runs of the applet. What is the​ P-value based this​ randomization?

​P-value= (Round to four decimal places as​ needed.)

2/ A researcher wishes to estimate the average blood alcohol concentration​ (BAC) for drivers involved in fatal accidents who are found to have positive BAC values. He randomly selects records from 82 such drivers in 2009 and determines the sample mean BAC to be 0.15 g/dL with a standard deviation of 0.080 g/dL.

Determine and interpret a​ 90% confidence interval for the mean BAC in fatal crashes in which the driver had a positive BAC. Select the correct choice below and fill in the answer boxes to complete your choice. (Use ascending order. Round to three decimal places as​ needed.)

The researcher is    ​% confident that the population mean BAC is between and    for drivers involved in fatal accidents who have a positive BAC value.

somebody pleaaaaaaase answer these two questions thanks alot!

Answer 1

Q1) Let P1 be the clinical trials of 3066 patients taking treatment​ A, 790 reported nausea as a side effect. then

$P_{1}=\frac{790}{3066}=0.257$

Let P2 be the clinical trials of 1013 patients taking treatment​ B, 216 reported nausea as a side effect then

$P_{2}=\frac{216}{1013}=0.213$

we will set up null hypothesis

$H_{0}:P_{1}=P_{2}$ Vs $H_{1}:P_{1}\neq P_{2}$

Under null hypothesis the test statistics is

$Z=\frac{P_{1}-P_{2}}{\sqrt{P(1-P)\left ( \frac{1}{n1} +\frac{1}{n2}\right )}}\sim Z(0,1)$

Where $P=\frac{790+216}{3066+1013}=0.246$

$Z=\frac{0.257-0.213}{\sqrt{0.246(1-0.246)\left ( \frac{1}{3066} +\frac{1}{1013}\right )}}=2.94$   

&

Since calculated P.Value is much less then 0.05 hence we reject our null hypothesis.

The​ P-value based this​ randomization using statchrunch

Q2) The Confidence Interval is calculated Via

$C.I=\bar{x}\pm \frac{Z_{\alpha }*\sigma }{\sqrt{n}}$ ,Where $Z_{\alpha}$ is critical value and $\sigma$ is standard deviation. n is sample size.

and the 90% confidence interval is (0.13547, 0.16453) respectively.