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A better drug? In a pilot study, a company's new cholesterol-reducing drug outperforms the currently available drug. If the data provide convincing

evidence that the mean cholesterol reduction with the new drug is more than 10 milligrams per deciliter of blood (mg/dl) greater than with the current drug, the company will begin the expensive process of mass-producing the new drug. For the 14 subjects who were assigned at random to the current drug, the mean cholesterol reduction was $54.1\mathrm{mg}/\mathrm{dl}$with a standard deviation of $11.93mg/dl.$For the 15 subjects who were randomly assigned to the new drug, the mean cholesterol reduction was $68.7\mathrm{mg}/\mathrm{dl}$with a standard deviation of$13.3mg/dl.$Graphs of the data reveal no outliers or strong skewness.

(a) Carry out an appropriate significance test. What conclusion would you draw? (Note that the null hypothesis is not$\left.H_{0:{渭}_1-{渭}_2}=0_{-}\right)$

(b) Based on your conclusion in part (a), could you have made a Type I error or a Type Il error? Justify your answer.

Step by step solution

01

Part (a) Step 1: Given Information

${\stackrel{炉}{x}}_1=68.7,{\stackrel{炉}{x}}_2=54.1$

$s_1=13.3,s_2=11.96$

$n_1=15,n_2=14$

02

Part (a) Step 2: Explanation

Test statistic formula is:

$t=\frac{{\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2}{\sqrt{\frac{s_1^2}{{}^{n_1}}+\frac{x_2^2}{n_2}}}$

The null and alternative hypotheses for the provided case are:

$H_0:{渭}_1-{渭}_2=10$

$H_1:{渭}_1-{渭}_2>10$

The test statistic is computed as:

$t=\frac{{\stackrel{炉}{x}}_1-{\stackrel{炉}{x}}_2-\left({渭}_1-{渭}_2\right)}{\sqrt{\frac{s_1^2}{{}^{n}1}+\frac{s_2^2}{n_2}}}$

$=\frac{68.7-54.1-\left(10\right)}{\sqrt{\frac{13.3^2}{15}+\frac{11.{93}^2}{14}}}$

$=0.982$

The degree of freedom is calculated as:

$df=\min \left(n_1-1,n_2-1\right)=\min (15-1,14-1)=3$

The p-value:

$P-\text{value}=0.828$

In this case,

$P\text{- value}=0.828>0.05$

The null hypothesis could not be rejected which is not showing sufficient evidence for the claim at a significant level of $5\backslash \%.$

03

Part(b) Step 1: Given Information

To determine the error that is committed using the result of the above part.

04

Part (b) Step 2: Explanation

From the above part, the null hypothesis has not been rejected. Thus, there is a possibility of committing the Type Il error.

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