Question

A certain drug can be used to reduce the acid produced by the body and heal damage to the esophagus due to acid reflux. The m

A certain drug can be used to reduce the acid produced by the body and heal damage to the esophagus due to acid reflux. The m

A certain drug can be used to reduce the acid produced by the body and heal damage to the esophagus due to acid reflux The ma

A certain drug can be used to reduce the acid produced by the body and heal damage to the esophagus due to acid reflux. The m

A certain drug can be used to reduce the acid produced by the body and heal damage to the esophagus due to acid reflux. The m



help with all parts please

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Answer #1

Because n*po*(1-po) = 234*0.94*0.06 = 13.2 > 10, the sample size is less than 5% of the population size, and the sample can be reasonably assumed to be normal (since, all the assumptions are met), the requirements for testing the hypothesis are satisfied.

Let the proportion of patients that get healed within 8 weeks of taking the drug be P,

Then,

Null hypothesis (Ho) : P \leq 0.94

Alternative hypothesis (H1) : P > 0.94

Test statistic is given by -

z = \frac{p - P_0}{\sqrt{\frac{P_0*(1-P_0)}{n}}}

where, p is the sample proportion = 223/234 = 0.95

n is the sample size = 234

Po is the specified value of Population proportion under the null hypothesis = 0.94

Hence, the value of the test statistic will be -

z = \frac{0.95-0.94}{\sqrt{\frac{0.95*(1-0.95)}{234}}}

= \frac{0.01}{0.014}

= 0.71

P value = P(z > 0.71)

= Area under the z curve on the right side of z = 0.71

= 1 - P(z < 0.71)

The P(z < 0.71) can be obtained from the z table by finding the area corresponding to the z = 0.7 at 0.01 probability and is equal to 0.7611

= 1 - 0.7611

= 0.2389

~ 0.239

Since, P value > level of significance (0.01), we may not reject the null hypothesis.

There is insufficient evidence to conclude that more than 94% of the patients taking drugs are healed within 8 weeks.

So, option (B) is the correct option.

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