Answer to Question #98501 in Statistics and Probability for Juliet Beglaryan

Question

4. A drug manufacturer claims that fewer than 10% of patients who take its new drug for treating Alzheimer’s disease will experience nausea. In a random sample of 250 patients, 23 experienced nausea. Perform a significance test at the 5% significance level to test this claim.
5. A manufacturer of sprinkler systems designed for fire protection claims that the average activating temperature is at least 135. To test this claim, you randomly select a sample of 32 systems and find the °Fmean activation temperature to be 133◦F with a standard deviation of 3.3 ◦F. At α = 0.10, do you have °Fenough evidence to reject the manufacturer’s claim? Let a= .0.05

Answer 1

1. Given that

$$p=10\%=0.1,x=23,n=250$$

Determine the hypotheses

$H_0:p=10\%=0.1$

$H_1:p<0.1$

Is the sample large enough?

$$np=250(0.1)=25\geq15$$ $$n(1-p)=250(1-0.1)=225\geq15$$

Test statistic

The sample proportion is the number of successes divided by the sample size

$$\hat{x}={x \over n}={23\over 250}=0.092$$

Determine the value of the z-statistic:

$$z={\hat{p}-p \over \sqrt{{p(1-p) \over n}}}$$ $$z={0.092-0.1 \over \sqrt{{0.1(1-0.1) \over 250}}}\approx-0.422$$

The significance level is $\alpha=0.05,$ and the critical value for a left-tailed test is $z_c=-1.64$

The rejection region for this left-tailed test is $R=\{{z:z<-1.64}\}$

Since it is observed that $z=-0.422\geq-1.64=z_c,$ it is then concluded that the null hypothesis is not rejected.

Determine the $p-$ value

$$p=P(Z<-0.421637)\approx0.336645$$

Since $p=0.336645\geq0.05,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not sufficient evidence to indicate that fewer than 10% of patients who take its new drug for treating for treating Alzheimer’s disease will experience nausea, at the 0.05 significance level.

5. The provided sample mean is $\bar{X}=133$ and the known population standard deviation is $\sigma=3.3,$ and the sample size

is $n=32.$

The following null and alternative hypotheses need to be tested:

$H_0: \mu\geq135$

$H_1:\mu<135$

This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha=0.1,$ and the critical value for a left-tailed test is

$z_c=-1.28$

The rejection region for this left-tailed test is $R=\{{z:z<-1.28}\}$

The z-statistic is computed as follows:

$$z={\bar{X}-\mu \over \sigma/\sqrt{n}}$$

$$z={133-135 \over 3.3/\sqrt{32}}\approx-3.4284$$

Since it is observed that $z=-3.4284<-1.28=z_c,$ it is then concluded that the null hypothesis is rejected.

The p-value is

$$p=P(Z<-3.4284)\approx0.0003$$

Since $p=0.0003<0.1,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is less than 135, at the 0.1 significance level.

There is enough evidence to reject the manufacturer’s claim, at the 0.1 significance level.

Let the significance level is $\alpha=0.05$ The critical value for a left-tailed test is $z_c=-1.64$

The rejection region for this left-tailed test is $R=\{{z:z<-1.64}\}$

The z-statistic is $z=-3.4284$

Since it is observed that $z=-3.4284<-1.64=z_c,$ it is then concluded that the null hypothesis is rejected.

The p-value is

$$p=P(Z<-3.4284)\approx0.0003$$

Since $p=0.0003<0.05,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$ is less than 135, at the 0.05 significance level.

There is enough evidence to reject the manufacturer’s claim, at the 0.05 significance level.