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brilliants [131]
3 years ago
6

An experimenter has prepared a drug dosage level that she claims will induce sleep for 80% of people suffering from insomnia. Af

ter examining the dosage, we feel that her claims regarding the effectiveness of the dosage are inflated. In an attempt to disprove her claim, we administer her prescribed dosage to 20 insomniacs and we observe Y , the number for whom the drug dose induces sleep. We wish to test the hypothesis H0 : p = 0.8 versus the alternative, Ha : p < 0.8. Assume that the rejection region {y ≤ 12} is used. (a) In terms of this problem, what is a type I error?
Mathematics
1 answer:
Digiron [165]3 years ago
8 0

Answer:

This type of error would lead to the conclusion that the dosage level is less effective than it really is.

Step-by-step explanation:

A Type I error happens when we reject the null hypothesis even when it is true.

In this case, this would happens when the sample taken gives results that end by pure chance in the rejection region, and nont because the null hypothesis is not true.

This type of error would lead to the conclusion that the dosage level is less effective than it really is.

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nadezda [96]

Answer:

a. There is evidence that the population mean is above $300.

b. There is no evidence that the population mean is above $300.

c. There is no evidence that the population mean is above $300.

d. The director could ask for cheaper similar books.

Step-by-step explanation:

Let X be the random variable that represents the cost of textbooks. We have observed n = 25 values, \bar{x} = 315.4 and s = 43.20. We suppose that X is normally distributed.

We have the following null and alternative hypothesis

H_{0}: \mu = 300 vs H_{1}: \mu > 300 (upper-tail alternative)

We will use the test statistic

T = \frac{\bar{X}-300}{S/\sqrt{25}} and the observed value is

t_{0} = \frac{315.4 - 300}{43.20/\sqrt{25}} = 1.7824.

If H_{0} is true, then T has a t distribution with n-1 = 24 degrees of freedom.

a. The rejection region is given by RR = {t | t > t_{0.9}} where t_{0.9} = 1.3178 is the 90th quantile of the t distribution with 24 df, so, RR = {t | t > 1.3178}. Because the observed value satisty 1.7824 > 1.3178, there is evidence that the population mean is above $300.

b. If s = 75, then the observed value is t_{0} = \frac{315.4 - 300}{75/\sqrt{25}} = 1.0267. The rejection region for a 0.05 level of significance is RR = {t | t > t_{0.95}} where t_{0.95} = 1.7108 is the 95th quantile of the t distribution with 24 df, so, RR = {t | t > 1.7108}. Because the observed value does not fall inside the rejection region, there is no evidence that the population mean is above $300.

c. If \bar{x} = 305.11 and s = 43.20, the observed value is t_{0} = \frac{305.11 - 300}{43.20/\sqrt{25}} =  0.5914. For RR = {t | t > 1.3178} we have that the observed value does not fall inside RR, therefore, there is no evidence that the population mean is above $300.

d. Because the director of admissions is concerned about the high cost of textbooks, and there is evidence that the population mean of costs is above $300, the director could ask for cheaper similar books.

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