Refer to Exercise 9.17. Explain what each of the following would mean.

(a) Type I error.

(b) Type II error.

(c) Correct decision.

Now suppose that the results of carrying out the hypothesis test lead to the rejection of the null hypothesis. Classify that conclusion by error type or as a correct decision if in fact the mean iron intake of all adult females under the age of 51 years.

(d) equals the RDA of 18 mg per day.

(e) is less than the RDA of 18 mg per day.

Short Answer

Expert verified

(a) Rejecting a Null Hypothesis, when it is true.

(b) Rejecting a Null Hypothesis, when H0is false.

(c) If the true null hypothesis is not rejected or a false null hypothesis is rejected.

(d) Type I error.

(e) Correct Decision.

Step by step solution

01

Step 1. Given Information.

The Null Hypothesis is,

H0:μ=18mg.

The Alternative Hypothesis is,

H0:μ<18mg.

02

Part (a). Type I Error.

According to the definition of the type I error it is to reject a null hypothesis when it is true. A type I error would occur in fact H0:μ=18mgtrue, that is the RDA of iron for adult females of age 51is 18mg but the result of the sampling lead to conclude that the mean RDA of females of age 51is less than 18mg.

03

Part (b). Type II error.

According to the definition of the type II error, it is to not reject a null hypothesis when it H0is false. A type II error would occur in fact localid="1651257139706" μ=18mgis not to be rejected, but the results of the sampling fall to lead to conclude that the mean RDA of females of age51is less than18mg.18

04

Part (c). Correct Decision.

A correct decision would occur if the true null hypothesis is not rejected or a false null hypothesis is rejected. Here, in the fact the mean RDA of adult females is μ=18mgand the results of the sampling do not lead to rejection, so is a correct decision; or the mean RDA of adult males isμ<18mg and the results of the sampling lead to that conclusion.

05

Part (d). Equals the RDA of 18 mg per day.

Here, the mean iron intake of adult females of 51years old is 18mgper day, and the results of a hypothesis test lead to rejection of the null hypothesis.

We are rejecting the true null hypothesis of μ=18mg, where it is also obtained as a sampling result. So, we are committing a Type I error.

06

Part (e). Is less than the RDA of 18mg per day.

Here, the mean iron intake of adult females of 51years old is less than18mgper day, and the results of a hypothesis test lead to rejection of the null hypothesis.

As a sampling result, we obtain the mean RDA of adult females of 51years old is less than 18mgper day, and we are rejecting the null hypothesis that the mean of adult females of 51years old is less than 18mgper day. Therefore, our decision is a correct decision.

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Most popular questions from this chapter

As we mentioned on page 378, the following relationship holds between hypothesis test and confidence intervals for one-mean z-procedures: For a two-tailed hypothesis test at the significance level α, the null hypothesis role="math" localid="1653038937481" H0:μ=μ0will be rejected in favor of the alternative hypothesis Ha:μμ0if and only if μ0lies outside the 1-α-level confidence interval for μ. In each case, illustrate the preceding relationship by obtaining the appropriate one-mean z-interval and comparing the result to the conclusion of the hypothesis test in the specified exercise.

Part (a): Exercise 9.84

Part (b): Exercise9.87

According to the Bureau of Crime Statistics and Research of Australia, as reported on Lawlink, the mean length of imprisonment for motor-vehicle-theft offenders in Australia is 16.7 months. One hundred randomly selected motor-vehicle-theft of-fenders in Sydney, Australia, had a mean length of imprisonment of 17.8 months. At the 5% significance level, do the data provide sufficient evidence to conclude that the mean length of imprisonment for motor-vehicle-theft offenders in Sydney differs from the national mean in Australia? Assume that the population standard deviation of the lengths of imprisonment for motor-vehicle-theft offenders in Sydney is6months.

9.128 Two-Tailed Hypothesis Tests and CIs. The following relationship holds between hypothesis tests and confidence intervals for one-mean t-procedures: For a two-tailed hypothesis test at the significance level α, the null hypothesis H0:μ=μ0will be rejected in favor of the alternative hypothesis H2:μ>μ0if and only if μ0lies outside the (1-α)-level confidence interval for μ. In each case, illustrate the preceding relationship by obtaining the appropriate one-mean t-interval (Procedure 8.2 on page 338 ) and comparing the result to the conclusion of the hypothesis test in the specified exercise.
a. Exercise 9.113
b. Exercise 9.116

Cadmium, a heavy metal, is toxic to animals. Mushrooms, however are able to absorb and accumulate cadmium at high concentrations. The Czech and Slovak governments have set a safety limit for cadmium in dry vegetables at \(0.5\) part per million M. Melgar et al. measured the cadmium levels in a random sample of the edible mushroom Boletus pinicola and published the results in the paper "Influence of Some Factors in Toxicity and Accumulation of Cd from Edible Wild Macrofungi in NW Spain. A hypothesis test is to be performed to decide whether the mean cadmium level in Boletus pinicola mushrooms is greater than the government's recommended limit.

a. determine the null hypothesis

b. determine the alternative hypothesis

c. classify the hypothesis test as two tailed, left tailed or right tailed.

Refer to Exercise 9.19. Explain what each of the following would mean.

(a) Type I error.

(b) Type II error.

(c) Correct decision.

Now suppose that the results of carrying out the hypothesis test lead to the rejection of the null hypothesis. Classify that conclusion by error type or as a correct decision if in fact the mean length of imprisonment for motor-vehicle-theft offenders in Sydney.

(d) equals the national mean of 16.7 months.

(e) differs from the national mean of 16.7 months.

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