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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

We have been provided a sample mean, sample size, and population standard deviation. In the given case, use the one-mean z-test to perform the required hypothesis test at the 5% significance level.

x¯=23,n=24,σ=4,H0:μ=22,Ha:μ22

The normal probability curve and stem-to-leaf diagram of the data are shown in figure; σis known.

Perform Hypothesis test for mean of the population from which data is obtained and decide whether to use z-test, t-test or neither. Explain your answer.

9.90 Hotels and Motels. The daily charges, in dollars, for a sample of 15 hotels and motels operating in South Carolina are provided on the WeissStats site. The data were found in the report South Caroline Statistical Abstract, sponsored by the South Carolina Budget and Control Board.
a. Use the one-mean z-test to decide, at the 5%significance level, whether the data provide sufficient evidence to conclude that the mean daily charge for hotels and motels operating in South Carolina is less than \(75. Assume a population standard deviation of \)22.40.

b. Obtain a normal probability plot, boxplot, histogram, and stem-and-leaf diagram of the data.
c. Remove the outliers (if any) from the data and then repeat part (a).
d. Comment on the advisability of using the z-test here.

Ankle Brachial Index. The ankle brachial index (ABI) compares the blood pressure of a patient's arm to the blood pressure of the patient's leg. The ABI can be an indicator of different diseases, including arterial diseases. A healthy (or normal) ABI is 0.9 or greater. In a study by M. McDermott et al. titled "Sex Differences in Peripheral Arterial Disease: Leg Symptoms and Physical Functioning" (Journal of the American Geriatrics Society, Vol. 51, No. 2, Pp. 222-228), the researchers obtained the ABI of 187 women with peripheral arterial disease. The results were a mean ABI of 0.64 with a standard deviation of 0.15 At the 1 % significance level, do the data provide sufficient evidence to conclude that, on average, women with peripheral arterial disease have an unhealthy ABI?

9.93 Cell Phones. The number of cell phone users has increased dramatically since 1987. According to the Semi-annual Wireless Survey, published by the Cellular Telecommunications & Internet Association, the mean local monthly bill for cell phone users in the United States was \(48.16in 2009 . Last year's local monthly bills, in dollars, for a random sample of 75 cell phone users are given on the WeissStats site. Use the technology of your choice to do the following.
a. Obtain a normal probability plot, boxplot, histogram, and stemand-leaf diagram of the data.
b. At the 5%significance level, do the data provide sufficient evidence to conclude that last year's mean local monthly bill for cell phone users decreased from the 2009 mean of \)48.16?Assume that the population standard deviation of last year's local monthly bills for cell phone users is $25.
c. Remove the two outliers from the data and repeat parts (a) and (b).
d. State your conclusions regarding the hypothesis test.

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