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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?

Short Answer

Expert verified

From the MINITAB output, the P-value is 0.000

Step by step solution

01

Given Information

Check whether the data provide sufficient evidence to conclude that on average, women with peripheral arterial disease have an unhealthy ABI.

State the null and alternative hypothesis:

Null hypothesis:

Ho:μ=0.9

That is, the data does not provide sufficient evidence to conclude that on average, women with peripheral arterial disease have an unhealthy ABI.

Alternative hypothesis:

Ho:μ<0.9

That is, the data provide sufficient evidence to conclude that on average, women with peripheral arterial disease have an unhealthy ABI

Decide a significance level

Here, the significance level is,α=0.01

02

Explanation

Compute the value of the test statistic and P-value by using MINITAB.

MINITAB procedure:

Step 1: Choose Stat > Basic Statistics > 1-Sample t.

Step 2: In Summarized data, enter the sample size 187 and mean 0.64.

Step 3: In Standard deviation, enter a value 0.15.

Step 4: In Perform hypothesis test, enter the test mean as 0.9

Step 5: Check Options, enter Confidence level as99

Step 6: Choose less than in alternative.

Step 7: Click OK in all dialogue boxes.

MINITAB output:

One-Sample T

Test of mu=0.9vs<0.9

99% Upper
NMeanStDevSE MeanBoundTP
1870.64000.15000.01100.6657-23.700.000

From the MINITAB output,

The value of test statistic is -23.70

The P-value is0.000

03

Conclusion

If P≤α, then reject the null hypothesis.

Here, the P-value is 0.000 which is less than the level of significance. That is,

P(=0.000)<α(=0.01).

Therefore, the null hypothesis is rejected at 1%level.

Thus, it can be conclude that the test results are statistically significant at 1%level of significance.

Interpretation:

Thus, the data provide sufficient evidence to conclude that on average, women with peripheral arterial disease have an unhealthy ABI

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Most popular questions from this chapter

In each of Exercises 9.41-9.46 ,determine the critical values for a one-mean z-test. For each exercise, draw a graph that illustrates your answer

A left-tailed test withα=0.05

In Exercise 8.146 on page 345,we introduced one-sided one mean-t-intervals. The following relationship holds between hypothesis test and confidence intervals for one-mean t-procedures: For a right-tailed hypothesis test at the significance level α, the null hypothesis H0:μ=μ0will be rejected in favor of the alternative hypothesis data-custom-editor="chemistry" Ha:μ>μ0if and only if μ0is less than or equal to the 1-α-level lower confidence bound for μ. In each case, illustrate the preceding relationship by obtaining the appropriate lower confidence bound and comparing the result to the conclusion of the hypothesis test in the specified exercise.

Part (a): Exercise 9.114 (both parts)

Part (b) Exercise9.115

Refer to Exercise 9.16. 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 lactation period of grey seals.

(d) equals 23 days.

(e) differs from 23 days.

9.95 Two-Tailed Hypothesis Tests and CIs. As we mentioned on page 378 , the following relationship holds between hypothesis tests and confidence intervals for one-mean z-procedures: For a two-tailed hypothesis test at the significance level α, the null hypothesis H0:μ=μ0 will be rejected in favor of the alternative hypothesis Ha:μ≠μ0if and only if μ0 lies outside the (1-α)-level confidence interval for μ. In each case, illustrate the preceding relationship by obtaining the appropriate one-mean z-interval (Procedure 8.1 on page 322 ) and comparing the result to the conclusion of the hypothesis test in the specified exercise.
a. Exercise 9.84
b. Exercise 9.87

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 Carolina Statistical Abstract, sponsored by the South Carolina Budget and Control Board.

Part (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.

Part (b): Obtain a normal probability plot, boxplot, histogram and stem-and-leaf diagram of the data.

Part (c): Remove the outliers (if any) from the data and then repeat part (a).

Part (d): Comment on the advisability of using thez-test here.

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