Early-Onset Dementia. Dementia is the loss of the in tual and social abilities severe enough to interfere with judí behavior, and daily functioning. Alzheimer's disease is the moon type of dementia. In the article "Living with Early Onsmentia: Exploring the Experience and Developing Evidence Guidelines for Practice" (Alzheimer's Care Quarterly, Vol. 5, I pp. 111-122), P. Harris and J. Keady explored the experience struggles of people diagnosed with dementia and their family simple random sample of 21 people with early-onset dementia the following data on age at diagnosis, in years.

60

58

52

58

59

58

51

61

54

59

55

53

44

46

47

42

56

57

49

41

43

At the 1% significance level, do the data provide sufficient ev| to conclude that the mean age at diagnosis of all people with onset dementia is less than 55 years old? Assume that the point standard deviation is 6.8 years. (Note: μ=52.5 years.)

Short Answer

Expert verified

The presented data is adequate evidence to infer that the mean age of diagnosis of all patients with early-onset dementia is 55 years old at the 1% significance level.

Step by step solution

01

Step 1. Given information.

The given table is

60

58

52

58

59

58

51

61

54

59

55

53

44

46

47

42

56

57

49

41

43

02

Step 2. Let the mean age at diagnosis of all people with early-onset dementia be μ ,

Population standard deviation is,

σ=6.8years

Now, the hypotheses test is,

H0:μ=55yearsHa:μ<55yearslocalid="1652289900230" H0:μ=55yearsHa:μ<55years

We must conduct the test at a 1% level of significance, i.e.,α=0.01

The sample size waslocalid="1652289909007" n=21

The sample mean is, x=52.5years

03

Step 3. Now Test statistic, 

Teststatistic:z=x-𝜇0σn=52.5-556.821=-1.68

04

Step 4. Solution

Because the test is a two-tailed test, α=0.01the critical value is-Za=-Z0.01, localid="1652289915878" -Za=-Z0.01.

Here, the region is,

localid="1652289920353" z<-z0.01z<-z0.01i.e.,

z<-2.33

05

Step 5. Here,

z=1.68>z0.01=2.33

Because we do not reject at a 1% threshold of significance as the test statistic's value.

Furthermore, z does not lie within the rejection zone.

The presented data is adequate evidence to infer that the mean age of diagnosis of all patients with early-onset dementia is 55 years old at the 1% significance level.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The Ericsson method is one of several methods claimed to increase the likelihood of a baby girl. In a clinical trial, results could be analysed with a formal hypothesis test with the alternative hypothesis of p>0.5, which corresponds to the claim that the method increases the likelihood of having a girl, so that the proportion of girls is greater than 0.5. If you have an interest in establishing the success of the method, which of the following P-values would you prefer: 0.999, 0.5, 0.95, 0.05, 0.01, and 0.001? Why?

In Exercises 1–4, use these results from a USA Today survey in which 510 people chose to respond to this question that was posted on the USA Today website: “Should Americans replace passwords with biometric security (fingerprints, etc)?” Among the respondents, 53% said “yes.” We want to test the claim that more than half of the population believes that passwords should be replaced with biometric security.

Requirements and Conclusions

a. Are any of the three requirements violated? Can the methods of this section be used to test the claim?

b. It was stated that we can easily remember how to interpret P-values with this: “If the P is low, the null must go.” What does this mean?

c. Another memory trick commonly used is this: “If the P is high, the null will fly.” Given that a hypothesis test never results in a conclusion of proving or supporting a null hypothesis, how is this memory trick misleading?

d. Common significance levels are 0.01 and 0.05. Why would it be unwise to use a significance level with a number like 0.0483?

Interpreting Power For the sample data in Example 1 “Adult Sleep” from this section, Minitab and StatCrunch show that the hypothesis test has power of 0.4943 of supporting the claim that μ<7 hours of sleep when the actual population mean is 6.0 hours of sleep. Interpret this value of the power, then identify the value of βand interpret that value. (For the t test in this section, a “noncentrality parameter” makes calculations of power much more complicated than the process described in Section 8-1, so software is recommended for power calculations.)

Critical Values. In Exercises 21–24, refer to the information in the given exercise and do the following.

a. Find the critical value(s).

b. Using a significance level of α= 0.05, should we reject H0or should we fail to reject H0?

Exercise 19

Estimates and Hypothesis Tests Data Set 3 “Body Temperatures” in Appendix B includes sample body temperatures. We could use methods of Chapter 7 for making an estimate, or we could use those values to test the common belief that the mean body temperature is 98.6°F. What is the difference between estimating and hypothesis testing?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free