Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Sanitarium administration of malaria cases. One of the most sedate health challenges in India is malaria. Accordingly, the Indian sanitarium director's must-have—the coffers to treat the high volume of admitted malaria cases. A study published in the National Journal of Community Medicine (Vol. 1, 2010) delved into whether the malaria admission rate is more advanced in months than in others. In a sample of 192 sanitarium cases admitted in January, 32 were treated for malaria.

In an independent sample of 403 cases admitted in May (4 months latterly), 34 were treated for malaria.

a. Describe the two populations of stake in this study.

b. Give a point estimate of the contrast in the malaria admission rates in January and May.

c. Find a 90% confidence interval for the contrast in the malaria admission rates in January and May.

d. Based on the interval, part c, can you conclude that contrast exists in the authentic malaria admission rates in January and May? Simplify.

Short Answer

Expert verified

Contrasts exist in the actual malaria admission rate in January and May.

Step by step solution

01

Step-by-Step SolutionStep 1: (a) State the two populations of interest

In the study, the first population of interest is all patients admitted in January, and the second population of interest is all patients admitted in May.

02

(b) Find the point estimate for the contrast in the malaria admission rates in January and May

Consider x1 = 32 and n1= 192.

The point estimate for the malaria admission rates in January is,

P_1=x1n1=3292

= 0.167

Consider x2 = 34 and n2 = 403.

The point estimate for the malaria admission rates in May is,

P_2=x2n2=34403=0.084

03

(c) The point estimate for the contrast in the malaria admission rates in January and May is

P1-P2=0.167-.084=0.083

Thus, the point estimate for the contrast in the malaria admission rates in January and May is 0.083.

04

(d) 90% confidence interval for the contrast in the malaria admission rates in January and May

The critical value for a two-tailed test is obtained below:

Here, the test is two-tailed, and the significance level is α=0.10.

The rejection region for the two-tailed test is|z|>za2.

The confidence coefficient is 0.90.

So,

(1-α) = 0.90

α =0.10

α2=0.05

From Appendix D, Table II, the critical value for the two-tailed test with α = 0.10 is za2(0.05)=±1.645 Hence, the rejection region is |z|> 1.645.

05

90% confidence interval

The 90% confidence interval is obtained below:

(P_1p_2)±z0.05P_1(1P_1)n1+P_2(1p2)n2=0.083±1.6450.167(10.167)192+0.084(10.084)403

=0.083±1.645(0.0303)=0.083±0.050=(0.033,0.133)

90% confidence interval for the contrast in the malaria admission rates in January and May is (0.033.0.133).

06

Conclusions

Yes, it can be concluded that the contrast exists in the authentic malaria admission rates in January and May.

Explanation

The 9 z- confidence interval for (p1 – p2) is (0.033.0.133), which doesn't contain the hypothecated value 0.

That is, the hypothecated value p0 = 0 falsehoods outside the interval (0.033,0.133)

So, by the condition, if the hypothecated value (p0) lies outside the corresponding 100 (1-α) Z- confidence interval for (P1 – P2), also reject the null hypothesis.

Therefore, it can be concluded that reject the null thesis H0 at α = -0.05.

07

Final answer

Hence, the contrast exists in the authentic malaria admission rates in January and May.

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

A random sample of n observations is selected from a normal population to test the null hypothesis that σ2=25. Specify the rejection region for each of the following combinations of Ha,αand n.

a.Ha:σ225;α=0.5;n=16

b.Ha:σ2>25;α=.10;n=15

c.Ha:σ2>25;α=.01;n=23

d. Ha:σ2<25;α=.01;n=13

e. Ha:σ225;α=.10;n=7

f. Ha:σ2<25;α=.05;n=25

Bankruptcy effect on U.S. airfares. Both Delta Airlines and USAir filed for bankruptcy. A study of the impact of bankruptcy on the fares charged by U.S. airlines was published in Research in Applied Economics (Vol. 2, 2010). The researchers collected data on Orlando-bound airfares for three airlines—Southwest (a stable airline), Delta (just entering bankruptcy at the time), and USAir (emerging from bankruptcy). A large sample of nonrefundable ticket prices was obtained for each airline following USAir’s emergence from bankruptcy, and then a 95% confidence interval for the true mean airfare was obtained for each. The results for 7-day advance bookings are shown in the accompanying table.

a. What confidence coefficient was used to generate the confidence intervals?

b. Give a practical interpretation of each of the 95% confidence intervals. Use the phrase “95% confident” in your answer.

c. When you say you are “95% confident,” what do you mean?

d. If you want to reduce the width of each confidence interval, should you use a smaller or larger confidence coefficient?

4.134 Refer to Exercise 4.133. Find the following probabilities:

a.P(20x30)b.P(20<x30)c.P(x30)d.P(x45)e.(x40)f.(x<40)g.P(15x35)h.P(21.5x31.5)

The gender diversity of a large corporation’s board of directors was studied in Accounting & Finance (December 2015). In particular, the researchers wanted to know whether firms with a nominating committee would appoint more female directors than firms without a nominating committee. One of the key variables measured at each corporation was the percentage of female board directors. In a sample of 491firms with a nominating committee, the mean percentage was 7.5%; in an independent sample of 501firms without a nominating committee, the mean percentage was role="math" localid="1652702402701" 4.3% .

a. To answer the research question, the researchers compared the mean percentage of female board directors at firms with a nominating committee with the corresponding percentage at firms without a nominating committee using an independent samples test. Set up the null and alternative hypotheses for this test.

b. The test statistic was reported as z=5.1 with a corresponding p-value of 0.0001. Interpret this result if α=0.05.

c. Do the population percentages for each type of firm need to be normally distributed for the inference, part b, to be valid? Why or why not?

d. To assess the practical significance of the test, part b, construct a 95% confidence interval for the difference between the true mean percentages at firms with and without a nominating committee. Interpret the result.

Question: Independent random samples n1 =233 and n2=312 are selected from two populations and used to test the hypothesis Ha:(μ1-μ)2=0against the alternative Ha:(μ1-μ)20

.a. The two-tailed p-value of the test is 0.1150 . Interpret this result.b. If the alternative hypothesis had been Ha:(μ1-μ)2<0 , how would the p-value change? Interpret the p-value for this one-tailed test.
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