Nigerian Spleen Length. O.Ehimwenma and M.Tagho, researchers in Nigeria, were interested in how characteristics of the spleen of residents in their tropical environment compare to those found elsewhere in te world. They published their findings in the article "Determination of Normal Dimensions of the Spleen by Ultrasound in an Endemic Tropical Environment" (Nige4ian Medical Journal, Vol.52, No. 3,pp. 198-203). The researchers randomly samples 91males and 109females in Nigeria. The mean and standard deviation of the spleen lengths for the males were 11.1cm and 0.9cm respectively, and those for the females were 10.1cm and 0.7cm, respectively. At the 1%significance level, do the data provide sufficient evidence to conclude that a difference exists in mean spleen lengths of male and female Nigerians?

Short Answer

Expert verified

The data provide sufficient evidence to conclude that a difference exists in mean spleen lengths of male and female Nigerians at1%significance level.

Step by step solution

01

Step 1. Given information 

Significance level=1%.

02

Step 2. Definition for Null and Alternative hypothesis:

Null Hypothesis:

H0:μ1=μ2

That is, the data does not provide sufficient evidence to conclude that a difference exists in mean spleen lengths of male and female Nigerians.

Alternative Hypothesis:

H0:μ1μ2

That is, the data provide sufficient evidence to conclude that a difference exists in mean spleen lengths of male and female Nigerians.

03

Step 3. The Given significance level is, α=0.01.

Now compute the value of the test statistic by using MINITAB.

MINITAB Procedure:

  • Choose Stat > Basic Statistics > 2-Sample t.
  • Choose Summarized data.
  • In first, enter Sample size as 91, Mean as 11.1, Standard deviation as 0.9.
  • In second, enter Sample size as 109, Mean as 10.1, Standard deviation as 0.7.
  • Select Assume equal variances.
  • Choose options.
  • In confidence level enter 99.
  • In Alternative select not equal.
  • Click OK in all the dialogue boxes.
04

Step 4. MINITAB output:

Two-Sample T-Test and CI

From the MINITAB output, the value of test statistic is8.83.

05

Step 5. Find the P-value:

From the MINITAB output, the P-value is, 0.0000.

If Pα, then reject the null hypothesis. Here, the P-value is 0.000which 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 at1%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

You know that the population standard deviations are not equal.

Suppose that you want to perform a hypothesis test to compare the means of two populations, using a paired sample. For each part, decide whether you would use the paired t-test, the paired Wilcoxon signed-rank test, or neither of these tests if preliminary data analyses of the sample of paired differences suggest that the distribution of the paired-difference variable is

a. uniform.

b. neither symmetric nor normal; the sample size is 132.

c. moderately skewed but otherwise roughly bell-shaped.

In each of Exercises 10.39-10.44, we have provided summary statistics for independent simple random samples from two populations. In each case, use the pooled t-test and the pooled t-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval.
10.39 x1=10,s1=2.1,n1=15,x2=12,s2=2.3,n2=15
a. Two-tailed test, α=0.05
b.95%confidence interval

In each of Exercises 10.75-10.80, we have provided summary statistics for independent simple random samples form non populations. In each case, use the non pooled t-fest and the non pooled t-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval.

x~1=20,s1=4,n1=10,x~2=18,s2=5,n2=15.

a. Right-tailed test,localid="1651298373729" α=0.05.

b. 90%confidence interval.

Stressed-Out Bus Drivers. An intervention program designed by the Stockholm Transit District was implemented to improve the work conditions of the city's bus drivers. Improvements were evaluated by G. Evans et al., who collected physiological and psychological data for bus drivers who drove on the improved routes (intervention) and for drivers who were assigned the normal routes (control). Their findings were published in the article "Hassles on the Job: A Study of a Job Intervention with Urban Bus Drivers" (Journal of Organizational Behavior, Vol. 20, pp. 199-208). Following are data, based on the results of the study, for the heart rates, in beats per minute, of the intervention and control drivers.

a. At the 5%significance level, do the data provide sufficient evidence to conclude that the intervention program reduces mean heart rate of urban bus drivers in Stockholm? (Note; x1=67.90, s1=5.49,x¯2=66.81and s2=9.04.

b. Can you provide an explanation for the somewhat surprising results of the study?

c. Is the study a designed experiment or an observational study? plain your answer.

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