A random sample of ten female sprinters were selected to participate in a intens

Question

A random sample of ten female sprinters were selected to participate in a

intensive training program under coach Jane Doe to improve their speed on 200

-meter dash run. After completing training program, each participate is measured and recorded the time of completing the run, and results are: 26, 30,

28, 29, 25, 28, 32, 35, 24, and 23 seconds.

Historical record shows the general population without any special training finishes the run on average in 30 seconds.

Using t test of single sample (p<0.05), what can the researcher conclude? Please also compute 95% confidence interval to see whether both processes come up with same conclusion.

Explanation / Answer

State the hypotheses:

H0:mu=30 (there is no difference in running time for trained sprinters)

H1:mu=/=30 (there is difference in running time for trained sprinters)

Compute 1-sample t test.

From information given, using technology,

xbar=28, s=3.71, n=10

t=(xbar-mu)/(s/sqrt n), where, xbar is sample mean, mu is population mean, s is sample standard deviation, and n is sample size.

=(28-30)/(3.71/sqrt 10)

=-1.70

p value at 9 degrees of freedom [df=n-1] is 0.1233. Per rejection rule based on p value, reject null hypothesis if p value is less than 0.05. Here, p value is not less than 0.05, therefore, fail to reject H0. There is insufficient sample evidence to suggest that there is significant difference in running time for trained sprinters.

The 95% c.i for population mean running tim efor trained sprinters is:xbar+-talpha/2, df=n-1 (s/sqrt n), where, t denote t critical at alpha/2 (alpha=0.05, alpha/2=0.025) and n-1=9 degrees of freedom.

=28+-2.26(3.71/sqrt 10)

=(25.34,30.66)

Per rejection rule based on confidence interval, reject H0, if c.i doesnot contain population mean. But, 95% c.i contains mu=30, hence, fail to reject H0.

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