Answer to Question #347490 in Statistics and Probability for jouuuuuu

Question #347490

A sports trainer wants to know whether the true average time of his athletes who do 100-meter sprint is 98 seconds. He recorder 18 trials of his team and found that the average time is 98.2 seconds with a standard deviation of 0.4 second


1
Expert's answer
2022-06-03T03:53:49-0400

The following null and alternative hypotheses need to be tested:

H0:μ=98H_0:\mu=98

H1:μ98H_1:\mu\not=98

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is α=0.05,\alpha = 0.05, df=n1=17df=n-1=17 and the critical value for a two-tailed test is tc=2.109816.t_c =2.109816.

The rejection region for this two-tailed test is R={t:t>2.109816}.R = \{t:|t|>2.109816\}.


The t-statistic is computed as follows:



t=xˉμs/n=98.2980.4/18=2.1213t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{98.2-98}{0.4/\sqrt{18}}=2.1213


Since it is observed that t=2.1213>2.109816=tc,|t|=2.1213>2.109816=t_c, it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for two-tailed, df=17df=17 degrees of freedom, t=2.1213t=2.1213 is p=0.048898,p=0.048898, and since p=0.048898<0.05=α,p= 0.048898<0.05=\alpha, it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean μ\mu

is different than 98, at the α=0.05\alpha = 0.05 significance level.


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