2. A sport trainer wants to know whether the true average time of his athlete who do 100 - meter sprint is 98 seconds. He recorded 18 trials of his team and found that the average time is 98.2 seconds with a standard deviation of 0.4 second. Is there sufficient evidence to reject the null hypothesis if µ = 98 seconds at the 0.05 level of significance? A Test statintio

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A sport trainer wants to know whether the true average time of his athlete who do 100 - meter
sprint is 98 seconds. He recorded 18 trials of his team and found that the average time is 98.2
seconds with a standard deviation of 0.4 second. Is there sufficient evidence to reject the null
hypothesis if µ = 98 seconds at the 0.05 level of significance?
A. Test statistic t = 2.1213 is a value greater than the critical value t, = + 2.110. Hence, the test
value is clearly in the critical region. Thus, we decide to reject the null hypothesis.
B. Test statistic t
value is clearly outside the critical region. Thus, we decide to reject the null hypothesis.
C. Test statistic t = 2.1213 is a value greater than the critical value ta = + 2.110. Hence, the test
value is clearly in the critical region. Thus, we decide to accept the null hypothesis.
D. Test statistic t = - 2.1213 is a value lesser than the critical value t = + 2.110. Hence, the test
value is clearly outside the critical region. Thus, we decide to accept the nuill hypothesis.
2.
%3D
%3D
- 2.1213 is a value lesser than the critical value t, = + 2.110. Hence, the test
Transcribed Image Text:A sport trainer wants to know whether the true average time of his athlete who do 100 - meter sprint is 98 seconds. He recorded 18 trials of his team and found that the average time is 98.2 seconds with a standard deviation of 0.4 second. Is there sufficient evidence to reject the null hypothesis if µ = 98 seconds at the 0.05 level of significance? A. Test statistic t = 2.1213 is a value greater than the critical value t, = + 2.110. Hence, the test value is clearly in the critical region. Thus, we decide to reject the null hypothesis. B. Test statistic t value is clearly outside the critical region. Thus, we decide to reject the null hypothesis. C. Test statistic t = 2.1213 is a value greater than the critical value ta = + 2.110. Hence, the test value is clearly in the critical region. Thus, we decide to accept the null hypothesis. D. Test statistic t = - 2.1213 is a value lesser than the critical value t = + 2.110. Hence, the test value is clearly outside the critical region. Thus, we decide to accept the nuill hypothesis. 2. %3D %3D - 2.1213 is a value lesser than the critical value t, = + 2.110. Hence, the test
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