Question

Does 10K running time change when the runner listens to music? Nine runners were timed as they ran a 10K with and without listening to music. The running times in minutes are shown below. Running Time With Music 38 47 45 51 40 34 38 51 44 Without Music 45 55 50 54 44 43 38 55 46 Assume a Normal distribution. What can be concluded at the the α = 0.05 level of significance? For this study, we should use The null and alternative hypotheses would be: H 0 : (please enter a decimal) H 1 : (Please enter a decimal) The test statistic = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is α Based on this, we should the null hypothesis. Thus, the final conclusion is that ... The results are statistically significant at α = 0.05, so there is sufficient evidence to conclude that the population mean running time with music is not the same as the population mean running time without music. The results are statistically insignificant at α = 0.05, so there is statistically significant evidence to conclude that the population mean running time with music is equal to the population mean running time without music. The results are statistically insignificant at α = 0.05, so there is insufficient evidence to conclude that the population mean running time with music is not the same as the population mean running time without music. The results are statistically significant at α = 0.05, so there is sufficient evidence to conclude that the nine runners finished with different times on average with music compared to running without music. Interpret the p-value in the context of the study. If the sample mean running time with music for the 9 runners is the same as the sample mean running time without music for these 9 runners and if another 9 runners are observed running the 10K with and without music then there would be a 0.14% chance of concluding that the mean running time with music for the 9 runners differs by at least 4.7 minutes from the mean running time for these 9 runners without music. There is a 0.14% chance of a Type I error. There is a 0.14% chance that the mean running time for the 9 runners with music differs by at least 4.7 minutes compared to the mean time for these 9 runners without music. If the population mean running time with music is the same as the population mean running time without music and if another 9 runners compete with and without music then there would be a 0.14% chance that the mean running time for the 9 runners with music would differ by at least 4.7 minutes compared to the 9 runners competing without music. Interpret the level of significance in the context of the study. There is a 5% chance that the runners aren't in good enough shape to run a 10K, so music is irrelevant. If the population mean running time with music is the same as the population mean running time without music and if another 9 runners compete in the 10K with and without music, then there would be a 5% chance that we would end up falsely concluding that the sample mean running times with music and without music for these 9 runners differ from each other. If the population mean running time with music is the same as the population mean running time without music and if another 9 runners compete with and without music then there would be a 5% chance that we would end up falsely concluding that the population mean running time with music is not the same as the population mean running time without music There is a 5% chance that the population mean running time is the same with and without music.

Answer 1

Given that,
mean(x)=43.1111
standard deviation , s.d1=6.0093
number(n1)=9
y(mean)=47.7778
standard deviation, s.d2 =6.0369
number(n2)=9
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.306
since our test is two-tailed
reject Ho, if to < -2.306 OR if to > 2.306
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =43.1111-47.7778/sqrt((36.11169/9)+(36.44416/9))
to =-1.6436
| to | =1.6436
critical value
the value of |t α| with min (n1-1, n2-1) i.e 8 d.f is 2.306
we got |to| = 1.6436 & | t α | = 2.306
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -1.6436 ) = 0.139
hence value of p0.05 < 0.139,here we do not reject Ho
ANSWERS
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a.
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -1.6436
critical value: -2.306 , 2.306
decision: do not reject Ho
p-value: 0.139 = 0.14
we do not have enough evidence to support the claim that The results are statistically significant at α = 0.05, so there is sufficient evidence to conclude that the population mean running time
with music is not the same as the population mean running time without music.
b.
Interpret the p-value in the context of the study.
If the sample mean running time with music for the 9 runners is the same as the sample mean running time without music for these 9 runners and if another 9 runners are observed running the 10K with and without music then there would be a 0.14% chance of concluding that
the mean running time with music for the 9 runners differs by at least 4.7 minutes from the mean running time for these 9 runners without music.
c.
Interpret the level of significance in the context of the study.
If the population mean running time with music is the same as the population mean running time without music and if another 9 runners compete with and without music then there would be a 5% chance that we would end up falsely concluding that
the population mean running time with music is not the same as the population mean running time without music.