A researcher claims that a post-lunch nap decreases the amount of time it takes
ID: 3200757 • Letter: A
Question
A researcher claims that a post-lunch nap decreases the amount of time it takes males to sprint 20 meters after a night with only 4 of sleep. The table shows the amounts of time (in seconds) it took for 10 males to sprint 20 meters after a night with only 4 hours of when they did not take a post-lunch nap and when they did take a post-lunch nap. At alpha = 0.01, is there enough evidence to support the researchers claim? Assume the samples are random and dependent, and the population is normally distributed. Complete parts (a) through (e) below. H_0: mu_d lessthanorequalto d bar H_a: mu_d > d bar H_0: mu_d lessthanorequalto 0 H_a: mu_d > 0 H_0: mu_d greaterthanorequalto d bar H_a: mu_dExplanation / Answer
Paired Samples Statistics
Mean N Std. Deviation Std. Error Mean
Pair 1 Sprinttimewhthoutnap 4.0067 9 .05362 .01787
Sprinttimewithnap 3.9100 9 .06083 .02028
Paired Samples Test
Mean Std. D Std. Error Mean t df Sig. (2-tailed)
Pair 1 Sprinttimewhthoutnap - Sprinttimewithnap .09667 .04272 .01424 6.788 8 .000
Null Hypothesis H0:Option E correct
Alternative Hypothesis Ha: Option E correct(Two tailed test)
test statistic t = d bar/SE(d bar) follwos n-1 degree of freedom
where Calculate the difference (di = yi xi) between the two observations on each pair, making sure you distinguish between positive and negative differences.
Calculate the mean difference, d bar
Calculate the standard deviation of the differences, sd, and use this to calculate the standard error of the mean difference, SE( ¯d) = sd / Sqrt(n). = 0.04272
therefore t = 6.788
from t tables the critical value of t at 0.01 level of significance with n-1 = 9 d.fis 3.25
therefore calculated value > table value (6.678>3.25)
therefore H0 is rejected i.e we may conclude that the given data is enough to spport the researchers claim
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