Chapter 5, Section 1, Exercise 026 Exercise and Gender The dataset ExerciseHours

Question

Chapter 5, Section 1, Exercise 026

Exercise and Gender

The dataset ExerciseHours contains information on the amount of exercise (hours per week) for a sample of statistics students. The mean amount of exercise was 9.4 hours for the 30 female students in the sample and 12.4 hours for the 20 male students. A randomization distribution of differences in means based on these data, under a null hypothesis of no difference in mean exercise time between females and males, is centered near zero and reasonably normally distributed. The standard error for the difference in means, as estimated from the standard deviation of the randomization differences, is SE=2.38. Use this information to test, at a 5% level, whether the data show that the mean exercise time for female statistics students is less than the mean exercise time of male statistics students.

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State the null and alternative hypotheses.

Let group 1 be the female participants and group 2 be the male participants.

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the absolute tolerance is +/-0.01

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the absolute tolerance is +/-0.001

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Chapter 5, Section 1, Exercise 026

Exercise and Gender

The dataset ExerciseHours contains information on the amount of exercise (hours per week) for a sample of statistics students. The mean amount of exercise was 9.4 hours for the 30 female students in the sample and 12.4 hours for the 20 male students. A randomization distribution of differences in means based on these data, under a null hypothesis of no difference in mean exercise time between females and males, is centered near zero and reasonably normally distributed. The standard error for the difference in means, as estimated from the standard deviation of the randomization differences, is SE=2.38. Use this information to test, at a 5% level, whether the data show that the mean exercise time for female statistics students is less than the mean exercise time of male statistics students.

Click here for the dataset associated with this question.

State the null and alternative hypotheses.

Let group 1 be the female participants and group 2 be the male participants.

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What is the test statistic?

Round your answer to two decimal places.

z=

the absolute tolerance is +/-0.01

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What is the p-value?

Round your answer to three decimal places.

p-value=

the absolute tolerance is +/-0.001

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What is the conclusion?

Reject H0 and find evidence that females exercise less. Reject H0 and find evidence that females do not exercise less. Do not reject H0 and do not find evidence that females exercise less. Do not reject H0 and find evidence that females do not exercise less.

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Explanation / Answer

Given that,
mean(x)=9.4
standard deviation , 1 =13.0357
number(n1)=30
y(mean)=12.4
standard deviation, 2 =10.6436
number(n2)=20
null, Ho: u1 = u2
alternate, H1: 1 < u2
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=9.4-12.4/sqrt((169.92947/30)+(113.28622/20))
zo =-0.891
| zo | =0.891
critical value
the value of |z | at los 0.05% is 1.645
we got |zo | =0.891 & | z | =1.645
make decision
hence value of | zo | < | z | and here we do not reject Ho
p-value: left tail - Ha : ( p < -0.891 ) = 0.186
hence value of p0.05 < 0.186,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: 1 < u2
test statistic: -0.891
critical value: -1.645
decision: do not reject Ho
p-value: 0.186

we do not have enough evidence that females do not exercise less

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