Question

Single data values (in hours): 3, 1, 4, 5, 5, 2, 2.5, 3.5, 4, 4.5, 0, 2, 2, 2.5, 3.5, 4, 4, 4, 4, 2, 3.5, 3.5, 2, 3, 4, 4, 3, 3, 3, 1

Claim: It was found that the mean time spent watching TV daily was 4 hours. A researcher claims that he believes the mean time spent watching TV is truly lower. The mean time spent watching TV daily is less than 4 hours.                                                                                         

Null Hypothesis: H₀: μ=4 hours Alternative Hypothesis: H_a: μ<4 hours

1. What is the t-statistic? What is the P-value? What is the conclusion:

2. Create a claim based off of the data below and state the claim.

Athletes	Non-athletes
3.7	2.5
3.1	4.1
4.2	4.2
4.2	4.2
3.4	3.0
3.7	3.8
2.1	2.1
3.5	2.7
3.6	1.8
4.0	2.0
2.9	3.6
3.2	3.9
2.9	2.6
3.5	2.8
3.6	3.1
3.4	3.5
2.9	3.5
3.9	3.6
2.8	2.9
3.1	2.7
3.1	3.2
3.6	3.3
2.6	3.4
3.8	1.9
3.4	1.9
3.4	3.1
3.4	2.7
3.2	1.8
2.8	3.0
3.7	2.2

Null Hypothesis

Alternative Hypothesis

Test Statistic

P-value

Conclusion – reject or fail to reject null.

Interpretation - final conclusion.

Answer 1

1.
Given that,
population mean(u)=4
sample mean, x =3.0833
standard deviation, s =1.196
number (n)=30
null, Ho: μ=4
alternate, H1: μ<4
level of significance, alpha = 0.05
from standard normal table,left tailed t alpha/2 =1.699
since our test is left-tailed
reject Ho, if to < -1.699
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =3.0833-4/(1.196/sqrt(30))
to =-4.1981
| to | =4.1981
critical value
the value of |t alpha| with n-1 = 29 d.f is 1.699
we got |to| =4.1981 & | t alpha | =1.699
make decision
hence value of | to | > | t alpha| and here we reject Ho
p-value :left tail - Ha : ( p < -4.1981 ) = 0.00012
hence value of p0.05 > 0.00012,here we reject Ho
ANSWERS
---------------
null, Ho: μ=4
alternate, H1: μ<4
test statistic: -4.1981
critical value: -1.699
decision: reject Ho
p-value: 0.00012
we have enough evidence to support the claim that The mean time spent watching TV daily is less than 4 hours

2.
Given that,
mean(x)=3.3566
standard deviation , s.d1=0.4753
number(n1)=30
y(mean)=2.97
standard deviation, s.d2 =0.7316
number(n2)=30
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.05
since our test is two-tailed
reject Ho, if to < -2.05 OR if to > 2.05
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =3.3566-2.97/sqrt((0.22591/30)+(0.53524/30))
to =2.427
| to | =2.427
critical value
the value of |t α| with min (n1-1, n2-1) i.e 29 d.f is 2.05
we got |to| = 2.4271 & | t α | = 2.05
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.4271 ) = 0.022
hence value of p0.05 > 0.022,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 2.427
critical value: -2.05 , 2.05
decision: reject Ho
p-value: 0.022
we have enough evidence to support the claim that difference of means of two samples