Question

Do the poor spend the same amount of time in the shower as the rich? The results of a survey asking poor and rich people how many minutes they spend in the shower are shown below.

Poor 26 20 29 15 26 33 35 23 36 31 36 39 31

Rich: 14 13 48 18 32 31 33 41 54 12 18 25 52

Assume both follow a Normal distribution. What can be concluded at the the αα = 0.05 level of significance level of significance?

For this study, we should use Select an answert-test for a population meanz-test for the difference between two population proportionst-test for the difference between two independent population meansz-test for a population proportiont-test for the difference between two dependent population means

1. The null and alternative hypotheses would be:   
2.   

H0:H0:  Select an answerμ1p1 ?≠<>= Select an answerp2μ2 (please enter a decimal)   

H1:H1:  Select an answerp1μ1 ?><≠= Select an answerp2μ2 (Please enter a decimal)

2. The test statistic ?tz = (please show your answer to 3 decimal places.)
3. The p-value = (Please show your answer to 4 decimal places.)
4. The p-value is ?>≤ αα
5. Based on this, we should Select an answeracceptrejectfail to reject the null hypothesis.
6. Thus, the final conclusion is that ...
   • The results are statistically significant at αα = 0.05, so there is sufficient evidence to conclude that the mean time in the shower for the thirteen poor people that were surveyed is not the same as the mean time in the shower for the thirteen rich people that were surveyed.
   • The results are statistically insignificant at αα = 0.05, so there is statistically significant evidence to conclude that the population mean time in the shower for the poor is equal to the population mean time in the shower for the rich.
   • The results are statistically insignificant at αα = 0.05, so there is insufficient evidence to conclude that the population mean time in the shower for the poor is not the same as the population mean time in the shower for the rich.
   • The results are statistically significant at αα = 0.05, so there is sufficient evidence to conclude that the population mean time in the shower for the poor is not the same as the population mean time in the shower for the rich.

Answer 1

Given that,
mean(x)=29.2308
standard deviation , s.d1=7.0018
number(n1)=13
y(mean)=30.0769
standard deviation, s.d2 =15.0026
number(n2)=13
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.179
since our test is two-tailed
reject Ho, if to < -2.179 OR if to > 2.179
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =29.2308-30.0769/sqrt((49.0252/13)+(225.07801/13))
to =-0.18
| to | =0.18
critical value
the value of |t α| with min (n1-1, n2-1) i.e 12 d.f is 2.179
we got |to| = 0.18426 & | t α | = 2.179
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 != -0.1843 ) = 0.857
hence value of p0.05 < 0.857,here we do not reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: u1 != u2
b.
test statistic: -0.18
critical value: -2.179 , 2.179
decision: do not reject Ho
c.
p-value: 0.857
d.
p value is greater than 0.05
e.
we do not have enough evidence support the claim that difference between two independent population means
f.
The results are statistically significant at α= 0.05, so there is sufficient evidence to conclude that the population mean time in the shower for the poor is not the same as the population mean time in the shower for the rich.