Question

This week, you will use two of the data sets that were posted during last week's discussion, as follows: 1) Refer to the data set that you posted last week (high temperatures for your area during the month of June 2019) and 2) Refer to the data set that one of your classmates posted last week (high temperatures for their area during the month of June 2019). Use these data sets to test the claim that the average high temperature in June for your area is the same as the average high temperature in June for your classmate's area. You must use a two-sample t-test with a 0.05 significance level, assuming independent samples, and assuming that the population variances are not equal. You must show all steps (hypotheses, degrees of freedom, critical values, test statistic, decision about the null hypothesis and final conclusion). Take a picture of your work and post it.

Guess: 85

79 77 76 84 86 87 86 85 84 80 7471 73 76 82 73 78 73 73 76 68 79 80 78 87 86 87 83 92 95

Mean: 80.26666667    Standard Dev. (S): 6.475062686

Guess: 85

94 91 95 93 78 88 90 90 91 90 81 87 88 89 90 90 86 90 91 92 93 92 92 83 80 89 92 93 81 87

88.86666667 Mean 4.344900565 Standard Deviation

Answer 1

Given that,
mean(x)=80.2666
standard deviation , s.d1=6.475
number(n1)=30
y(mean)=88.866
standard deviation, s.d2 =4.3449
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 =2.045
since our test is two-tailed
reject Ho, if to < -2.045 OR if to > 2.045
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =80.2666-88.866/sqrt((41.92563/30)+(18.87816/30))
to =-6.0404
| to | =6.0404
critical value
the value of |t α| with min (n1-1, n2-1) i.e 29 d.f is 2.045
we got |to| = 6.04037 & | t α | = 2.045
make decision
hence value of | to | > | t α| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -6.0404 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -6.0404
critical value: -2.045 , 2.045
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that difference of means between the average high temperature in June for your area is the same as the average high temperature in June for your classmate's area.