I ONLY NEED HELP WITH QUESTIONS F THROUGH I.. Thank you! Use the values in the t
ID: 3170702 • Letter: I
Question
I ONLY NEED HELP WITH QUESTIONS F THROUGH I.. Thank you!
Use the values in the table below, which represent the heights, in inches, of 36 randomly-selected students enrolled in QMETH 201, this quarter, to answer/complete Parts a through j: For the 36 students in the sample, the mean height equals 68 inches and the standard deviation equals 4.44 inches. State the null and alternative hypotheses you would use to test whether the mean height for all students enrolled in QMETH 201 this quarter differs from 70 inches. If the level of significance equals 0.05, what is the critical value of the associated test statistic? What is the calculated value of the associated test statistic? State your decision about the null hypothesis by comparing the critical and calculated values of the test statistic (Parts b and c). State your conclusion (meaning, describe what the decision means in this problem). State the null and alternative hypotheses you would use to test whether the proportion of all students enrolled in QMETH 201 this quarter who are at least 70 inches tall differs from 0.5. What is the calculated value of the associated test statistic? What is the p-VALUE for the associated test? If the level of significance equals 0.10, state your decision about the null hypothesis by comparing the level of significance to the p-VALUE (from Part h). State your conclusion (meaning, describe what the decision means in this problem).Explanation / Answer
Given that,
population mean(u)=70
sample mean, x =68
standard deviation, s =4.44
number (n)=36
null, Ho: =70
alternate, H1: !=70
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.03
since our test is two-tailed
reject Ho, if to < -2.03 OR if to > 2.03
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =68-70/(4.44/sqrt(36))
to =-2.703
| to | =2.703
critical value
the value of |t | with n-1 = 35 d.f is 2.03
we got |to| =2.703 & | t | =2.03
make decision
hence value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -2.7027 ) = 0.0105
hence value of p0.05 > 0.0105,here we reject Ho
ANSWERS
---------------
a.
null, Ho: =70
alternate, H1: !=70
b.
critical value: -2.03 , 2.03
c.
test statistic: -2.703
d.
decision: reject Ho
e.
the mean height for all
students enrolled in QMETH 201 this quarter is not different from 70 inches
h.
p-value: 0.0105
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