Section 8-3 t Test for a Mean 449 franchise thinks that its cost is below that f

Question

Section 8-3 t Test for a Mean 449 franchise thinks that its cost is below that figure. A random sample of 28 patients across the country had an average cost of $5250 with a standard deviation of $629 At = 0.025, can it be concluded that the mean is less than $5400? Hoights of Tall Buildings A researcher estimates that he average height of the buildings of 30 or more stories in a large city is at least 700 feet. A random sample of 10 buildings is selected, and the heights in feet are shown. At = 0.025, is there enough evidence to reject the claim? 485 511 841 725 615 520 535 635 616 582 Source: Pittsburgh Tribune-Review. Number of Words in a Novel The National Novel Writing Association states that the average novel is at least 50,000 words. A particularly ambitious writing 0. h at a callege-preparatory high school had randomly

Explanation / Answer

9.

Given that,
population mean(u)=700
sample mean, x =606.5
standard deviation, s =109.0792
number (n)=10
null, Ho: =700
alternate, H1: !=700
level of significance, = 0.025
from standard normal table, two tailed t /2 =2.685
since our test is two-tailed
reject Ho, if to < -2.685 OR if to > 2.685
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =606.5-700/(109.0792/sqrt(10))
to =-2.711
| to | =2.711
critical value
the value of |t | with n-1 = 9 d.f is 2.685
we got |to| =2.711 & | t | =2.685
make decision
hence value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -2.7106 ) = 0.024
hence value of p0.025 > 0.024,here we reject Ho
ANSWERS
---------------
null, Ho: =700
alternate, H1: !=700
test statistic: -2.711
critical value: -2.685 , 2.685
decision: reject Ho
p-value: 0.024

we have enough evidence to support the claim

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