Question

In: Statistics and Probability

Suppose the National Transportation Safety Board (NTSB) wants to examine the safety of compact cars, midsize...

Suppose the National Transportation Safety Board (NTSB) wants to examine the safety of compact cars, midsize cars, and full-size cars. It collects a sample of cars each of the cars types. The data below displays the frontal crash test performance percentages. Test whether there are statistical differences in the frontal crash test performance for each type of car. Compact Cars Midsize Cars Full-Size Cars 95 95 93 98 98 97 87 98 92 99 89 92 99 94 84 94 88 87 99 93 88 98 99 89

What is the critical value for the follow-up t-test for Midsize cars vs. Full-size cars (different from question above)? (Round to 2 decimal places.)

Solutions

Expert Solution

Assumed data,
population mean(u)=93.1 because not given in the data.
Given that,
sample mean, x =93.541
standard deviation, s =4.564
number (n)=24
null, Ho: μ=93.1
alternate, H1: μ!=93.1
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.069
since our test is two-tailed
reject Ho, if to < -2.069 OR if to > 2.069
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =93.541-93.1/(4.564/sqrt(24))
to =0.47
| to | =0.47
critical value
the value of |t α| with n-1 = 23 d.f is 2.069
we got |to| =0.47 & | t α | =2.069
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 != 0.4734 ) = 0.6404
hence value of p0.05 < 0.6404,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=93.1
alternate, H1: μ!=93.1
test statistic: 0.47
critical value: -2.069 , 2.069
decision: do not reject Ho
p-value: 0.6404
we do not have enough evidence to support the claim that population mean for Midsize cars vs. Full-size cars.


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