(1 pt) Traffic police monitor the speed of vehicles as they travel over a new br
ID: 3313685 • Letter: #
Question
(1 pt) Traffic police monitor the speed of vehicles as they travel over a new bridge. The average speed for a sample of 27 vehicles was 91.29 km/h, with the sample standard deviation being 4.94 km/h. We will assume that the speeds are Normally distributed, and the police are interested in the mean speed. Part a) Since the variance of the underlying Normal distribution is not known, inference here would involve the t distribution. How many degrees of freedom would the relevant t distribution have? Part b) Create a 95 % confidence interval for the mean speed of vehicles crossing the bridge. Give the upper and lower bounds to your interval, each to 2 decimal places. ( Part c The police hypothesized that the mean speed of vehicles over the bridge would be the speed limit 80 kmh. Taking a significance eve of 5 % what should infer about this hypothesis? A. We should reject the hypothesis since 80 km/h is in the interval found in (b). B. We should not reject the hypothesis since the sample mean is in the interval found in (b) C. We should not reject the hypothesis since 80 km/h is in the interval found in (b) D. We should reject the hypothesis since 80 km/h is not in the interval found in (b) E. We should reject the hypothesis since the sample mean was not 80 km/h. Part d) Decreasing the significance level of the hypothesis test above would (select all that apply) A. increase the Type l error probability. B. either increase or decrease the Type I error probability C. not change the Type ll error probability. D. not change the Type l error probability. E. decrease the Type l error probability.Explanation / Answer
a.
from standard normal table, two tailed value of |t /2| with n-1 = 26 d.f is 2.056
degree of freedom = 2.056
b.
TRADITIONAL METHOD
given that,
sample mean, x =91.29
standard deviation, s =4.94
sample size, n =27
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 4.94/ sqrt ( 27) )
= 0.95
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 26 d.f is 2.056
margin of error = 2.056 * 0.95
= 1.95
III.
CI = x ± margin of error
confidence interval = [ 91.29 ± 1.95 ]
= [ 89.34 , 93.24 ]
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DIRECT METHOD
given that,
sample mean, x =91.29
standard deviation, s =4.94
sample size, n =27
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 26 d.f is 2.056
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 91.29 ± t a/2 ( 4.94/ Sqrt ( 27) ]
= [ 91.29-(2.056 * 0.95) , 91.29+(2.056 * 0.95) ]
= [ 89.34 , 93.24 ]
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interpretations:
1) we are 95% sure that the interval [ 89.34 , 93.24 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
c.
Given that,
population mean(u)=80
sample mean, x =91.29
standard deviation, s =4.94
number (n)=27
null, Ho: =80
alternate, H1: !=80
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.06
since our test is two-tailed
reject Ho, if to < -2.06 OR if to > 2.06
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =91.29-80/(4.94/sqrt(27))
to =11.875
| to | =11.875
critical value
the value of |t | with n-1 = 26 d.f is 2.06
we got |to| =11.875 & | t | =2.06
make decision
hence value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 11.8754 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: =80
alternate, H1: !=80
test statistic: 11.875
critical value: -2.06 , 2.06
decision: reject Ho
p-value: 0
option :A
we have enough evidence to support the claim that mean speed of vehicles over bridge would be limit 80kmph
d.
level of significance is decreases then type1 error is not change but it will effect the type 2 error
option :D
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