The data were collected to verify the claim that California drivers are more agg

Question

The data were collected to verify the claim that California drivers are more aggressive (drive faster than drivers from other states). 1. Construct a 95% confidence interval for California drivers (copy the data to a new worksheet) 2. Perform a hypothesis test, with =0.05, to see if California drivers drive faster than the speed limit of 65 miles per hour California Cars 66 66 58 68 63 63 63 63 70 76 69 54 70 62 80 61 65 61 47 77 75 52 77 86 58 53 74 54 58 62 78 81 67 54 68 54 54 76 65 62 69 56 66 58 70 57 48 74 52 49 79 70 71 63 67 68 68 65 63 61 56 68 69 68 72 68 57 52 66 63 57 65 56 71 55 64 62 76 66 62 64 59 72 54 75 63 59 42 64 65 70 55 63 58 63 69 66 65 62 66 56 66 67 79 74 67 68 57 75 76 59 60 64 59 69 69 76 82 68 56 59 62 53 57 65 56 69 61 70 56 72 62 81 71 67 50 63 67 63 61 The data were collected to verify the claim that California drivers are more aggressive (drive faster than drivers from other states). 1. Construct a 95% confidence interval for California drivers (copy the data to a new worksheet) 2. Perform a hypothesis test, with =0.05, to see if California drivers drive faster than the speed limit of 65 miles per hour California Cars 66 66 58 68 63 63 63 63 70 76 69 54 70 62 80 61 65 61 47 77 75 52 77 86 58 53 74 54 58 62 78 81 67 54 68 54 54 76 65 62 69 56 66 58 70 57 48 74 52 49 79 70 71 63 67 68 68 65 63 61 56 68 69 68 72 68 57 52 66 63 57 65 56 71 55 64 62 76 66 62 64 59 72 54 75 63 59 42 64 65 70 55 63 58 63 69 66 65 62 66 56 66 67 79 74 67 68 57 75 76 59 60 64 59 69 69 76 82 68 56 59 62 53 57 65 56 69 61 70 56 72 62 81 71 67 50 63 67 63 61

Explanation / Answer

a.
Confidence Interval
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
Mean(x)=64.45
Standard deviation( sd )=8.018257
Sample Size(n)=140
Confidence Interval = [ 64.45 ± t a/2 ( 8.018257/ Sqrt ( 140) ) ]
= [ 64.45 - 1.977 * (0.678) , 64.45 + 1.977 * (0.678) ]
= [ 63.11,65.79 ]
b.

Given that,
population mean(u)=64.45
sample mean, x =65
standard deviation, s =8.018257
number (n)=140
null, Ho: µ=64.45
alternate, H1: µ>64.45
level of significance, a = 0.05
from standard normal table,right tailed t a/2 =1.656
since our test is right-tailed
reject Ho, if to > 1.656
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =65-64.45/(8.018257/sqrt(140))
to =0.812
| to | =0.812
critical value
the value of |t a| with n-1 = 139 d.f is 1.656
we got |to| =0.812 & | t a | =1.656
make decision
hence value of |to | < | t a | and here we do not reject Ho
p-value :right tail - Ha : ( p > 0.8116 ) = 0.2092
hence value of p0.05 < 0.2092,here we do not reject Ho

ANSWERS
---------------
null, Ho: µ=64.45
alternate, H1: µ>64.45
test statistic: 0.812
critical value: 1.656
decision: do not reject Ho
p-value: 0.2092

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