Can someone please help? I need to get this doe so I can do something important
1 answer:
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Answer:
Given:
Sample mean, n = 12
Sample mean of crashes per year, X' = 4.76
Std deviation, s.d = 0.47
a) Given:
u = 4.56
Significance level = 20% = 0.2
For null and alternative hypothesis, we have:
H0 : u = 4.56
Ha : u ≠ 4.56
This is a two-tailed test.
Given a small sample size and standard deviation was not given, we test if the crash data at the 12 intersections is different from the city average of 4.56 using the t-test.
= 1.4741
Degree of freedom, df = n - 1
df = 12 - 1 = 11
Therefore from the distribution table at
T.DIST.2T( 1.4741, 11)
p-value = 0.1685
Since the p-value(0.1685) is less than level of significance (0.2) we reject the null hypothesis, H0.
We can say there is enough evidence to show that the crash data at the 12 intersections at 20% significance level is different from the city average of 4.56
b) Given:
u = 4.56
Significance level = 5% = 0.05
This is the same as part (a), apart from a different level of significance.
H0 : u = 4.56
Ha : u ≠ 4.56
= 1.4741
Degree of freedom, df = n - 1
df = 12 - 1 = 11
Therefore from the distribution table at
T.DIST.2T( 1.4741, 11)
p-value = 0.1685
Since p-value(0.1685) is greater than level of significance (0.05), we fail to reject the null hypothesis.
We can say there is not enough statistical evidence that ctash data at the 12 intersections at 5% significance level is different from the city average (4.56)