A simple random sample of 40 recorded speeds (in mi/h) is obtained from cars traveling on a specific highway. The sample has a m
ean of 68.4mi / h and a standard deviation of 5.7mi / h . Use a 0.05 significance level to test the claim that the mean speed of all cars is greater than the posted speed limit of 65mi / h . State the initial and final conclusion.
2 answers:
We need to reject the null hypothesis; there is sufficient evidence to support the claim that the mean speed is greater than 65 miles / hour.
<h3>What is normal distribution?</h3>'Normal distribution is a continuous probability distribution wherein values lie in a symmetrical fashion mostly situated around the mean.'
According to the given problem,
Since the sample size is fairly large (n > 30), we use normal distribution.
The null hypothesis tested is
Mean speed of all cars ≤ 65 miles/hour. (µ ≤ 65)
The alternative hypothesis is
Mean speed of all cars > 65 miles/hour. (µ > 65)
Significance level = 0.05
The test statistic used is Z = - µ/σ / √ n, where
= 68.4, n = 40, σ = 5.7
Therefore, Z = 68.4 - 65 / 5.7 /√40 = 3.77254177
If the calculated value of test statistic is greater than the critical value at the 0.05 significance level,
Upper critical value = 1.644853627
P-value = P (Z > 3.77254177) = 0.000080796
Hence, we can conclude that we should reject the null hypothesis since there is enough evidence to support the claim that the mean speed is greater than 65 miles / hour.
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Answer:
Step-by-step explanation:
There are simply 5 possible values in the given set. Out of these, only one of these is the number 1. Therefore, the probability a 1 is drawn (P(1)) is .