Break downs occur on a 20-years-old car with rate λ= 0.5 breakdowns/week. The owner of the car is planning to have a trip on his
1 answer:
Answer: There is probability of 0.367 that there will be no breakdown on his car in the trip.
Step-by-step explanation:
Since we have given that
Mean (λ) = 0.5 breakdown per week
Number of weeks the owner of the car is planning to have a trip on his car for = 2 weeks
So, mean for 2 weeks would be
We need to find the probability that there will be no breakdown on his car in the trip.
Probability that there will be no breakdown on his car in the trip is given by
P(X=0) is given by
Hence, there is probability of 0.367 that there will be no breakdown on his car in the trip.
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Answer:
a) Null hypothesis:
Alternative hypothesis:
b) t =-7.44
c)
d) Reject Null hypothesis. The is enough evidence to conclude that the mean prediction error is not equal to 0.
We reject the null hypothesis because the . So we can conclude that the difference is significantly different from 0 at 5% of significance.
Step-by-step explanation:
Assuming this output:
t(observed value) =-7.44
t(Critical value )= 1.984
DF = 98
p value (two tailed) < 0.0001
a) What are the null and alternative hypothesis
Null hypothesis:
Alternative hypothesis:
The reason is because the output says a bilateral test so for this case w eselect this option.
b) Identify the statistic
The correct formula for the statistic is given by:
Based on the output the calculated value its t =-7.44
c) Identify the p value
We have the degrees of freedom given 98
Based on the alternative hypothesis the p value is given by:
d) State the final conclusion that addresses the original claim
Reject Null hypothesis. The is enough evidence to conclude that the mean prediction error is not equal to 0.
We reject the null hypothesis because the . So we can conclude that the difference is significantly different from 0 at 5% of significance.