**** please it you are not going to solve all of it , leave it for some who will
ID: 3225944 • Letter: #
Question
**** please it you are not going to solve all of it , leave it for some who will
**** please provide step by step solution .
Hypothesis testing predicted, it was inevitable that the world of Reality TV As you possibly could have primary example would eventually find way to penetrate the Game Show genre. A is Wipeout', where the complex competitive events are now intertwined with expansive self-explo ation and antics. You are starting to believe that the producers intentionally making the events easier to allow competitors additional air time. Looking back at the last season, you find that on average, the overall event time per individual is 17 min. This season, the first 5 competitors lasted 21 minutes into the competition with a standard deviation of2.2 minutes. Based on this information, are Wipeout competitors lasting longer before dropping out this year as opposed to the previous 6 seasons? Use a level of significance of 0.01. 90 Hypothesis Testing machine is designed to fill jars with 16 ounces of coffee. A consumer suspects that the machine is not filling the jars completely. A sample of 35 jars has a mean of 15.8 ounces and a standard deviation of0.6 ounces. At a 0.10, would you reject or accept the consumer's claim?Explanation / Answer
Q-8
We have to test the null hypothesis H0: µ=17 versus the alternative Ha: µ>17 ,where µ is the true mean time lasted by the competitor.
We are given sample mean xbar=21, sample standard deviations=2.2 and sample size n=5
So test statistic t=(xbar-17)/(s/sqrt(n))
=(21-17)/(2.2/sqrt(5))
=4.07
Degree of freedom =n-1=5-1=4
Right tailed Critical t at 0.01 level and df=4 is t=3.75
As calculated t>3.75, we reject the null hypothesis and conclude that Wipeout competitors lasting longer before dropping out this year as opposed to the previous 6 seasons
Q-9
We have to test the null hypothesis H0: µ=16 versus the alternative Ha: µ17 ,where µ is the true mean amount filled in jars.
We are given sample mean xbar=15.8, sample standard deviations=0.6 and sample size n=35
So test statistic t=(xbar-16)/(s/sqrt(n))
=(15.8-16)/(0.6/sqrt(35))
=-1.97
Degree of freedom =n-1=55-1=34
Two tailed Critical t at 0.1 level and df=34 is t=1.69
As calculated |t|>1.69, we reject the null hypothesis and conclude that machne is not filling the jars completely.
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