According to a survey conducted by the department of...

According to a survey conducted by the Department of Statistics Malaysia (DOSM), university students get a mean of 7 hours of sleep per night. A random sample of 50 students at IIUM revealed the mean length of time slept last night was 6 hours and 48 minutes (6.8 hours). The standard deviation of the sample was 0.9 hours. At the 5% level of significance, is it reasonable to conclude that IIUM students sleep less than the normal university students?

Transcript

00:01 Question we are given here according to survey according to survey of the department of the statistics malaysia that is dosm so university that mean of we are given here mean of seven hour sleep mean of seven hours sleep seven hours sleep per night seven hours sleep per night and the random sample of the 50 students select in a 50 student so we are given a random sample of the 50 students i i u m that revealed that mean length of the time mean length of sleep last night is 6 hour and 48 minutes so mean length here is the 6 hours and 8 say here 6 .8 hours 6 .8 hours and just under deviation of the sample as to this equal to we are given here is the 0 .9 hours the value significance we are given here 5 percent which is equal we can write 0 .0 5 so let's solve this question and finding the answer so first we can write the value of mu we are given here which is 7 hours this is the value of me which is equal to 7 hours and the x bar sample mean we are given here x bar which is equal to 6 .8 hours 6 .8 hours and here standard deviation is 0 .9 howards so first we can write not have for this so h0 which is equal to we can write mu which is equal to 7 and the alternative 5h and that we are given in mu is less than 7 now let's let's find out taste statistics so taste statistics so taste statistics t which is equal we can write x -war minus mu divided by we can write s divided by square root or n so we can write t which is equal to here 6 .6 .8 minus 7 divided by we can write value of 0 .9 divided by square root of 50 now we just simplify and we can write here answer is so answer is 1 .57 here 5 .57 this is our correct answer we can write p value so in this question we can wear here p value and the degree of freedom we are given a df which is equal to n minus 1 and df which is equal we can write 50 minus 1 which is equal to answer is 49 and let's find out the we know that the value of p in this question we are given here the p value here the p value which is equal we can write p of t and d t and d f so here we can write the value of p of t we are given here minus 1 .571 and the df we are given here is 49 now we just simplify and we can write your answer is so answer here is 0 .0 6139 6139 and the value here this is our p value the level of significance alpha which is equal to we can write is 5 % which means we can write 0 .0 5 and we can say that from here the p value our p value is greater than 0 .0 .0...

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