00:01
From the data which is given to us in the question, we say that the mean is going to be equal to 180, x is equal to 179 .1, s, that is the standard deviation is 10 .9.
00:15
Alpha is equal to 0 .05 and n is equal to 1 ,500 multiplied by 0 .02, which is equal to 30.
00:26
Now the degree of freedom, that is df, that is the number of observations that are free to vary, calculated by the formula that is n minus 1 so degree of freedom is n minus 1 which is equal to 30 minus 1 hence 29 that is a degree of freedom in the first part the null an alternative hypothesis can be stated as h0 is going to be first of all the null hypothesis h0 mu greater than equal to 180 then we will have h alpha where mu is less than 180 so this is the null and alternative hypothesis, right? so moving forward, the second one, we say the test statistics used in this case will be the t test.
01:14
So here what is the test statistic used? it is going to be t test.
01:17
Why? because here what can happen is the population standard deviation is unknown.
01:23
The population standard deviation is unknown.
01:40
Hence we use the t test.
01:41
Now the critical value, which is equal to minus 1 .699.
01:45
So critical value is equal to minus 1 .699.
01:53
So using t critical value with table with alpha is equal to 0 .005 and degrees of freedom as 29, we are going to get the t value as this one because in the table for alpha is equal to 0 .05 and df is equal to 29.
02:09
The t test value from the table is going to be 1 .699.
02:13
That is a critical value for us.
02:14
Then in the third subpart, if t is less than 1 .699, then we can say we can reject the null hypothesis...