00:01
So our question says what do you think is the ideal number of children for a family to have? in group a, the 71 women who responded had a mean of 2 .65 and the standard deviation of 1 .09.
00:13
We're supposed to click the pay value and test the null hypothesis at 5 % level of significance.
00:19
In group b, we have 66 women in the same age range and answered 2 .9, which is standard deviation of 1 .3.
00:28
We are supposed to perform our hypothesis test so let's try to extract out of details we have group a and we have group b so for group a we have our 71 women so let's call group a also group one and group b group two so n1 is across to 71 our x our x bar 1 is 2 .65 and our s one is actually 1 .09.
01:09
N2 is equal to 66.
01:13
So we have the mean to be equal to 2 .9.
01:17
The x2 bar is equal to 2 .9.
01:21
And the standard deviation is to be equals to 1 .3.
01:27
So the null hypothesis h not is the fact that m1 is across to m .2, both groups are actually similar.
01:34
And the alternative hypothesis is that m1 is not equals to mute so that is both groups are not similar so the next step is for us to state the decision rule that is going to either make us accept or reject the null hypothesis and we have it here as the decision rule and our decision rule states that if the p value is greater than alpha we fail to reject the null hypothesis and if the p value is lesser than alpha we actually reject the null hypothesis please not study alpha is equals to 5%.
02:10
Now that we've stated, the session rule is time for us to get the test statistics because the test statistics is going to provide the p value.
02:19
Now, judging by the central limit theorem when the sample size of a sampling distribution is greater than or equals to 30, we can definitely use a, we can definitely say that the data set is normally distributed.
02:31
So at a sample size of 71 and 66 for each of this group, for each of the group, then we can simply say that our data set is normally distributed and our test statistics is going to be a z test so we have that z is equal to x1 bar minus x2 bar divided by the square root of s1 squared divided by n1 plus s2 squared divided by n2 so let's substitute the parameter we have 2 .65 minus 2 .9 divided by the square root of 1 .09 squared squared divided by 7 to 1 plus 1 .3 squared divided by 66.
03:15
So 2 .65 minus 2 .9, that gives us minus 0 .25 divided by, so we have the square root of 1 .09 squared divided by 71 plus 1 .3 squared divided by 66, and we have that to be 0 .2057...