The Weather Underground reported that the mean amount of...

The Weather Underground reported that the mean amount of summer rainfall for the northeastern US is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the $\alpha=0.05$ level, can it be concluded that the mean rainfall was below the reported average? What if $\alpha=0.01 ?$ Assume the amount of summer rainfall follows a normal distribution.

Transcript

00:01 So in this case of us, we are supposed to use an hypothesis test to test for the mean rainfall in a specific region and conclude whether the mean rainfall is actually below the average or is above the average.

00:12 So the initial belief is that the amounts of rainfall in the summertime in the north -eastern usa is at least 11 .52 inches.

00:24 So that means we are saying our population mean that's h -nought.

00:27 It's based on the fact that mule is at least 11 .52 inches like that's the least amount of frame so the alternative hypothesis is going to be mu is lesser than 11 .52 so the first step is to state our decision we are going to be using the p value mentor to get our decision and conclusion so that means if your p value is greater than your alpha then you are going to accept the null hypothesis if your p value is lesser than your alpha you are going to reject the null hypothesis so so before you can be able to get up a value we need our test statistics and our test statistics in this case is going to be a t test and the reason being the fact that our sample size which is 10 is actually lesser than 30 and also we are making the sample standard deviation instead of the population standardization which is unknown here so that means you are making some test so our t test formula is equal to x bar minus m.

01:30 Divided by zigma divided by the square root of n so our x bar is 7 .42 minus our population mean is 11 .52 divided by sorry it this is not zigma this is s rather pardon me for that so our s is going to be 1 .3 divided by square root of n our n is 30 so we have that t is equal to seven points so i'll be using my calculator to do uh 7 .42 minus 11 .52 that gives us minus 4 .1 divided by so we have 1 .3 divided by square root of 30 so we have 0 .2373 so if we use our calculator to do this we have minus 4 .1 divided by 0 .2 373 so we have our our test statistics to be minus 17 .277 so this is our test statistics here so the next thing is to get a pay value now whenever i want to get the pay value of any kind of distribution we can decide to use the distribution is of the distribution table in this case a t table or we can decide to use calculator but because of the fact that the t table is not accurate when it comes to checking pay value i'll be making use of software to get a p value of this question of this data set or distribution so we come to a calculator here.

03:09 So this is my calculator.

03:11 So i press refresh...

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