The mean age of women in the US giving birth for the first t

The mean age of women in the US giving birth for the first time is believed to be 25.6 years old with a standard deviation of 5 years. With 95% confidence, we look at a random sample of 100 first time mothers to test the theory that the mean age might be older than 25.6 years. (14 Points)

State the null and alternative hypotheses:

Calculate the value(s) of the sample mean that would cause us to reject H_o

Suppose the actual mean age is 27.2 years. Calculate

Solution

the null hypothesis is always going to be that the variable mean equals what we think it is, and then the alternative hypothesis is going to always be the mean is either less than, equal to or greater than the mean we thought. So in this instance the null hypothesis is going to be that the mean age of women in the US giving birth for the first time is 25.6 years old. and the alternative hypothesis is going to be that the mean age of women in the US giving birth for the first time is going to be older (greater than) 25.6 years old.

Now that we know our null and alternative hypothesis, the next question is asking us to find the value of the sample that would cause us to reject the null in favor of the alternative. Since we are looking at the mean age being greater than the value we have a 1 sided test, so for 95% confidence we want 95% of the data (0.95) to be below the value that we are going to try and find. Therefore we would use a z table like this (http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf) to find the z score that corresponds to a probability of 0.95 and we see that this is 1.645. Now we want to use the formula: z = (x - mean) / (SD/sqrt(sample size)) to find our x value. So we sub in what we know and solve. so we have: 1.645 = (x - 25.6) / (5/sqrt(100)) and get that x = 26.4225 years old.

Now the final part of this question tells us that the actual mean is 27.2 years old and is asking for us to find beta. Beta is the probability of getting a type II error or the probability of accepting the null when it is false. this site explains all about what beta and a type II error is and how to calculate it: (https://www.easycalculation.com/statistics/learn-beta-error.php). The simple way to go about calculating this is to solve it like a normal z function and then the p value of getting greater than this is going to be our result. this is because the p value greater than is saying the probability that we get something that is greater than our mean (should reject) and we assume that we do accept it. So we solve: (27.2 - 25.6) / (5/sqrt(100)) = 3.2 and using the z table from above we see that this gives us a p value of 6.872 * 10^-4. this would be our beta value.