the mean weight of a 6 year old child is 49.3 lbs. another study indicated that in a sample of 196 six year olds the mean weight was 51.5 lbs. Assume a population standard deviation of 14. use the significance level 0.01 to determine z score and corresponding p value.
I've been stuck on this question for quite some time now, can anyone help me out?
The null hypothesis is that the mean weight is 49.3 lbs, with 14 lbs. standard deviation. The question is, what is the probability that you observed a mean of 51.5 lbs. in a sample of 196 children, if the true mean and standard deviation followed the null hypothesis?
We'll assume that a great many random factors are involved in the weight of any particular child, and that the sampling of children was random, and that their weights are independent; then by appealing to the central limit theorem (and observation), we'd expect the weights of children to be distributed as a normal distribtuion.
Let $X_n\sim\mathcal{N}(\mu,\sigma^2)$ be independent, indentically distributed random variables, for $n=1,2,\ldots N$, each representing the weight of a child. $N=196$, $\mu=49.3$, and $\sigma=14$. What we observed was the sample mean
$$ \hat\mu = \frac{1}{N}\sum_{n=1}^N X_n $$
By the properties of normal random variables, the sample mean is also distributed as a normal, with $\hat\mu\sim\mathcal{N}(\mu,\sigma^2/N)$. Now our question is, what were the chances that we observed a sample mean different than the true mean? How often could we have expected to see this value, or a more extreme value? This is the definition of the $p$-value. $$ p = \Pr[\hat\mu\geq51.5] = \Pr[\mathcal{N}(\mu,\sigma^2/N)\geq51.5] = $$ $$ \Pr\left[\mathcal{N}(0,1)\geq \frac{51.5-49.3}{14/\sqrt{196}}\right] = 1-\Phi(2.2) \approx 1-0.9861 = 0.0139 = 1.39\% $$ So if the null hypothesis was true to begin with, there is slightly more than a $1\%$ chance that we'd observe a sample mean of $51.5$ in a sample of $196$. That's a rare-sounding outcome; I'll leave it to you to figure out if this is significant at the $1\%$ level, and what the z-score is above.