Question

Mar in fish were int in AL The level of so mercury in seafood allowed by FDA is 1ppm (1 part of mercury a million parts seafood). A sample of n=5 caught particular region, the sample mean level of mercury is found to be x =.995 pom, and sample standard deviation is found to be ss=.01. Let r be amount of mercury in all such fish, and assume that the distribution of levels of all such fish is normal. the true mean the mercury

Answer 1

Let $\mu$ be the true mean amount of Mercury in all fish caught in a particular region

We want to test if $\mu<1$

The hypotheses are

$\begin{align*} &H_0:\mu=1\\ &H_a:\mu<1\\ \end{align*}$

We have the following sample information

n=5 is the sample size

ppm is the sample mean amount of Mercury for the fish in the sample

T= 0.995

s=0.01 is the sample standard deviation of amount of Mercury in all fish caught in a particular region

1) This is a left tailed test (The alternative hypothesis has "<")

The p-value is the area under the left tail

$\text{p-value}=P(\bar{X}<0.995)$, given that the hypothesized value of the mean of is 1 (that is the null hypothesis is true)

X

p-value is given as 0.167. This indicates that if the null hypotheses were true, the chance of obtaining a sample of size 5 with a mean of 0.995 or less is 0.167

ans:

actual took D) If the mean amount of mercury in all such fish is 1ppm, and we many random samples of size n=5 under the same conditions as as the ones we na have observed, then about 16.7% of the time, we would obtain data with sample mean less than or equal to 995ppm.

2) We will reject the null hypothesis, if the p-value is less than the significance level $\alpha=0.1$

Here, the p-value is 0.167 and it is greater than 0.1. Hence we do not reject the null hypothesis.

ans:

A) fail to reject Ho since the p-value greater than a.