Question

NO LINKS!! Please help me with this problem 1g​

Answer 1

Answer: Choice D

Do not reject the null hypothesis; these fish are not larger than usual.

=================================================

Step-by-step explanation:

Here are the common variables we run into with stats problems

• mu = population mean
• sigma = population standard deviation
• xbar = sample mean
• n = sample size

The status quo is that mu = 57, which is implied to have been found through previous studies. Also through prior studies, researchers found that sigma = 6.3

Then a new study is conducted which found that xbar = 58.5 on a sample size of n = 45

The question your teacher asks: is mu = 57 too small? Should mu be larger? That's what the hypothesis test below will try to find out.

----------------------

Let's set up the hypotheses

• Null: mu = 57
• Alternate: mu > 57

The claim is that the weight seems to be larger than 57 lbs. We have a right tailed test.

Since we know sigma, we can use the standard normal Z distribution.

Let's find the z score

z = (xbar - mu)/(sigma/sqrt(n))

z = (58.5 - 57)/(6.3/sqrt(45))

z = 1.59719141249984

z = 1.60 approximately

Convention usually has the z scores rounded to two decimal places, so we can use them in a table. Such a table is found in the back of many stats textbooks in the appendix section. If you don't have a textbook in front of you, then search out "z table" online.

When using a Z table, we find that

P(Z < 1.60) = 0.94520

so,

P(Z > 1.60) = 1 - P(Z < 1.60)

P(Z > 1.60) = 1 - 0.94520

P(Z > 1.60) = 0.0548

This value is approximate representing the area under the standard normal distribution curve to the right of z = 1.60

This is the p-value of our hypothesis test.

A significance level (alpha) was never mentioned, so I'll assume the default of alpha = 0.05 is implied.

Rule: If the p-value is smaller than alpha, then reject the null.

We found the p-value was roughly 0.0548 which is not smaller than alpha = 0.05, so we fail to reject the null.

Meaning we cannot overturn the null and we must accept it for now until further evidence comes along to disprove it.

Conclusion: Do not reject the null hypothesis; these fish are not larger than usual. This points us to the answer of choice D

---------------------

Extra info:

The xbar = 58.5 is larger than mu = 57, but it's not "large enough" to be considered statistically significant. Recall that we're dealing with a random variable so xbar won't always land on the mu value. The xbar will fluctuate around mu. The xbar value might be a little high or a little low compared to mu. If xbar was large enough to be considered significant (causing the p-value to be very small) then we would reject the null at that point.

We're always testing the null. The two outcomes are either "reject the null" or "fail to reject the null".

Answer 2

D. Do not reject the null hypothesis; these fish are not larger than usual.

Step-by-step explanation:

Let μ be the population mean weight of the salmon.

We are testing to see if the mean weight has increased, therefore the hypotheses are:

`\boxed{\textsf{H}_0: \mu = 57 \quad \quad \textsf{H}_1: \mu > 57}`

So this is a one-tailed test.

There is no significance level stated in the question, so use 5%.

`\implies \alpha= 0.05`

`\boxed{\begin{minipage}{10 cm}If $X \sim \sf N(\mu,\sigma^2)$, then $\overline{X} \sim \left(\mu,(\sigma^2)/(n)\right) \implies Z=\frac{\overline{X}-\mu}{\sigma / √(n)} \sim \textsf{N}(0,1)$\\\\\\Then the value of the test statistic will be: \quad $z=\frac{\overline{x}-\mu}{\sigma / √(n)}$\\\end{minipage}}`

Given:

• 
  `\mu=57`
• 
  `\sigma=6.3`
• 
  `\overline{x}=58.5`
• 
  `n=45`

Find the value of the test statistic (z):

`\overline{x}=58.5\implies z=\frac{\overline{x}-\mu}{\sigma / √(n)}=(58.5-57)/(6.3/√(45))=1.5971...`

Finding the Critical Region

This is a one-tailed test and the critical region will be at the upper end of the distribution.

So the critical value is z such that P(Z > z) = 0.05.

Using a calculator or statistical tables, the critical value is z = 1.6449 (4 d.p.) meaning the critical region is Z > 1.6449.

Since z = 1.5971... < 1.6449, the observed value of the test statistic lies outside the critical region. So the result is not significant.

Therefore, there is insufficient evidence at the 5% level of significance to reject H₀ and to support the alternative hypothesis that the mean weight of the salmon has increased.