b = 2
x = 4
c = 4
k =7
I'm sorry, I think you have accidentally left out the variable in the question. I can't solve it because there's no variable to solve.
v = 1

There is no specific instruction in your question. I try to answer it based on what I understand from the equations. Hope this helps!

Step-by-step explanation:

Just rearrange the equation to solve the variable.

8 0

3 years ago

A random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 11 and the sample stand

Tpy6a [65]

Answer:

We conclude that the population mean is different from 10.5.

Step-by-step explanation:

We are given that a random sample of 16 values is drawn from a mound-shaped and symmetric distribution. The sample mean is 11 and the sample standard deviation is 2.

We have to test the claim that the population mean is 10.5.

Let, NULL HYPOTHESIS, $H_0$ : $\mu$ = 10.5 {means that the population mean is 10.5}

ALTERNATE HYPOTHESIS, $H_a$ : $\mu \neq$ 10.5 {means that the population mean is different from 10.5}

The test statistics that will be used here is One-sample t-test;

T.S. = $\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = 11

s = sample standard deviation = 2

$\mu$ = population mean

n = sample of values = 16

So, test statistics = $\frac{11-10.5}{\frac{2}{\sqrt{16} } }$ ~ $t_1_5$

= 1

Now, at 0.05 significance level, t table gives a critical value of 2.131 at 15 degree of freedom. Since our test statistics is way less than the critical value of t so we have insufficient evidence to reject null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the population mean is different from 10.5.

6 0

4 years ago