HALP! EASY MATH (not for me obviously) <br>show work plz<br>halp!

A random sample of n = 40 observations from a quantitative population produced a mean x = 2.2 and a standard deviation s = 0.29.

zmey [24]

Answer:

$t=\frac{2.2-2.1}{\frac{0.29}{\sqrt{40}}}=2.18$

$df=n-1=40-1=39$

Since is a one side test the p value would be:

$p_v =P(t_{(39)}>2.18)=0.0177$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and then the population mes seems to be higher than 2.1 at 5% of significance

Step-by-step explanation:

Data given and notation

$\bar X=2.2$ represent the sample mean

$s=0.29$ represent the sample standard deviation

$n=40$ sample size

$\mu_o =2.1$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 2,1, the system of hypothesis would be:

Null hypothesis: $\mu \leq 2.1$

Alternative hypothesis: $\mu > 2.1$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{2.2-2.1}{\frac{0.29}{\sqrt{40}}}=2.18$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=40-1=39$

Since is a one side test the p value would be:

$p_v =P(t_{(39)}>2.18)=0.0177$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and then the population mes seems to be higher than 2.1 at 5% of significance

7 0

3 years ago

In a class 20 of students, 7 have a cat and

blsea [12.9K]

Answer:

$\frac{3}{20}$

Step-by-step explanation:

Since there are 20 students, this will be the denominator.

Now, to find how many students have a dog and a cat, you need to subtract all the numbers with 20. The leftover number will be the amount of students.

7+8+8=23

23-20=3

So, there are 3 students who have a dog and a cat.

So, the fraction will be $\frac{3}{20}$

8 0

2 years ago

Look at the table. Make a conjecture about the sum of the first 10 positive even numbers.

inysia [295]

The answer is 10•11, if you look at the pattern, you don't have to solve the addition.