Evaluate the following

What best summarizes the data?

yanalaym [24]

Answer:

what the question there is no picture

Step-by-step explanation:

8 0

3 years ago

A manufactured lot of buggy whips has 20 items, of which 5 are defective. A random sample of 5 items is chosen to be inspected.

dimulka [17.4K]

Answer:

a. 39.55%

b. 44.02%

Step-by-step explanation:

We have the following data:

n = 5

x = 1

p = 5/20 = 0.25

to. If the sampling is done with replacement.

We apply the binomial distribution formula, which is as follows:

P = nCx * (p ^ x) * ((1-p) ^ (n-x))

Where nCx, is a combination, and is equal to:

nCx = n! / x! * (n-x)!

replacing we have:

5C1 = 5! / 1! * 4! = 5

replacing in the main formula:

P = 5 * (0.25 ^ 1) * ((1- 0.25) ^ (5-1))

P = 0.3955

that is, without replacing the probability is 39.55%

b. if the sampling is done without replacement.

Here it is a little different from the previous one, but what you should do is calculate three cases,

the first was the one at point a, when n = 5 and x = 1

5C1 = 5! / 1! * 4! = 5

the second is when n = 20 and x = 5, this is all possible scenarios.

20C5 = 20! / 5! * 15! = 15504

and the third is when n = 15 (20-5) and x = 4 (5-1), which corresponds to the cases when none were damaged

15C4 = 15! / 4! * 11! = 1365

In the end, it would be:

P = (5C1 * 15C4) / 20C5

Replacing:

P = 5 * 1365/15504

P = 0.4402

Which means that without replacing the probability is 44.02%

7 0

3 years ago

Consider the function represented by the equation y-6x-9=0. Which answer shows the equation written in function notation with x

hichkok12 [17]

Y=6X+9 the A is #1
all you need to do is simp the equation

4 0

3 years ago

Read 2 more answers

Solutions

Paraphin [41]

D. none of the above.

3 0

3 years ago

In the past, the average age of employees of a large corporation has been 40 years. Recently, the company has been hiring older

Viktor [21]

Answer:

$p_v =P(t_{(63)}>2.5)=0.0075$

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 fail reject the null hypothesis, so we can conclude that the mean age is significantly higher than 45 years at 5% of significance.

Step-by-step explanation:

1) Data given and notation

$\bar X=45$ represent the mean height for the sample

$s=16$ represent the sample standard deviation for the sample

$n=64$ sample size

$\mu_o =40$ 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 age is higher than 40 years, the system of hypothesis would be:

Null hypothesis: $\mu \leq 40$

Alternative hypothesis: $\mu > 40$

If we analyze the size for the sample is < 30 and 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:

$t=\frac{45-40}{\frac{16}{\sqrt{64}}}=2.5$

P-value

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

$df=n-1=64-1=63$

Since is a one right tailed test the p value would be:

$p_v =P(t_{(63)}>2.5)=0.0075$

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 fail reject the null hypothesis, so we can conclude that the mean age is significantly higher than 45 years at 5% of significance.

6 0

2 years ago