Did I put the letters in the right places?

A random sample of 25 ACME employees showed the average number of vacation days taken during the year is 18.3 days with a standa

Norma-Jean [14]

Answer:

a) Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

b) $df=n-1=25-1=24$

For this case the p value is given $p_v = 0.0392$

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, so we can conclude that the true mean is higher than 15 at 5% of signficance.

c) Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

Step-by-step explanation:

Data given and notation

$\bar X=18.3$ represent the sample mean

$s=3.72$ represent the sample standard deviation

$n=25$ sample size

$\mu_o =15$ 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)

Part a: State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean for vacation days is higher than 15, the system of hypothesis would be:

Null hypothesis: $\mu \leq 15$

Alternative hypothesis: $\mu > 15$

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".

Part b: P-value and conclusion

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

$df=n-1=25-1=24$

For this case the p value is given $p_v = 0.0392$

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, so we can conclude that the true mean is higher than 15 at 5% of signficance.

Part c

Type I error, also known as a “false positive” is the error of rejecting a null hypothesis when it is actually true. Can be interpreted as the error of no reject an alternative hypothesis when the results can be attributed not to the reality.

So for this case a type I of error would be reject the hypothesis that the true mean is less or equal than 15 and is actually true.

3 0

3 years ago

What is the volume of a cone (in cubic inches) with a radius of 6 inches and a height of 3 inches? Use 3.14 for π. Round your an

ICE Princess25 [194]

When you square the radius = 6×6= 36. Multiply 36×3×3.14=339.12. Using the formula, you then multiply 1/3×339.12=339.12/3= 113.04 in^3. Final answer is 113.04^3

4 0

3 years ago

Your plant runs two assembly lines. The first line produces 1,250 units a day and the second produces 2,825 units a day. What is

Step2247 [10]

Answer:

$\frac{50}{113}$

Step-by-step explanation:

The ratio is simply the division of one by another (and simplification).

Hence, <u>the ratio of production of the first line to the second line</u> is 1250 divided by 2825. So,

Ratio = $\frac{1250}{2825}=\frac{50}{113}$

5 0

3 years ago

Read 2 more answers

What Is The Prime Exponent Form Of 192

Mandarinka [93]

Answer:

2^6 * 3

Step-by-step explanation:

3 0

2 years ago

An icicle drips at a rate that can be represented by the function f(x) = −x2 + 8x − 7, where 0 ≤ x ≤ 10 and x is the number of h

mart [117]

Find when f(x) = 0. By factorizing, we have

-x² + 8x - 7 = - (x - 7) (x - 1) = 0

so either x - 7 = 0 or x - 1 = 0, or simply x = 1 or x = 7.

Split the domain into three intervals, and check the sign of f(x) over each interval at some test point in the interval. The sign will tell you when the icicle is dripping (positive) or not (negative).

• [0, 1) : at x = 0, we have f(0) = -7 < 0

• (1, 7) : at x = 2, we have f(2) = 5 > 0

• (7, 10] : at x = 8, we have f(8) = -7 < 0

This tells us that the icicle is not dripping when 0 ≤ x < 1, begins to drip when x = 1, drips for the duration of 1 < x < 7, stops dripping when x = 7, and keeps on not dripping when 7 < x ≤ 10.

3 0

2 years ago