Do employees perform better at work with music playing? The music was turned on during the working hours of a business with 45 employees. Their productivity level averaged 5.2. On a different day the music was turned off and there were 40 workers. The workers' productivity level averaged 4.8. What can we conclude at the .01 level, if computed p-value is 0.12?\

We can conclude that there is no enough evidence that the employees perform better with music playing.

Step-by-step explanation:

In this test, where it is claimed that the employees perform better with music playing, the null hypothesis has to be that there is no difference in performance.

If the P-value is 0.12, at a significance level of α=0.01, the effect is not significant and the null hypothesis failed to be rejected.

If the null hypothesis is not rejected, we can conclude that there is no enough evidence that the employees perform better with music playing.

Performing the hypothesis test at a significance level of 0.01, a p-value of 0.12 is greater than the significance level of 0.01, Hence, we fail to reject the null hypothesis and conclude that there is no difference in the productivity of the workers when music is on and when it is off.

Step-by-step explanation:

For hypothesis testing, the first thing to define is the null and alternative hypothesis.

The null hypothesis plays the devil's advocate and is always about the absence of significant difference between two variables being compared.

While, the alternative hypothesis takes the other side of the hypothesis; that there is indeed a significant difference between two proportions being compared.

For this question, we are comparing the effect of music on productivity of the employees of the firm.

We want to check if there is a significant difference between the productivity when the music is on and when the music is off.

If the productivity when the music is on = p₁

And

The productivity when the music is off = p₂

And the difference between them is given as

μ₀ = p₁ - p₂

The null hypothesis and alternative hypothesis can be expressed mathematically as

The null hypothesis that there is no significant difference between the productivity of the workers in the firm when music is on and when the music is off.

H₀: μ₀ = 0

or

H₀: p₁ = p₂

And the alternative hypothesis that there is a significant difference between the productivity of the workers in the firm when music is on and when the music is off.

Hₐ: μ₀ ≠ 0

or

Hₐ: p₁ ≠ p₂

So, to interpret the p-value now,

The p-value is 0.12, and they conducted this test at a 1% level of significance. What does a p-value of 0.12 mean?

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 1% = 0.01

p-value = 0.12

0.12 > 0.01

Hence,

p-value > significance level

This means that we fail to reject the null hypothesis & conclude that there is no significant difference between the productivity of the workers in the firm when music is on and when the music is off.

Hope this Helps!!!

8 apples are left.

step-by-step explanation:

10-2=8

10 apples

take 2

now u have 8

hello from mrbilldoesmath!

answer:   x (x+6)^2, choice b

discussion:

factor x out of the equation:

x^3 + 12x^2 + 36x = x (x^2 + 12x + 36).

note that 6+6 = 12 and 6*6 = 36 so the above factors into:

x (x+6)^2

which is choice b

you,

mrb