Mathematics, 19.03.2021 06:00 squawk1738

The director of training for an equipment manufacturing company is interested in determining whether different training methods have an affect on the productivity of assembly line employees. She randomly assigns 42 newly hired employees into two groups of 21. The first group receives an on-line training program and the second receives a team based training program. At the end of the training, the employees are evaluated on the time (in seconds) that it takes to assemble a part. The results follow. A hypothesis test is made to see if there is a significant difference in the mean assembly times. What is the null hypothesis?On-Line19.4 16.7 20.7 19.3 21.8 16.8 14.1 17.7 16.1 19.8 16.819.3 14.7 16.0 16.5 17.7 16.2 17.4 16.4 16.8 18.5Team22.4 13.8 18.7 18.0 19.3 20.8 15.6 17.1 18.0 28.2 21.720.8 30.7 24.7 23.7 17.4 23.2 20.1 12.3 15.2 16.0

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The null hypothesis is that there is no difference between the mean time it takes an online trained employee or a team-based trained employee to assemble the given part

Step-by-step explanation:

The information the director of the equipment manufacturing company is interested in determining is weather the productivity of assembly line employees is affected by the method used in their training

The total number newly hired employees in the sample = 42

The number of newly hired employees that receive training online = 21

The number of newly hired employees that receive training in a team = 21

The given data of the result of the time it takes an employee to assemble a part is presented as follows;

On-Line 19.4, 16.7, 20.7, 19.3, 21.8, 16.8, 14.1, 17.7, 16.1, 19.8, 16.8, 19.3, 14.7, 16.0, 16.5, 17.7, 16.2, 17.4, 16.4, 16.8, 18.5

Team; 22.4, 13.8, 18.7, 18.0, 19.3, 20.8, 15.6, 17.1, 18.0, 28.2, 21.7, 20.8, 30.7, 24.7, 23.7, 17.4, 23.2, 20.1, 12.3, 15.2, 16.0

The mean time for the of the on-line trained employee, $\overline x_1$ = 17.55714

The standard deviation of the time for the of the on-line trained employee, s₁ = 1.93328

The mean time for the of the team based trained employee, $\overline x_2$ = 19.89048

The standard deviation of the time for the of the team based trained employee, s₂ = 4.5667

The null hypothesis, H₀: $\overline x_1$ = $\overline x_2$

The alternative hypothesis, Hₐ: $\overline x_1$ ≠ $\overline x_2$

$t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}-\dfrac{s _{2}^{2}}{n_{2}}}}$

Therefore, we have;

$t=\dfrac{(19.89048 - 17.55714)}{\sqrt{\dfrac{4.57667^{2}}{21} - \dfrac{1.93328^{2} }{21}}} \approx 2.5776$

The degrees of freedom, df = n₁ - 1 = 21 - 1 = 20

At 95% confidence level, we have α = 1 - 0.95 = 0.05, and t = 2.086

Therefore, given that the test statistic is larger than the critical 't' value, we reject the null hypothesis. There is sufficient statistical evidence to show that there is a difference between the mean time of assembly value for the on-line trained and team-based employee

Answer from: Quest

the simplified form is

$\frac{x^2(2x+3)}{2(4x-3)}$

step-by-step explanation:

we want to simplify the expression,

$\frac{\frac{6x+9}{15x^2} }{\frac{16x-12}{10x^4} }$

let us change the middle bar to a normal division sign to obtain,

$\frac{6x+9}{15x^2}\div\frac{16x-12}{10x^4}$

we multiply the first fraction by the reciprocal of the second fraction to obtain,

$\frac{6x+9}{15x^2}\times \frac{10x^4}{16x-12}$

we factor to obtain,

$\frac{3(2x+3)}{3\times5\times x^2}\times \frac{2\times5\times x^2\times x^2}{4(4x-3)}$

we cancel the common factors to get,

$\frac{(2x+3)}{1}\times \frac{ x^2}{2(4x-3)}$

we simplify to get,

$\frac{x^2(2x+3)}{2(4x-3)}$

the correct answer is b

Answer from: Quest

2pounds will cost $1.96

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