Please help me. Please.

It is desired to compare the hourly rate of an entry-level job in two fast-food chains. Eight locations for each chain are rando

notka56 [123]

Answer:

It can be concluded that at 5% significance level that there is no difference in the amount paid by chain A and chain B for the job under consideration

Step by Step Solution:

The given data are;

Chain A 4.25, 4.75, 3.80, 4.50, 3.90, 5.00, 4.00, 3.80

Chain B 4.60, 4.65, 3.85, 4.00, 4.80, 4.00, 4.50, 3.65

Using the functions of Microsoft Excel, we get;

The mean hourly rate for fast-food Chain A, $\overline x_1$ = 4.25

The standard deviation hourly rate for fast-food Chain A, s₁ = 0.457478

The mean hourly rate for fast-food Chain B, $\overline x_2$ = 4.25625

The standard deviation hourly rate for fast-food Chain B, s₂ = 0.429649

The significance level, α = 5%

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

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

The pooled variance, $S_p^2$ , is given as follows;

$S_p^2 = \dfrac{s_1^2 \cdot (n_1 - 1) + s_2^2\cdot (n_2-1)}{(n_1 - 1)+ (n_2 -1)}$

Therefore, we have;

$S_p^2 = \dfrac{0.457478^2 \cdot (8 - 1) + 0.429649^2\cdot (8-1)}{(8 - 1)+ (8 -1)} \approx 0.19682$

The test statistic is given as follows;

$t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{S_{p}^{2} \cdot \left(\dfrac{1 }{n_{1}}+\dfrac{1}{n_{2}}\right)}}$

Therefore, we have;

$t=\dfrac{(4.25-4.25625)}{\sqrt{0.19682 \times \left(\dfrac{1 }{8}+\dfrac{1}{8}\right)}} \approx -0.028176$

The degrees of freedom, df = n₁ + n₂ - 2 = 8 + 8 - 2 = 14

At 5% significance level, the critical t = 2.145

Therefore, given that the absolute value of the test statistic is less than the critical 't', we fail to reject the null hypothesis and it can be concluded that at 5% significance level that chain A pays the same as chain B for the job under consideration

3 0

2 years ago

5x-6y=-15 5x+6y=12 use elimination

Vlad [161]

<span>5x-6y=-15
5x+6y=12
---------------add
10x = -3
x = -3/10
x = -0.3

</span>5x+6y=12
5(-0.3) + 6y = 12
6y = 12 + 1.5
6y = 13.5
y = 13.5 / 6

y = 2.25

4 0

2 years ago

The time spent dancing (minutes) and the amount of calories burned can be modeled by the equation c = 5.5t. Which table of value

pishuonlain [190]

To build the table you just have to give values to t and then calculate the corresponding c a per the model.

If the model is c = 5.5 t this is the table

t    c
   5.5 t

0 5.5(0) = 0

1 5.5(1) = 5.5

2      5.5(2) = 11.0

3 5.5(3) = 16.5

4 5.5(4) = 22.0

5 5.5(5) = 27.5

The viables solutions are all where t is equal or greater than 0. You can even use decimal values for t. You cannot use negative values for t.

5 0

3 years ago

Read 2 more answers

PLEASE HELP AND PLEASE DON'T GUESS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Sindrei [870]

The 8th term is 33 and I don't think any of the options are there

7 0

3 years ago

. Using the Binomial Theorem explicitly, give the 15th term in the expansion of (-2x + 1)^19

timofeeve [1]

Let's rewrite the binomial as:
$(1 - 2x)^{19}$

$\text{Binomial expansion:} (1 + x)^{n} = \sum_{r = 0}^n\left(\begin{array}{ccc}n\\r\end{array}\right) (x)^{r}$

Using the binomial expansion, we get:
$\text{Binomial expansion: } (1 - 2x)^{19} = \sum_{r = 0}^{19}\left(\begin{array}{ccc}19\\r\end{array}\right) (-2x)^{r}$

For the 15th term, we want the term where r is equal to 14, because of the fact that the first term starts when r = 0. Thus, for the 15th term, we need to include the 0th or the first term of the binomial expansion.

Thus, the fifteenth term is:
$\text{Binomial expansion (15th term):} \left(\begin{array}{ccc}19\\14\end{array}\right) (-2x)^{14}$

3 0

3 years ago