Solve for the missing information, given that the two triangle in each question are similar

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

3 years ago

student randomly receive 1 of 4 versions(A, B, C, D) of a math test. What is the probability that at least 3 of the 5 student te

alexdok [17]

Answer:

1.2%

Step-by-step explanation:

We are given that the students receive different versions of the math namely A, B, C and D.

So, the probability that a student receives version A = $\frac{1}{4}$ .

Thus, the probability that the student does not receive version A = $1-\frac{1}{4}$ = $\frac{3}{4}$ .

So, the possibilities that at-least 3 out of 5 students receive version A are,

1) 3 receives version A and 2 does not receive version A

2) 4 receives version A and 1 does not receive version A

3) All 5 students receive version A

Then the probability that at-least 3 out of 5 students receive version A is given by,

$\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{3}{4}$ + $\frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}\times \frac{1}{4}$