Which pairs of numbers are opposites? Check all that

A professor wishes to discover if seniors skip more classes than freshmen. Suppose he knows that freshmen skip 2% of their class

KIM [24]

Answer:

We conclude that seniors skip more than 2% of their classes at 0.01 level of significance.

Step-by-step explanation:

We are given that a professor wishes to discover if seniors skip more classes than freshmen. Suppose he knows that freshmen skip 2% of their classes.

He randomly samples a group of seniors and out of 2521 classes, the group skipped 77.

Let p = percentage of seniors who skip their classes.

So, Null Hypothesis, $H_0$ : p $\leq$ 2% {means that seniors skip less than or equal to 2% of their classes}

Alternate Hypothesis, $H_A$ : p > 2% {means that seniors skip more than 2% of their classes}

The test statistics that will be used here is One-sample z proportion statistics;

T.S. = $\frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of seniors who skipped their classes = $\frac{77}{2521}$

n = sample of classes = 2521

So, test statistics = $\frac{\frac{77}{2521} -0.02}{{\sqrt{\frac{\frac{77}{2521}(1-\frac{77}{2521})}{2521} } } } }$

= 3.08

The value of the test statistics is 3.08.

Now at 0.01 significance level, the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is more than the critical value of z as 2.3263 < 3.08, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that seniors skip more than 2% of their classes.

6 0

3 years ago

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

Evaluate the expression: v ? w Given the vectors: r = <8, 1, -6>; v = <6, 7, -3>; w = <-7, 5, 2> . If the answ

Snezhnost [94]

Answer:

-13

Step-by-step explanation:

Given are two vectors v and w.

v=(6,7,-3) and

w=(-7,5,2)

We are to find the dot product of v.w

We have

Dot product is obtained by multiplying corresponding pairs and adding them

Here we have

v.w=6(-7)+7(5)-3(2)

=-13

7 0

3 years ago

3x - 8 = 13 + -2(x - 8)

valentinak56 [21]

Answer:

37/5 or 7.4 or 7 2/5

Step-by-step explanation:

I hope this helps! Have a great day!!

Skye~

3 0

2 years ago

The striton family had a meal catered for a wedding rehearsal dinner.the cost of the dinner was 476.there was a 5% sales tax and

Leokris [45]

476.00×5%=23.80 tax=476+23.80=499.80
499.80×15%=74.97
499.80+74.97=574.77

3 0

3 years ago