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3 years ago

Random/probability sampling is when everyone in the group you are studying has an equal chance of participating.

1 answer:

7 0

Probability sampling is when everyone in the group you are studying has an equal chance of participating.

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PLEASE HELP, I NEED THIS OR I WILL FAIL MY CLASS PLZZZZ.

irga5000 [103]

Answer:

Step-by-step explanation:

i dont know how to show the work i did it in my head

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2 years ago

Which model represents the factors of x^2 + 9x+8?

grandymaker [24]

Answer:

Step-by-step explanation:

x² + 9x+8 = (x+1)(x+8)

6 0

3 years ago

Plsss help (Cuz i'm dumb:D) Calculate the surface area of the prism below, shape below

marin [14]

Surface area of the the rectangles:
8 * 5 + 8 * 4 + 8 * 3 = 96

Surface area of the two triangles:
Use bh/2 base time height over 2
(4 * 3)/2 = 6
There are 2 triangles, 2 * 6 = 12

12 + 96 = 108

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3 years ago

Guided Practice

Fittoniya [83]

y=17,000 · 0.90xy equals 17,000 times 0.90 superscript x baseline

Step-by-step explanation:

5 0

3 years ago

2. The National Safety Council routinely analyzes the benefit of seat belt use on driver safety. Their data showed that among 28

JulijaS [17]

Answer:

We conclude that there is difference in the proportion of deaths between the 2 groups.

Step-by-step explanation:

We are given that among 2823 drivers not wearing seat belts, 31 died as a result of injuries, and among 7765 drivers wearing seat belts 16 were killed.

Let $p_1$ = proportion of deaths when drivers were not wearing seat belts.

$p_2$ = proportion of deaths when drivers were wearing seat belts.

So, Null Hypothesis, $H_0$ : $p_1=p_2$ {means that there is no difference in the proportion of deaths between the 2 groups}

Alternate Hypothesis, $H_A$ : $p_1\neq p_2$ {means that there is difference in the proportion of deaths between the 2 groups}

The test statistics that would be used here Two-sample z test for proportions;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of deaths when drivers were not wearing seat belts = $\frac{31}{2823}$ = 0.011

$\hat p_2$ = sample proportion of deaths when drivers were wearing seat belts = $\frac{16}{7765}$ = 0.002

$n_1$ = sample of drivers not wearing seat belts = 2823

$n_2$ = sample of drivers wearing seat belts = 7765

So, the test statistics = $\frac{(0.011-0.002)-(0)}{\sqrt{\frac{0.011(1-0.011)}{2823}+\frac{0.002(1-0.002)}{7765} } }$

= 4.438

The value of z test statistics is 4.438.

Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistic doesn't lie within the range of critical values of z, 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 there is difference in the proportion of deaths between the 2 groups.

Also, Margin of error (E) = $1.96 \times \sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} }$

= $1.96 \times \sqrt{\frac{0.011(1-0.011)}{2823}+\frac{0.002(1-0.002)}{7765} }$

= 0.00397

5 0

3 years ago