Which table of values would you use to graph y = 3X^2?

A nationwide sample of influential Republicans and Democrats was asked as a part of a comprehensive survey whether they favored

pickupchik [31]

Answer:

We conclude that there is an equal or larger proportion of Republicans in favor of lowering the standards.

Step-by-step explanation:

The complete question is: A nationwide sample of influential Republicans and Democrats was asked as a part of a comprehensive survey whether they favored lowering environmental standards so that high-sulfur coal could be burned in coal-fired power plants. The results were:

Number sampled: 1,000 (republican) , 800 (democrats)

Number in favor: 200 (republican) , 168 (democrats)

At the 0.02 level of significance, can we conclude that there is a larger proportion of Democrats in favor of lowering the standards? Determine the p-value.

Let $p_1$ = proportion of Republicans in favor of lowering the standards.

$p_2$ = proportion of Democrats in favor of lowering the standards.

SO, Null Hypothesis, $H_0$ : $p_1 \geq p_2$ {means that there is an equal or larger proportion of Republicans in favor of lowering the standards}

Alternate Hypothesis, $H_A$ : $p_1 < p_2$ {means that there is a larger proportion of Democrats in favor of lowering the standards}

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 Republicans in favor of lowering the standards = $\frac{200}{1000}$ = 0.20

$\hat p_2$ = sample proportion of Democrats in favor of lowering the standards = $\frac{168}{800}$ = 0.21

$n_1$ = sample of Republicans = 1000

$n_2$ = sample of Democrats = 800

So, the test statistics = $\frac{(0.20-0.21)-(0)}{\sqrt{\frac{0.20(1-0.20)}{1000}+\frac{0.21(1-0.21)}{800} } }$

= -0.52

The value of z test statistics is -0.52.

Now, P-value of the test statistics is given by the following formula;

P-value = P(Z < -0.52) = 1 - P(Z $\leq$ 0.52)

= 1 - 0.6985 = 0.3015

Now, at 0.02 significance level, the z table gives a critical value of -2.054 for left-tailed test.

Since our test statistic is more than the critical value of z as -0.52 > -2.054, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is an equal or larger proportion of Republicans in favor of lowering the standards.

7 0

3 years ago

Jackson is at a collectible fair.He has exactly 8$ and will purchase a collectible card for 5$.What percent of his money will he

makkiz [27]

Answer:

62.5% or 63%

Step-by-step explanation:

He has $8. To find the percent of money that he spent on the purchase you put the amount spent over the total amount.

In this case, he spent $5. You would put that $5 over the $8, 5/8. When you divide, you get 0.625. To get a percent, you would multiply this number by 100. You end up with 62.5, or 63 if you round up.

5 0

3 years ago

Which description best fits the distribution of

Marta_Voda [28]

C is the correct answer I worked it out

5 0

3 years ago

A number written using only words is written using a

Blababa [14]

Answer:

i genuinlly dont know what you mean

Step-by-step explanation:

5 0

3 years ago

PLEASE HELP

Eva8 [605]

Answer: 51 Children and 36 Adults.

Step-by-step explanation: Let's call x the number of children admitted and call z the number of adults admitted.

We know that x+z= 87

We also know that 3.25x+3.5z= 291.75.

We want to find the value of x and z. Then we solve the system of equations:

Multiply the first equation by -3.5 and add it to the second equation:

-3.5x-3.5z= -304.5

3.25x+3.5z= 291.75

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-0.25x=-12.75

x= -42.75 ÷ -0.25

x=51

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Now we substitute the value of x in the first equation and solve for the variable z

51+z=87

z=87-51

z= 36

Omg took me like 10 minutes to calculate would appreciate brainliest. Have a wonderful day.

6 0

3 years ago