Evaluate.<br><br>1,000 squared 5 =

A newspaper wants to estimate the proportion of Americans who will vote for Candidate A. A random sample of 1000 voters is selec

masya89 [10]

Answer:

the excel that would be used to construct a confident interval estimate is (A) NORM.S.INV

Step-by-step explanation:

NORM>S>INV returns the of the standard normal cumulative distribution.

4 0

3 years ago

Please helpppppp

melomori [17]

Given Information:

Years = t = 35

Semi-annual deposits = P = $2,000

Compounding semi-annually = n = 2

Interest rate = i = 6.5%

Required Information

Accumulated amount = A = ?

Answer:

Accumulated amount = $515,827

Step-by-step explanation:

The future value of amount earned over period of 35 years and interest rate 6.5% with semi-annual deposits is given by

FV = PMT * ((1 + i/n)^nt - 1)/(i/n))

Where

n = 2

i = 0.065

t = 35

FV = 2000*((1 + 0.065/2)^2*35 - 1)/(0.065/2))

FV = 2,000*(257.91)

FV ≈ $515,827

Therefore, Anthony will have an amount of $515,827 when he retires in 35 years.

6 0

3 years ago

Select the correct answer.

xeze [42]

The answer is (D.50%)

6 0

3 years ago

A TV station claims that 38% of the 6:00 - 7:00 pm viewing audience watches its evening news program. A consumer group believes

Elena L [17]

Answer:

$z=\frac{0.340 -0.38}{\sqrt{\frac{0.38(1-0.38)}{830}}}=-2.374$

$p_v =P(Z$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that regularly watch the TV station’s news program is significantly less than 0.38 .

Step-by-step explanation:

1) Data given and notation

n=830 represent the random sample taken

X=282 represent the people that regularly watch the TV station’s news program

$\hat p=\frac{282}{830}=0.340$ estimated proportion of people that regularly watch the TV station’s news program

$p_o=0.38$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the proportion is less than 0.38:

Null hypothesis: $p\geq 0.38$

Alternative hypothesis: $p < 0.38$

When we conduct a proportion test we need to use the z statisitc, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.340 -0.38}{\sqrt{\frac{0.38(1-0.38)}{830}}}=-2.374$

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(Z$

So the p value obtained was a very low value and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that regularly watch the TV station’s news program is significantly less than 0.38 .

3 0

3 years ago

1. It is a three-digit whole number.

taurus [48]

Answer:

660

Step-by-step explanation:

4x5x11x3 = 660

8 0

3 years ago

Evaluate.1,000 squared 5 =