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WITCHER [35]
3 years ago
11

The results of a survey indicate that the true proportion of households who want a park in their neighborhood is likely in the i

nterval (0.52, 0.8) . What is the point estimate of the proportion of households who want a park in their neighborhood? Enter your answer, as a decimal, in the box.
Mathematics
2 answers:
GrogVix [38]3 years ago
4 0

Answer:

0.66

Step-by-step explanation:

What are you giving there is a confidence interval. You can obtain a confidence interval based on a sample you got. The length of the confidence interval is determined on how much confidence do you want for your interval (the probability of the real value being inseide the interval) and how big is the sample: the bigger the sample, the smaller the length of the confidence interval. Independently of the sample length, all intervals are centered on the average value you got for the sample, and that is your estimate. In this case, the center of the interval is 0.52+0.8/2 = 0.66.

Diano4ka-milaya [45]3 years ago
4 0

Answer: The point estimate is 0.66

Step-by-step explanation:

The point estimate is the best estimation in a given range. It is usually the midpoint of the range.

Here, we have that the range is (0.52, 0.8) so we need to find the mid point between 0.52 and 0.8

for this, we calculate the difference between these numbers, we divide it by two and then we add it to the smallest number:

d = 0.8 - 0.52 = 0.28

d/2 = 0.28/2 = 0.14

midpoint = 0.52 + 0.14 = 0.66

so the point estimate of the proportion of households that want a park in their neighborhood is 0.66

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Answer:

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⟹cos23x=−sin3x

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⟹sin23x−sin3x−1=0

This is a quadratic equation in sin3x.

sin3x=−(−1)±(−1)2−4×1×(−1)−−−−−−−−−−−−−−−−−√2×1

sin3x=1±5–√2

If x takes real values, the upper sign must be rejected.

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Step-by-step explanation:

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Step-by-step explanation:

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Read 2 more answers
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 .  

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3 years ago
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