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fgiga [73]
3 years ago
11

Please I need help I'm bad at math

Mathematics
2 answers:
Aleonysh [2.5K]3 years ago
5 0
I think the answer is  -44n+2
Mariana [72]3 years ago
5 0
This is what I get

-44n+2
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Step-by-step explanation:

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1. What is the difference between 6 and 3<br> A. 6+3<br> B. 6-3<br> C. 6/3<br> D. 6x3
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⦁ In a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football. Construct a 95% confiden
fomenos

Answer:

95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

Step-by-step explanation:

We are given that in a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football.

Firstly, the pivotal quantity for 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is given by;

        P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } ~ N(0,1)

where, \hat p = proportion of adults in the United States whose favorite sport to watch is football in a sample of 1219 adults = \frac{354}{1219}

           n = sample of US adults  = 1291

           p = population proportion of adults

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5%

                                                    significance level are -1.96 & 1.96}

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } ) = 0.95

<u>95% confidence interval for p</u> = [ \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } ]

                         = [ \frac{354}{1219}-1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} } , \frac{354}{1219}+1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} } ]

                         = [0.265 , 0.316]

Therefore, 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

5 0
3 years ago
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