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stealth61 [152]
3 years ago
6

A town council wants to estimate the proportion of residents who are in favor of a proposal to upgrade the computers in the town

library. A random sample of 100 residents was selected, and 97 of those selected indicated that they were in favor of the proposal. Is it appropriate to assume that the sampling distribution of the sample proportion is approximately normal?
A.No, because the sample is not large enough to satisfy the normality conditions.
B.No, because the size of the population is not known.
C.Yes, because the sample was selected at random.
D.Yes, because sampling distributions of proportions are modeled with a normal model.
E.Yes, because the sample is large enough to satisfy the normality conditions.
Mathematics
1 answer:
tester [92]3 years ago
7 0

Answer:

E.Yes, because the sample is large enough to satisfy the normality conditions

Step-by-step explanation:

Given that  a random sample of 100 residents was selected, and 97 of those selected indicated that they were in favor of the proposal then;

The percent in favor of the proposal is 97/100 *100 =97%

Thus from this information ;

p=0.97 and q=1-p =1-0.97 =0.03 where p=population proportion

Finding mean and standard deviation will be;

μp =mean of sample proportion

μp=p= 0.97

δp= standard deviation of sample proportion

δp=√{pq/n }= √{(0.97*0.03)/100} where n=100 (sample size)

δp= 0.0171

3δp = 3*0.0171 =0.051

check using ;

{p- 3δp, p+3δp} = { 0.97-0.05,0.97+0.05} ={0.92,1.02}

compare if {0.92,1.02} lies in {0,1)

A sample is large enough if the interval {p- 3δp, p+3δp} lies wholly within the interval {0,1}

This sample wholly lie in the interval of {0,1} thus it is safe to assume p' is approximately normally distributed

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Step-by-step explanation:

8•7.5=60

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3 years ago
If all the sides of a prism are multiplied by 5, by what factor does the volume increase?
NISA [10]
I'd say increasing each side by 5 increases the volume by a factor of 125 because 5*5*5 = 125.

Example:
A 4*5*6 prism has a volume of 120

Increasing each side by a factor of five
(20*25*30) = 15,000

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3 years ago
What type of data is the following: “There were four dogs in the group."
BlackZzzverrR [31]

Answer: B.) Quantitative data

Step-by-step explanation:

  • Quantitative data is basically defined as a data with numbers that measure quantities.
  • Category data have different categories or groups in it.

In the given statement, we do not have categories but the count of dogs as "four" in a group.

Hence, the given statement represents "Quantitative data" .

So, the correct option is B) .

7 0
4 years ago
Please help me with this..
Paraphin [41]
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7 0
3 years ago
In a recent poll of 1,200 homeowners in the United States, one in five homeowners reports having a home equity loan that he or s
Aleks04 [339]

Answer:

Interval estimate for the proportion of all homeowners in the United States that hold a home equity loan lie between (.17033 , .22967 ) or 17% to 22.9%

Step-by-step explanation:

Given -

In a recent poll of 1,200 homeowners in the United States, one in five homeowners reports having a home equity loan that he or she is currently paying off.

If one in five homeowners reports having a home equity loan then for 1200 homeowners = \frac{1200}{5} = 240 homeowners reports having a home equity loan.

Sample proportion (\widehat{p}) = \frac{240}{ 1200} = 0.2

confidence coefficient = 0.99

(\alpha) = 1 - confidence coefficient  = 1 - 0.99 = .01

z_{\frac{\alpha}{2}} = z_{\frac{.01}{2}}  =  2.58

interval estimate for the proportion of all homeowners in the United States that hold a home equity loan

=  \widehat{p}\pm z_{\frac{\alpha}{2}}\sqrt\frac{{\widehat{p}( 1 - \widehat{p})}}{n}

     =  0.2\pm z_{\frac{.01}{2}}\sqrt\frac{{0.2( 1 - 0.2)}}{1200}

    = 0.2\pm 2.58\times \sqrt\frac{{0.2( 0.8)}}{1200}

    = 0.2\pm .02967

    = (0.2 - .02967 ) , (0.2 + .02967)

    = (.17033 , .22967 )

6 0
4 years ago
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