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Vesnalui [34]
1 year ago
11

g Students conducted a survey and found out that 36% of their peers on campus had tattoos but only 4% of their peers were smoker

s. If 100 students were surveyed, can these students use the Normal approximation to construct a confidence interval for the proportion of students in the population who are smokers? No, because either n p np or n ( 1 − p ) n(1−p) are greater than 15. Yes, because both n p np and n ( 1 − p ) n(1−p) are greater than 15. Yes, because both n p np and n ( 1 − p ) n(1−p) are less than 15. No, because either n p np or n ( 1 − p ) n(1−p) are less than 15.
Mathematics
1 answer:
IgorC [24]1 year ago
5 0

Answer:

D) No, because either np or n(1−p) are less than 15.

Step-by-step explanation:

Percentage of students who had tattoos = 36%

Percentage of students who were smokers = 4%

Sample size = n = 100

The condition to use the Normal distribution as an approximation to construct the confidence interval for population proportion is:

Both np and n(1-p) must be equal to or greater than 15.

Since, we are interested in smokers only, so p = 4% = 0.04

np = 100  x 0.04 = 4

n (1 - p) = 100 x 0.96 = 96

Since, np < 15, we cannot use the Normal distribution as an approximation here.

Therefore, the correct answer is:

No, because either np or n(1−p) are less than 15.

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Answer:


the science fiction books and five mystery books in each package. This is because 6 is the highest common factor of both 48 and 30, and 6 x 8 = 48 and 6 x 5 = 30.




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A wheat farmer cuts down the stalks of wheat and gathers them in 200 piles. The 200 gathered piles will be put on a truck. The t
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Answer:

Check the explanation

Step-by-step explanation:

We want to estimate the total weight of grain on the field based on the data on a simple random sample of 5 piles out of 200. The population and sample sizes are N=200 & n=5 respectively.

1) Let Y_1,Y_2,...,Y_{200} be the weight of grain in the 200 piles and y_1,y_2,...,y_{5} be the weights of grain in the pile from the simple random sample.

We know, the sample mean is an unbiased estimator of the population mean. Therefore,

\widehat{\mu}=\overline{y}=\frac{1}{5}\sum_{i=1}^{5}y_i=\frac{1}{5}(3.3+4.1+4.7+5.9+4.5)=4.5

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Hence, the total grain weight of the population is

\widehat{Y}=Y_1+Y_2+...+Y_{200}

=200\times \widehat{\mu}\: \: \: =200\times 4.5\: \: \: =900\, lbs

2) To calculate a bound on the error of estimates, we need to find the sample standard deviation.

The sample standard deviation is

 S=\sqrt{\frac{1}{5-1}\sum_{i=1}^{5}(y_i-\overline{y})^2}\: \: \: =0.9486

Then, the standard error of \widehat{Y} is

\sigma_{\widehat{Y}} =\sqrt{\frac{N^2S^2}{n}\bigg(\frac{N-n}{N}\bigg)}\: \: \:=83.785

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[\pm z_{0.025}\times \sigma_{\widehat{Y}}]\: \: \: =[\pm 1.96\times 83.875]\: \: \: =[\pm 164.395]

3) Let x_1,x_2,...,x_5 denotes the total weight of the sampled piles.

Mean total weight of the sampled piles is

\overline{x}=\frac{1}{5}\sum_{i=1}^{5}x_i=45

The sample ratio is

r=\frac{\overline{y}}{\overline{x}}=\frac{4.5}{45}=0.1 , this is also the estimate of the population ratio R=\frac{\overline{Y}}{\overline{X}} .

Therefore, the estimated total weight of grain in the population using ratio estimator is

\widehat{Y}_R\: \: =r\times 8800\: \: =0.1\times 8800\: \: =880\, lbs

4) The variance of the ratio estimator is

var(r)=\frac{N-n}{N}\frac{1}{n}\frac{1}{\mu_x^2}\frac{\sum_{i=1}^{5}(y_i-rx_i)^2}{n-1}   , where \mu_x=8800/200=44lbs

=\frac{200-5}{200}\, \frac{1}{5}\: \frac{1}{44^2}\, \frac{0.2}{5-1}=0.000005

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\sigma_R=\sqrt{X^2 \: var(r)}\: \: \: =\sqrt{8800^2\times 0.000005}\: \: \:=21.556

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[\pm z_{0.025}\times \sigma_{R}]\: \: \: =[\pm 1.96\times 21.556]\: \: \: =[\pm 42.25]

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