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Elden
1 month ago
7

Alma is estimating the proportion of students in her school district who, in the past month, read at least 1 book. From a random

sample of 50 students, she found that 32 students read at least 1 book last month. Assuming all conditions for Inference are met, which of the following defines a 90 percent confidence interval for the proportion of all students in her district who read at least 1 book last month? 32 +1.645,8908 32 + 1.9620 0.64 + 1.282,16.620.) © 264 265 CHAT 0.04 + 1.00, layanan
Mathematics
1 answer:
zzz [12.3K]1 month ago
5 0

Answer:

0.64 ± 0.1117 or

0.64\pm 1.645*\sqrt{\frac{0.64*(0.36)}{50}}

Step-by-step explanation:

Sample size (n) = 50

Z-score for a 90% confidence level (z) = 1.645

Proportion of students who read at least one book (p):

p=\frac{32}{50}=0.64

The confidence interval is calculated as:

p\pm z*\sqrt{\frac{p*(1-p)}{n} }

By applying the provided information:

0.64\pm 1.645*\sqrt{\frac{0.64*(1-0.64)}{50}}\\ 0.64\pm 0.1117

The confidence interval results in 0.64 ± 0.1117

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A rectangular prism must have a base with an area of no more than 27 square meters. The width of the base must be 9 meters less
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Answer:

The upper limit for the height of the prism is 12\ m

Step-by-step explanation:

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L*W\leq 27 -------> inequality A

W=x-9 ------> equation B

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Substituting equations B and D into inequality A

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The resultant solution for x lies in the interval---------->[0,12]

consult the attached figure

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the solution for x is confined to the interval ------> (9,12]

The maximum height of the prism equals 12\ m

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1 month ago
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The Insure.com website reports that the mean annual premium for automobile insurance in the United States was $1,503 in March 20
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Answer:

a) Null and alternative hypothesis

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Accordingly, the null and alternative hypotheses are:

H_0: \mu=1503\\\\H_a:\mu< 1503

The significance level is set at 0.05.

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The sample mean equates to M=1425.

A point estimate of the difference between the Pennsylvania mean premium and the national average can be computed using the sample mean:

d=M-\mu=1425-1503=-78

Given that the standard deviation of the population is unknown, we approximate it using the sample standard deviation, which is s=160.

The estimated standard error of the mean is determined with the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{160}{\sqrt{25}}=32

Then, we can calculate the t-statistic as:

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The degrees of freedom for this sample size stand at:

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\text{P-value}=P(t

As the P-value (0.0113) falls below the significance level (0.05), the results prove significant.

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