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tensa zangetsu [6.8K]
2 years ago
12

You are told that a random sample of 150 people from Manchester New Hampshire have been given cholesterol tests, and 60 of these

people had levels over the "safe" count of 200.
Using Excel, construct a 95% confidence interval for the population proportion of people in New Hampshire with cholesterol levels over 200.
Mathematics
1 answer:
Alenkasestr [34]2 years ago
5 0

Answer:

40%

Step-by-step explanation:

60/150 = 40% (0.4 * 100)

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The correct answer is D, 6x
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True or false...the segment AB is congruent to the segment BC​
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Answer:

true

Step-by-step explanation:

the segment ¯AB¯  is congruent to the segment ¯BC¯

5 0
3 years ago
In a simple random sample of 14001400 young​ people, 9090​% had earned a high school diploma. Complete parts a through d below.
ratelena [41]

Answer:

(a) The standard error is 0.0080.

(b) The margin of error is 1.6%.

(c) The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d) The percentage of young people who earn high school diplomas has ​increased.

Step-by-step explanation:

Let <em>p</em> = proportion of young people who had earned a high school diploma.

A sample of <em>n</em> = 1400 young people are selected.

The sample proportion of young people who had earned a high school diploma is:

\hat p=0.90

(a)

The standard error for the estimate of the percentage of all young people who earned a high school​ diploma is given by:

SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}

Compute the standard error value as follows:

SE_{\hat p}=\sqrt{\frac{\hat p(1-\hat p)}{n}}

       =\sqrt{\frac{0.90(1-0.90)}{1400}}\\

       =0.008

Thus, the standard error for the estimate of the percentage of all young people who earned a high school​ diploma is 0.0080.

(b)

The margin of error for (1 - <em>α</em>)% confidence interval for population proportion is:

MOE=z_{\alpha/2}\times SE_{\hat p}

Compute the critical value of <em>z</em> for 95% confidence level as follows:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the margin of error as follows:

MOE=z_{\alpha/2}\times SE_{\hat p}

          =1.96\times 0.0080\\=0.01568\\\approx1.6\%

Thus, the margin of error is 1.6%.

(c)

Compute the 95% confidence interval for population proportion as follows:

CI=\hat p\pm MOE\\=0.90\pm 0.016\\=(0.884, 0.916)\\\approx (88.4\%,\ 91.6\%)

Thus, the 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

(d)

To test whether the percentage of young people who earn high school diplomas has​ increased, the hypothesis is defined as:

<em>H₀</em>: The percentage of young people who earn high school diplomas has not​ increased, i.e. <em>p</em> = 0.80.

<em>Hₐ</em>: The percentage of young people who earn high school diplomas has not​ increased, i.e. <em>p</em> > 0.80.

Decision rule:

If the 95% confidence interval for proportions consists the null value, i.e. 0.80, then the null hypothesis will not be rejected and vice-versa.

The 95% confidence interval for the percentage of all young people who earned a high school diploma is (88.4%, 91.6%).

The confidence interval does not consist the null value of <em>p</em>, i.e. 0.80.

Thus, the null hypothesis is rejected.

Hence, it can be concluded that the percentage of young people who earn high school diplomas has ​increased.

8 0
3 years ago
Oxerville middle school is going to rent some vans to take students on a field trip. each van can hold 8 students. if a total of
Natalija [7]
11 vans would be needed
7 0
3 years ago
Suppose that shoe sizes of American women have a bell-shaped distribution with a mean of 8.47 and a standard deviation of 1.5. U
alex41 [277]

Answer:

2.5% of American women have shoe sizes that are greater than 11.47.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 8.47

Standard deviation = 1.5

The normal distribution is symmetric, which means that half the measures are below the mean, and half are above the mean.

What percentage of American women have shoe sizes that are greater than 11.47?

11.47 = 8.47 + 2*1.5

This means that 11.47 is two standard deviations above the mean.

Of those measures below the mean, none are greater than 11.47.

Of those measures above the mean, 95% are between the mean and 11.47, and 5% are above. So

0.5*0.05 = 0.025

2.5% of American women have shoe sizes that are greater than 11.47.

8 0
3 years ago
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