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3 years ago

HELP ME PLS ASAP!!!!!!!!!!!

Mathematics

1 answer:

mina [271]3 years ago

7 0

Answer:

$\sigma^2=73.96 \ kg^2$

Step-by-step explanation:

Standard Deviation and Variance

If we have a data set of measured values the variance is defined as the average of the squared differences that each value has from the mean. The formula to calculate the variance is:

$\displaystyle \sigma^2=\frac{\sum(x_i-\mu)^2}{n}$

Where $\mu$ is the mean of the measured values xi (i running from 1 to n), and n is the total number of values.

$\displaystyle \mu=\frac{\sum x_i}{n}$

The standard deviation is known by the symbol $\sigma$ and is the square root of the variance. We know the standar deviation of the weight in kg of a group of teenagers to be 8.6 kg. Thus, the variance is

$\sigma^2=8.6^2=73.96 \ kg^2$

$\boxed{\sigma^2=73.96 \ kg^2}$

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In a simple random sample of 14001400 young people, 9090% had earned a high school diploma. Complete parts a through d below.

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(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 p = proportion of young people who had earned a high school diploma.

A sample of n = 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 - α)% confidence interval for population proportion is:

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

Compute the critical value of z 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:

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

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

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3 years ago