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3 years ago

14

PLZZZZZZ NEED HELP!!!!!!!

1 answer:

Sladkaya [172]3 years ago

4 0

The correct answer is C. Hope this helps~

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Cloud [144]

Answer:

the correct answer is the frequency

3 0

2 years ago

Two different simple random samples are drawn from two different populations. The first sample consists of 20 people with 9 havi

QveST [7]

Answer: $(-0.48,\ -0.04)$

Step-by-step explanation:

The confidence interval for the difference of two population proportion is given by :-

$(p_1-p_2)\pm z_{\alpha/2}\sqrt{\dfrac{p_1(1-p_1)}{n_1}+\dfrac{p_2(1-p_2)}{n_2}}$

Given : The first sample consists of 20 people with 9 having a common attribute.

Here, $n_1=20$ , $p_1=\dfrac{9}{20}=0.45$

$n_2=2100$ , $p_1=\dfrac{1492 }{2100}\approx0.71$

Significance level : $\alpha=1-0.95=0.05$

Critical value : $z_{\alpha/2}=1.96$

A 95% confidence interval for the difference of two population proportion will be :-

$(0.45-0.71)\pm z_{\alpha/2}\sqrt{\dfrac{0.45(1-0.45)}{20}+\dfrac{0.71(1-0.71)}{2100}}\\\\\approx -0.26\pm0.22\\\\=(-0.26-0.22,-0.26+0.22)\\\\=(-0.48,\ -0.04)$

3 0

4 years ago

Do You Understand?

Trava [24]

Answer:

An equation is basically a way to show a relationship of variables (x,y,a,b, etc) and numbers.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

The table showing the stock price changes for a sample of 12 companies on a day is contained in the Excel file below.

AfilCa [17]

Answer:

(a) The sample variance for the daily price change is 0.2501.

(b) The sample standard deviation for the daily price change is 0.5001.

(c) The 95% confidence interval estimates of the population variance is (0.1255, 0.7210).

Step-by-step explanation:

Let the random variable <em>X</em> denote the stock price changes for a sample of 12 companies on a day.

The data provided is:

<em>X</em> = {0.82 , 1.44 , -0.07 , 0.41 , 0.21 , 1.33 , 0.97 , 0.30 , 0.14 , 0.12 , 0.42 , 0.15}

(a)

The formula to compute the sample variance for the daily price change is:

$s^{2}=\frac{1}{n-1}\sum\limits^{12}_{i=1}{(X_{i}-\bar X)^{2}}$

The sample mean is computed using the formula:

$\bar X=\frac{1}{n}\sum\limits^{12}_{i=1}{X_{i}}$

Consider the Excel output attached below.

In Excel the formula to compute the sample mean and sample variance are:

$\bar X$ =AVERAGE(A2:A13)

$s^{2}$ =VAR.S(A2:A13)

Thus, the sample variance for the daily price change is 0.2501.

(b)

The formula to compute the sample standard deviation for the daily price change is:

$s=\sqrt{\frac{1}{n-1}\sum\limits^{12}_{i=1}{(X_{i}-\bar X)^{2}}}$

Consider the Excel output attached below.

In Excel the formula to compute the sample standard deviation is:

$s$ =STDEV.S(A2:A13)

Thus, the sample standard deviation for the daily price change is 0.5001.

(c)

The (1 - <em>α</em>)% confidence interval for population variance is:

$CI=[\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2} } \leq \sigma^{2}\leq \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2} } ]$

Compute the critical value of Chi-square for <em>α</em> = 0.05 and (n - 1) = (12 - 1) = 11 degrees of freedom as follows:

$\chi^{2}_{\alpha/2, (n-1)}=\chi^{2}_{0.05/2,11}=21.920$

$\chi^{2}_{1-\alpha/2, (n-1)}=\chi^{2}_{(1-0.05/2),11}=\chi^{2}_{0.975,11}=3.816$

*Use a Chi-square table.

Compute the 95% confidence interval estimates of the population variance as follows:

$CI=[\frac{(n-1)s^{2}}{\chi^{2}_{\alpha/2} } \leq \sigma^{2}\leq \frac{(n-1)s^{2}}{\chi^{2}_{1-\alpha/2} } ]$

$=[\frac{(12-1)\times 0.2501}{21.920 } \leq \sigma^{2}\leq \frac{(12-1)\times 0.2501}{3.816} ]$

$=[0.125506\leq \sigma^{2}\leq 0.720938]\\\approx [0.1255, 0.7210]$

Thus, the 95% confidence interval estimates of the population variance is (0.1255, 0.7210).

7 0

4 years ago

For each sequence, determine whether it appears to be

Airida [17]

Answer:

I think the answer is neither

7 0

3 years ago