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Harlamova29_29 [7]
3 years ago
6

Please help. 10 points. Please do not guess. Guesses will be reported. Thanks

Mathematics
1 answer:
Allushta [10]3 years ago
8 0

The intersecting secant-tangent theorem says that

AP^2=PT\cdot TO

where T is the point at which line segment PO touches the circle. Similarly,

BP^2=PT\cdot TO

so we have

AP^2=BP^2\implies AP=BP

OP is of course as long as itself, and AO and BO are radii of the same circle so they have the same lengths.

This means triangles APO and BPO are congruent, which means angles APO and BPO are congruent, so angle APB is bisected by PO.

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

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