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IRINA_888 [86]
3 years ago
13

An investor buys 400 shares of stock in a fund when the net asset value is $23.79 and the offer price is $23.89. The investor se

lls these shares when the net asset value is $50.34 and the offer price is $50.46. Determine the amount of profit the investor earns. a. $10,580 b. $10,620 c. $10,628 d. $20,136 Please select the best answer from the choices provided
Mathematics
2 answers:
lawyer [7]3 years ago
7 0

Answer:

The answer is A. Have a splendid day fellow students!!!

Step-by-step explanation:


FrozenT [24]3 years ago
5 0
C 10,628 because 400 times offer price (23.89) = $9,556, 400 times (50.46)= $20,184. $20,184-$9,556= $10,628profit.
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An automotive part must be machined to close tolerances to be acceptable to customers. Production specifications call for a maxi
Vilka [71]

Answer:

H0: \sigma \leq 0.0004

H1: \sigma >0.0004

t=(30-1) [\frac{0.0224}{0.02}]^2 =36.25

For this case since we have a right tailed test the p value is given by:

p_v = P(\Chi^2_{29}>36.25)=0.166

If we compare the p value and the significance level given we see that p_v >\alpha so then we have enough evidence to FAIL to reject the null hypothesis that the true population variance it's lower or equal  than 0.0004 at 5% of significance.  

Step-by-step explanation:

Previous concepts and notation

The chi-square test is used to check if the standard deviation of a population is equal to a specified value. We can conduct the test "two-sided test or a one-sided test".

n = 30 sample size

s^2 =0.0005 represent the sample variance

s= 0.0224 represent the sample deviation

\sigma_o =0.095 the value that we want to test

p_v represent the p value for the test

t represent the statistic

\alpha=0.05 significance level

State the null and alternative hypothesis

On this case we want to check if the population variance is higher than 0.0004 (the specification), so the system of hypothesis are:

H0: \sigma \leq 0.0004

H1: \sigma >0.0004

In order to check the hypothesis we need to calculate th statistic given by the following formula:

t=(n-1) [\frac{s}{\sigma_o}]^2

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

t=(30-1) [\frac{0.0224}{0.02}]^2 =36.25

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a right tailed test the critical zone it's on the right tail of the distribution. On this case we need a quantile on the chi square distribution with df=30-1=19 degrees of freedom that accumulates 0.05 of the area on the right tail and 0.95 on the left tail.  

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.95,29)". And our critical value would be \Chi^2 =42.56

Since our calculated value is lower than the critical value we to reject the null hypothesis.

What is the approximate p-value of the test?

For this case since we have a right tailed test the p value is given by:

p_v = P(\Chi^2_{29}>36.25)=0.166

If we compare the p value and the significance level given we see that p_v >\alpha so then we have enough evidence to FAIL to reject the null hypothesis that the true population variance it's lower or equal  than 0.0004 at 5% of significance.  

6 0
3 years ago
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