Mathematics, 08.07.2020 01:01 chackey

According to creditcard , the mean outstanding credit card debt of college undergraduate was $3173 in 2010. A researcher believes that this amount has decreased since then. Required:
a. Determine the null and alternative hypotheses.
b. Explain what it would mean to make a Type I and Type Il error.

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Answer from: Candieboo4006

a. The null and alternative hypothesis can be written as:

$H_0: \mu=3173\\\\H_a:\mu< 3173$

b. A Type I error is made when a true null hypothesis is rejected. In this case, it would happen if it is concluded that the actual mean outstanding credit card debt of college undergraduate is significantly less than $3173, when in fact it does not.

A Type II error is made when a false null hypothesis is failed to be rejected. In this case, the actual mean outstanding credit card debt of college undergraduate is in fact less than $3173, but the test concludes there is no enough evidence to claim that.

Step-by-step explanation:

We have a prior study of the mean outstanding credit card debt of college undergraduate that states that it was $3173 in 2010.

A researcher believes that this amount has decreased since then.

Then, he has to perform a hypothesis test where the null hypothesis states that the mean is still $3173 and an alternative hypothesis that states that the actual credit card debt is significantly smaller than $3173.

The null and alternative hypothesis can be written as:

$H_0: \mu=3173\\\\H_a:\mu< 3173$

Answer from: Quest

This does not make sense

Answer from: Quest

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12

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