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PtichkaEL [24]
3 years ago
15

company is developing a new​ high-performance wax for cross country ski racing. In order to justify the price marketing​ wants,

the wax needs to be very fast.​ Specifically, the mean time to finish their standard test course should be less than 55 seconds for a former Olympic champion. To test​ it, the champion will ski the course 8 times. The​ champion's times are 59.5​, 64.7​, 46.6​, 54.8​, 47.1​, 46.1​, 52.2​, and 41.6 seconds to complete the test course. Complete parts a and b below. ​a) Should they market the​ wax? Assume the assumptions and conditions for appropriate hypothesis testing are met for the sample. Assume alphaequals0.05. What are the null and alternative​ hypotheses? Choose the correct answer below. A. H0​: muless than55 vs. HA​: muequals55 B. H0​: muequals55 vs. HA​: mugreater than55 C. H0​: mugreater than55 vs. HA​: muequals55 D. H0​: muequals55 vs. HA​: muless than55'
Mathematics
1 answer:
lukranit [14]3 years ago
4 0

Answer:

Correct option: (D) <em>H₀</em>: <em>μ</em> =  55 vs. <em>Hₐ</em>: <em>μ</em> < 55.

Step-by-step explanation:

In this case we need to test whether the new​ high-performance wax price is justifiable or not.

It is provided that the wax needs to be very fast specifically the mean time to finish their standard test course should be less than 55 seconds for a former Olympic champion, for the price to justifiable.

So, to test this a sample of <em>n</em> = 8 champion's times are recorded.

A one-sample <em>t</em>-test will be used to perform the test.

The hypothesis to test this is defined as follows:

<em>H₀</em>: The mean time to finish their standard test course is 55 seconds, i.e. <em>μ</em> =  55.

<em>Hₐ</em>: The mean time to finish their standard test course is less than 55 seconds, i.e. <em>μ</em> < 55.

Thus, the correct option is (D).

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\begin{vmatrix} x & y & z \\ -8 & 2 & -12 \\ u & v & w \end{vmatrix} = - \begin{vmatrix} x & y & z \\ u & v & w \\ -8 & 2 & -12 \end{vmatrix}

Multiplying a single row by a scalar scales the determinant by the same amount:

\begin{vmatrix} x & y & z \\ u & v & w \\ -8 & 2 & -12 \end{vmatrix} = -2 \begin{vmatrix} x & y & z \\ u & v & w \\ 4 & -1 & 6 \end{vmatrix}

Then

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