A company is developing a new high performance wax for cross country ski racing.

Question

A 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. Complete parts a and b below. a) The champion's times (selected at random) are 54.4 , 65.9 , 48.5 , 54.9 , 49.9 , 49.2 , 53.6 , and 41.3 seconds to complete the test course. Should they market the wax? Assume the assumptions and conditions for appropriate hypothesis testing are met for the sample. Assume alphaequals0.05.

Calculate the test statistic. (Round to three decimal places as needed.)

Calculate the P-value. (Round to four decimal places as needed.)

Choose the correct conclusion below.

A. the wax as there is insufficient evidence to conclude the meantime is less than 55 seconds.

B. Yes, market the wax as there is insufficient evidence to conclude the meantime is less than 55 seconds.

C. Yes market the wax as there is sufficient evidence to conclude the mean time is less than 55 seconds.

D. No, do not market the wax as there is sufficient evidence to conclude the mean time is less than 55 seconds.

b) Suppose they decide not to market the wax after the test, but it turns out that the wax really does lower the champion's average time to less than 55 seconds. What kind of error have they made? Explain the impact to the company of such an error.

A. They have made a type I error and will lose the potential profit from selling the wax.

B. They have made a type II error and will lose the potential profit from selling the wax.

C. They have made a type I error and customers might demand their money back.

D. They have made a type II error and customers might demand their money back.

Explanation / Answer

Hypothesis:
Null hypothesis: mu = 55
Alternative hypothesis : mu < 55

a)

Test statistics:

x = 52.2125 , s = 7.0566 , n = 8
t = ( x - mu) / ( s/sqrt(n))
= ( 52.2125 - 55)/( 7.0566/sqrt(8))
= -1.1173
P value is calculated using t = -1.1173 with df = 7
P value = .1504
we do not reject the null hypothessis

A. N0 , market the wax as there is insufficient evidence to conclude the meantime is less than 55 seconds.

b)

B. They have made a type II error and will lose the potential profit from selling the wax.

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