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

5555

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

59.259.2​,

63.563.5​,

51.851.8​,

53.153.1​,

46.746.7​,

45.145.1​,

52.152.1​,

and

40.240.2   

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

alphaαequals=0.05.

Choose the correct null and alternative hypotheses below.

Answer 1

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

59.2

63.5

51.8

53.1

46.7

45.1

52.1

40.2  

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 α equals=0.05.

Choose the correct null and alternative hypotheses below.

Lower tail test.

calculated test statistic t =-1.3270            

Table value of t with 7 DF at 0.05 level = -1.8946

Rejection Region: Reject Ho if t < -1.8946

Calculated t = -1.3270 , not in the rejection region

The null hypothesis is not rejected.

There is not enough evidence to support that the mean time to finish their standard test course is less than 55 seconds.

t Test for Hypothesis of the Mean
Data
Null Hypothesis                m=	55
Level of Significance	0.05
Sample Size	8
Sample Mean	51.4625
Sample Standard Deviation	7.540166254
Intermediate Calculations
Standard Error of the Mean	2.6659
Degrees of Freedom	7
t Test Statistic	-1.3270
Lower-Tail Test
Lower Critical Value	-1.8946
p-Value	0.1131
Do not reject the null hypothesis