Question

Friendly Cabinet Makers is a company that builds cabinets for houses. The company recently employed a new building technique that is supposed to save time over the old method which had a mean of 6.5 hours. Friendly Cabinet Makers wants to determine if the new method actually saves time. A random sample was taken of number of hours to build a cabinet. At the .05 significance level, can Friendly Cabinet Makers conclude that the new method is faster? Using the data in the range A10:A59, run a hypothesis test and make a concluding statement..

9.25
4.31
5.87
8.04
5.63
3.2
4.09
7.12
6.17
6.23
4.33
8.56
7.87
13.62
6.36
6.76
4.82
7.69
3.83
7.51
4.08
6.61
5.01
3.69
5.93
4.88
4.4
6.39
2.99
4.91
6.29
7.89
2.47
8.81
10.31
5.12
4.31
5.67
7.86
9.19
7.41
5.22
5.66
10.58
7.23
5.93
4.79
9.79
9.13
8.3
9.25
4.31
5.87
8.04
5.63
3.2
4.09
7.12
6.17
6.23
4.33
8.56
7.87
13.62
6.36
6.76
4.82
7.69
3.83
7.51
4.08
6.61
5.01
3.69
5.93
4.88
4.4
6.39
2.99
4.91
6.29
7.89
2.47
8.81
10.31
5.12
4.31
5.67
7.86
9.19
7.41
5.22
5.66
10.58
7.23
5.93
4.79
9.79
9.13
8.3
4.31
5.87
8.04
5.63
3.2
4.09
7.12
6.17
6.23
4.33
8.56
7.87
13.62
6.36
6.76
4.82
7.69
3.83
7.51
4.08
6.61
5.01
3.69
5.93
4.88
4.4
6.39
2.99
4.91
6.29
7.89
2.47
8.81
10.31
5.12
4.31
5.67
7.86
9.19
7.41
5.22
5.66
10.58
7.23
5.93
4.79
9.79
9.13
8.3

Answer 1

Here we have given that,

Claim: To check whether the mean amount of time save using new method is faster than the old method which had a mean of 6.5 hours.

= population mean = 6.5 hours

The Hypothesis is as follows

hours

v/s

hours

i.e. This is left (one) tailed test because our interest is in to see the new method is save more time i.e it is faster then old method.

We have given that,

Xi (time hours)
9.25
4.31
5.87
8.04
5.63
3.2
4.09
7.12
6.17
6.23
4.33
8.56
7.87
13.62
6.36
6.76
4.82
7.69
3.83
7.51
4.08
6.61
5.01
3.69
5.93
4.88
4.4
6.39
2.99
4.91
6.29
7.89
2.47
8.81
10.31
5.12
4.31
5.67
7.86
9.19
7.41
5.22
5.66
10.58
7.23
5.93
4.79
9.79
9.13
8.3
9.25
4.31
5.87
8.04
5.63
3.2
4.09
7.12
6.17
6.23
4.33
8.56
7.87
13.62
6.36
6.76
4.82
7.69
3.83
7.51
4.08
6.61
5.01
3.69
5.93
4.88
4.4
6.39
2.99
4.91
6.29
7.89
2.47
8.81
10.31
5.12
4.31
5.67
7.86
9.19
7.41
5.22
5.66
10.58
7.23
5.93
4.79
9.79
9.13
8.3
4.31
5.87
8.04
5.63
3.2
4.09
7.12
6.17
6.23
4.33
8.56
7.87
13.62
6.36
6.76
4.82
7.69
3.83
7.51
4.08
6.61
5.01
3.69
5.93
4.88
4.4
6.39
2.99
4.91
6.29
7.89
2.47
8.81
10.31
5.12
4.31
5.67
7.86
9.19
7.41
5.22
5.66
10.58
7.23
5.93
4.79
9.79
9.13
8.3

n= Number of observation =149

= sample mean = 6.42

s= Sample standard deviation =2.21

Now, we can find the test statistic

t-statistics=

=

=0.181

we get,

the Test statistic is 0.181

Now we find the P-value

= level of significance=0.05

This is left (one) tailed test

degrees of freedom = n-1 = 149 - 1= 148

Now, we can find the P-value

P-value =0.4283 using EXCEL = TDIST( | t-statistics| = 0.181, D.F= 49, tail = 1)

we get the P-value is 0.0404

Decision:

P-value > 0.05 ()

That is we fail to reject Ho (Null Hypothesis)

Conclusion

There is the not sufficient evidence that the mean amount of time save using new method is faster than the old method which had a mean of 6.5 hours.