Question

In: Statistics and Probability

Skagen-Tvosky Clock Manufacturers, LLC claim that their wall clocks on average will neither gain nor lose...

Skagen-Tvosky Clock Manufacturers, LLC claim that their wall clocks on average will neither gain nor lose time throughout the week. A sample of 18 wall clocks showed the following gains (+) or losses (-) in seconds per week. Is it reasonable to conclude that the mean gain or loss in time for the watches is 0?

0.10

0.10

0.40

-0.32

-0.30

-0.23

-0.20

0.25

-0.10

-0.37

-0.61

0.20

-0.30

-0.64

0.40

-0.20

-0.68

-0.40

A.)

State the decision rule for 0.05 significance level. (Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.)

  Reject H0: μ = 0 if t < ____________ or t > ________
B.)

Compute the Value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)

  Value of the test statistic:

_____________

C.) What is the P-Value?

Solutions

Expert Solution

The sample size is n=18. The provided sample data along with the data required to compute the sample mean Xbar and sample variance s2 are shown in the table below:

Gain Gain2
0.10 0.01
0.10 0.01
0.40 0.16
-0.32 0.1024
-0.30 0.09
-0.23 0.0529
-0.20 0.04
0.25 0.0625
-0.10 0.01
-0.37 0.1369
-0.61 0.3721
0.20 0.04
-0.30 0.09
-0.64 0.4096
0.40 0.16
-0.20 0.04
-0.68 0.4624
-0.4 0.16
Sum = -2.9 2.409

The sample mean Xbar is computed as follows:

Now applying T-test for the analysis

The provided sample mean is \bar X = -1.61Xˉ=−1.61 and the sample standard deviation is s = 0.338s=0.338, and the sample size is n = 18n=18.

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 00

Ha: μ ≠ 0

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a two-tailed test is tc​=2.11.

The rejection region for this two-tailed test is R={t:∣t∣>2.11}

Here critical value is calculated using t distribution with 17 degrees of freedom with two-tailed distribution.

So reject H0 if

t < -2.11 or t>0.211

Test Statistics

The decision about the null hypothesis

Since it is observed that |t| = 20.209 ∣t∣=20.209>tc​=2.11, it is then concluded that the null hypothesis is rejected.

Using the P-value approach

The p-value is p=0.0, and since p=0.00 <0.05, it is concluded that the null hypothesis is rejected.

Here p-value is calculated using t distribution using 17 df

p-value = P[∣t∣>20.209]

p-value = 2*P[t>20.209]

p-value = 2*0.00..

p-value = 0.00..

Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is different than 0, at the 0.05 significance level. Hence, it is reasonable to conclude that the mean gain or loss in time for the watches is 0?


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