55 people are randomly selected and the accuracy of their wristwatches is checked, with positive errors representing watches that are ahead of the correct time and negative errors representing.watches that are behind the correct time. The 55 values have a mean of $95 \mathrm{sec}$ and a poptulation standard deviation of $240 \mathrm{sec}$. Use a 0.02 significance level to test the claim that the population of all watches has a mean of 0 (use a two-sided alternative).

The test statistic is $\square$
The P-Value is $\square$
The final conclustion is
A. There is not sufficient evidence to warrant rejection of the claim that the mean is equal to 0
B. There is sufficient evidence to warrant rejection of the claim that the mean is equal to 0

Step 1 ：Given in the problem: \(\bar{X} = 95\) sec, \(\mu = 0\) sec, \(\sigma = 240\) sec, and \(n = 55\).

Step 2 ：We use the formula for the test statistic in a hypothesis test for a population mean: \(Z = \frac{\bar{X} - \mu}{\sigma/\sqrt{n}}\).

Step 3 ：Substitute the given values into the formula: \(Z = \frac{95 - 0}{240/\sqrt{55}} = \frac{95}{240/\sqrt{55}}\).

Step 4 ：Calculate the above expression to get the test statistic: \(Z \approx 1.41\).

Step 5 ：Next, we need to find the P-value. The P-value is the probability that a Z-score is more extreme than the observed Z-score of 1.41, assuming the null hypothesis is true.

Step 6 ：Since this is a two-tailed test, we need to find the two-tailed P-value. Looking up the Z-score of 1.41 in a standard normal distribution table or using a calculator, we find that the one-tailed P-value is approximately 0.0793.

Step 7 ：Since this is a two-tailed test, we double this value to get the two-tailed P-value: \(P-value = 2 * 0.0793 = 0.1586\).

Step 8 ：Finally, we compare the P-value to the significance level. The significance level is 0.02, and our P-value is 0.1586.

Step 9 ：Since the P-value is greater than the significance level, we do not reject the null hypothesis.

Step 10 ：\(\boxed{\text{There is not sufficient evidence to warrant rejection of the claim that the mean is equal to 0.}}\)