Question

(1 point) 45 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 45 values have a mean of 107 sec and a standard deviation of 190 sec. Use a 0.1 significance level to test the claim that the population of all watches has a mean of sec. The test statistic is Round off your answer to three decimal places. The P-Value Is Round off your answer to four decimal places. 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

Answer 1

The null and alternate hypothesis are:

H0: $\mu=0$

Ha: $\mu\neq 0$

The test statistic is given by:

$Z=\frac{\overline{x}-\mu}{\frac{s}{\sqrt{n}}}=\frac{107-0}{\frac{190}{\sqrt{45}}}=3.778$

Since this is a two-tailed test, so the p-value is given by:

2 x P(Z > 3.778) = 2 x 0.000079 = 0.000158 0.0002

Since p-value is less than 0.1, so we have sufficient evidence to reject the null hypothesis H0.

Hence option B.