Question

The test statistic in a two-tailed test is z=-2.06. Determine the P-value and decide whether, at the 10% significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the altermative hypothesis. Click here to view a partial table of areas under the standard normal curve The P-value is (Round to four decimal places as needed.) This P-value IS Vsufficient evidence to reject the null hypothesis in favor of the alternative hypothesis because it is the significance level Partial table of areas under the standard normal curve Second decimal place in z 0.09 0.08 0.07 0.060.05 0.04 0.03 0.02 0.0 0.00 0.0110 0.0113 0.0116 0.0119 0.0122 0.0125 0.0129 0.0132 0.0136 0.0139 -2.2 0.0143 0.0146 0.0150 0.0154 0.0158 0.0162 0.0166 0.0170 0.0174 0.0179-2.1 0.0183 0.0188 0.0192 0.0197 0.0202 0.0207 0.0212 0.0217 0.0222 0.0228-2.0 0.0233 0.0239 0.0244 0.0250 0.0256 0.0262 0.0268 0.0274 0.0281 0.028기 -1.9 0.0294 0.0301 0.0307 0.0314 0.0322 0.0329 0.0336 0.0344 0.0351 0.0359-1.8 Enter your answer in the a Print Done

fill in blank is 1. does not provide/provide or 2. greater than/less than

Answer 1

The test statistics is z=-2.06. This is a 2 tailed test. Hence the p-value is

$\begin{align*} \text{P-value}&=P(Z<-2.06) &plus;P(Z>2.06)\end{align*}$

From the partial table given we can get P(Z<-2.06) as

P(Z<-2.06)=0.0197

Due to the symmetry of standard normal distribution, we know that P(Z>2.06) = P(Z<-2.06) =0.0197

that means p-value is 0.0197+0.0197=0.0394

ans: the P-value is 0.0394

We will reject the null hypothesis if the p-value is less than the level of significance alpha=0.1 (10%). Here the P-value of 0.0394 is less than the significance level 0.1. Hence we will reject the null hypothesis

ans: This P-value provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis because it is less than the significance level.