(c) Based on your answer to part (b), choose what the researcher can conclude, at the 0.10 level of significance.O Since the value of the test statistic lies in the rejection region, the null hypothesis is rejected. So, there is enough evidenceto conclude that the mean distance hiked by each visitor is now greater than 30 kilometers.O Since the value of the test statistic lies in the rejection region, the null hypothesis is not rejected. So, there is not enoughevidence to conclude that the mean distance hiked by each visitor is now greater than 30 kilometers.O since the value of the test statistic doesn't lie in the rejection region, the null hypothesis is rejected. So, there is enoughevidence to conclude that the mean distance hiked by each visitor is now greater than 30 kilometers.O Since the value of the test statistic doesn't lie in the rejection region, the null hypothesis is not rejected. So, there is notenough evidence to conclude that the mean distance hiked by each visitor is now greater than 30 kilometers. According to previous studies, the mean distance each visitor in Greenspan National Park hikes during their visit is 30 kilometers. The park recently closed itsshuttle system, which used to transport hikers to many of the park's most popular hiking trails. Because of this, an administrator at the park suspects the meandistance, u, is now less than 30 kilometers. The administrator chooses a random sample of 45 visitors. The mean distance hiked for the sample is 27.2kilometers. Assume the population standard deviation is 9.9 kilometers.Can the administrator conclude that the mean distance hiked by each visitor is now less than 30 kilometers? Perform a hypothesis test, using the 0.10 level ofsignificance.(a) State the null hypothesis H, and the alternative hypothesis H,.AaOOH: ID=D(b) Perform a hypothesis test. The test statistic has a normal distribution (so the test is a "Z-test"). Here is some other information to help you with your test.Z0 10 is the value that cuts off an area of 0.10 in the right tail.Z0.10

The random variable distance follows normal distribution. The sample size is 45. The sample mean is 27.2. The population standard deviation is 9.9. We have to test whether the mean distance hiked by each visitor is less than 30 kilometres or not. This is one-sample z-test with left-tailed.The null and alternative hypotheses areThe value of the test statistic is z =-1.90. From Z-table, the critical value at 0.10 level is -1.28. Critical value = -1.28.Since z (=-1.90) < -1.28, then reject the null hypothesis. Since the test statistic lies in the rejection region, the null hypothesis is rejected. So, there is enough evidence to conclude that the mean distance hiked by each visitor is less than 30 kilometers.