Standard Normal Distribution04Step 1: Select one-tailed or two-tailed.O One-tailedO Two-tailed0.3Step 2: Enter the test statistic.(Round to 3 decimal places.)0.2+Step 3: Shade the area represented bythe p-value.0.1+Step 4: Enter the p-value.(Round to 3 decimal places.) At a nearby college, there is a school-sponsored website that matches people looking for roommates. According to the school's reports, 38% of students will finda match their first time using the site. A writer for the school newspaper tests this claim by choosing a random sample of 165 students who visited the sitelooking for a roommate. Of the students surveyed, 48 said they found a match their first time using the site.Complete the parts below to perform a hypothesis test to see if there is enough evidence, at the 0.05 level of significance, to reject the claim that theproportion, p, of all students who will find a match their first time using the site is 38%.(a) State the null hypothesis H, and the alternative hypothesis H, that you would use for the test.OSOH: ID=D0(b) For your hypothesis test, you will use a Z-test. Find the values of np and n (1-p) to confirm that a Z-test can be used. (One standard is that npz 10and n (1-p) > 10 under the assumption that the null hypothesis is true.) Here n is the sample size and p is the population proportion you are testing.3D

The sample size is 165 in which 48 said they found a match their first time using the site. The population proportion of all students who will find a match their first time using the site is 38%.(a) The hypothesis is, Null hypothesis: H0:p=0.38 Alternative hypothesis: H1:p≠0.38(b) Checking conditions: np=165×0.38=62.7>10 n1-p=165×1-0.38=165×0.62=102.3>10(c) The sample proportion is, p^=xn=48165=0.2909 The test statistic is, z=p^-pp1-pn=0.2909-0.380.381-0.38165=-0.08910.03778728=-2.3579 The test statistic is -2.358. The p-value for the two tailed test is, p-value=2Pz<-2.358 The probability of z less than –2.358 can be obtained using the excel formula “=NORM.S.DIST (–2.358, TRUE)”. The probability value is 0.009. The required probability value is, p-value=2Pz<-2.358=2×0.009=0.018 Thus, the p-value is 0.018. Step 1: Choose Two-tailed. Step 2: Enter test statistic as 2.358 Step 3: Choose shaded area, and Step 4: Enter p-value as 0.018. The sketch is,Since, the p-value is less than level of significance indicating that the null hypothesis is rejected. Thus, there is sufficient evidence to reject the claim that 38% of students will find a match their time using the site.