The p-value is...O less than (or equal to) aO greater than aThis test statistic leads to a decision to...O reject the nullaccept the nullO fail to reject the nullAs such, the final conclusion is that...O There is sufficient evidence to warrant rejection of the claim that the proportion of stocks thatwent up is less than 0.3.O There is not sufficient evidence to warrant rejection of the claim that the proportion of stocksthat went up is less than 0.3.O The sample data support the claim that the proportion of stocks that went up is less than 0.3.O There is not sufficient sample evidence to support the claim that the proportion of stocks thatwent up is less than 0.3. Many investors and financial analysts believe the Dow Jones Industrial Average (DJIA) gives a goodbarometer of the overall stock market. On January 31, 2006, 9 of the 30 stocks making up the DJIAincreased in price (The Wall Street Journal, February 1, 2006). On the basis of this fact, a financialanalyst says that maybe we can assume that 30% of the stocks traded on the New York StockExchange (NYSE) went up the same day. You have decided to do your own test.A sample of 78 stocks traded on the NYSE that day showed that 15 went up.You are going to complete your test to see if the proportionstocks that went up is significantlyless than 0.3.You use a significance level of a =0.001.What is the test statistic for this sample? (This is called the Z-Stat in your StatCrunch output. Reportanswer accurate to three decimal places.)test statistic =What is the p-value for this sample? (Report answer accurate to four decimal places.)p-value =

The sample of 78 stocks traded on the NYSE that day showed that 15 went up. That is, n=78,x=15.Let p be the population proportion of stocks that went up. The claim is that to check whether the proportion of stocks that went up is significantly less than 0.3. The hypothesis is, Null hypothesis: H0:p≥0.3 Alternative hypothesis: H1:p<0.3 The sample proportion of stocks that went up is, p^=xn=1578=0.1923The test statistic is, z=p^-pp1-pn=0.1923-0.30.31-0.378=-0.10770.051887=-2.0755≈-2.076 Therefore, the test statistic is -2.076.The hypothesis is left tail. The p-value for left tailed test is, p-value=Pz<-2.0755. The probability of z less than -2.0755 can be obtained using the excel formula “=NORM.S.DIST(-2.0755,TRUE)”. The probability value is 0.0190. Therefore, the p-value is 0.0190. Decision rule: If p-value≤α, then reject the null hypothesis. Conclusion: The p-value (0.0190) is greater than the level of significance (0.001). Based on decision rule, fail to reject the null hypothesis. Thus, there is sufficient evidence to warrant rejection of the claim that the proportion of stocks that went up is significantly less than 0.3.