Question

uises 7.64-7.81 ning the Mechanics NW 7.64 Suppose a random sample of 100 observations from a binomial population gives a value of p = .63 and you wish to test the null hypothesis that the population parameter p is equal to 70 against the alternative hypothesis that p is less than .70. a. Noting that Ø = .63, what does your intuition tell you? Does the value of p appear to contradict the null hypothesis? b. Use the large-sample z-test to test Ho : p= .70 against the alternative hypothesis, H:p <.70. Use a = .05. How do the test results compare with your intuitive decision from part a? c. Find and interpret the observed significance level of the test you conducted in part b. . 7.65 Suppose the sample in Exercise 7.64 has produced Ộ = .83 and we wish to test Ho : p= .9 against the alternative He: p <.9. a. Calculate the value of the z-statistic for this test. b. Note that the numerator of the z-statistic (@n - Po = .83 - .90 = -.70) is the same as for Exercise 7.64 0. Considering this, why is the absolute value of z for this exercise larger than that calculated in Exercise 7.64 O? c. Complete the test using a = .05 and interpret the result. d. Find the observed significance level for the test and interpret its value. 7.66 A statistics student used a computer program to test the null hypothesis HARD 15 against the one-tailed alternative, HAN> 5. A

Answer 1

7.64: a. Since $\hat{p}$ =0.63<0.70 the value of $\hat{p}$ appears to contradict null hypothesis.

b.

$z=\frac{\hat{p}-0.70}{\sqrt{\frac{0.7*0.3}{100}}}=-1.5275\\\\ Critical~value=-z_{0.05}=-1.6449$

Since value of z> Critical value, we fail to reject H₀ at 5% level of significance.

c.

$Observed~significance~level=p-value=P(Z<-1.5275|Z\sim N(0,1))\\\\ =0.0633~(Use~R~code:~round(pnorm(-1.5275),4)) \\\\ Since~p-value>0.05~(=pre-assigned~significance~level),~we~fail~to\\ reject~H_0~at~5\%~level~of~significance~and~conclude~that~there~is~insufficient~\\ evidence~that~population~proportion~is~significantly~less~than~0.70.$

7.65: a.

P-0.90 2 = = -2.3333 0.9 0.1 100 6. Standard error of p in Q7.64 under Ho 0.7* 0.3 100 0.0458.... (1) Standard error of p in Q7.65 under Ho 0.9 * 0.1 100 0.03.....(2) Hence (1) > (2), 2 – value of 07.65 > 2 - value of 07.64 even if value of p- po = 0.07 is same for both questions. c. Since value of 2 < Critical value of significance. -1.6449, we reject Ho at 5% level

d.

$Observed~significance~level=p-value=P(Z<-2.3333|Z\sim N(0,1))\\\\ =0.0098~(Use~R~code:~round(pnorm(-2.3333),4)) \\\\ Since~p-value<0.05~(=pre-assigned~significance~level),~we~\\ reject~H_0~at~5\%~level~of~significance~and~conclude~that~there~is~sufficient~\\ evidence~that~population~proportion~is~significantly~less~than~0.9.$