Question

Q5: In a random sample of 2000 families in a city, it was found that 700 have a plasma TV set. Test the hypothesis that the proportion of all families having a plasma TV set is greater than 0.3 using the p-value approach at a 1% level of significance. 20.0003 = 3.45 Q6: A group of doctors is deciding whether or not to perform an operation. Suppose the null hypothesis, H, is "the surgical procedure will go well". State the Type I and the Type II errors in complete sentences, and decide which error has the greater consequence.

Answer 1

Given that, x = 700,n = 2000 The sample proportion is, X 700 p=-= = 0.35 2000 Null and alternative hypothesis: n H:P <0.3 H:P>0.3 (Right tailed test) Significance level: a = 0.01 Test statistic: Z= P-P P(1-P) n 0.35 -0.3 = 0.3(1-0.3) 2000 -4.88 =

P-value: p = P(Z > 2) =1-P(Z < 4.88) = 1-0(4.88) =1-1 (Standard normal tables) = 0.0000 Decision rule: Reject H , if p sa otherwise do not reject H. Since the p-value is less than the level of significance, so we reject the null hypothesis at the specified significance level. Conclusion: There is a sufficient evidence to support the claim that proportion of all families having a plasma tv set is greater than 0.3 Note: If the post contains multiple questions only allow to do first question as per HomeworkLib rules