6..24. Jones and Smith separately conduct studies to test Ho: it = 0.50 against Ho'- tt # 0.50, each with n = 400. Jones gets tt = 220/400 = 0.550. Smith gets tt = 219/400 = 0.5475.

(a) Show that z = 2.00 and P-value = 0.046 for Jones. Show that z — 1.90 and P-value =

0.057 for Smith.

(b) Using a = 0.05, indicate in each case whether the result is "statistically significant." Interpret.

(c) Use this example to explain why important information is lost by reporting the result of a test as "P-value ^ 0.05" versus "P-value >

0.05," or as "reject //q" versus "Do not reject

//o", without reporting the P-valuc.

(d) The 95% confidence interval for tt is (0.501, 0.599) for Jones and (0.499, 0.596) for Smith.

Explain how this method shows that, in practical terms, the two studies had very similar results.