Question

Question 2

ny s batteries will fail Procedures. A study • The advertised claim for batterie Wres. A study of 5000lbatteries is carrie Perimental results support the clair wuring the advertised time-period, will during heries for cell phones is set at 48 operating hours with proper a carried out and 15 stop operating prior to 48 hours. Do these at less than 0.002 (0.2 percent of the company's batteries w od, with proper charging procedures? (a) (5 points) Test the hypo appropriate hypotheses, test statistics est the hypothesis Ho: p = 0.002 against 02 against H:p < 0.002 at a =0.01. Be sure to include ision (reject or not reject) and why you made the decision. (b) (4 point) Use an upper 99% bound for p conclusion. Tp to answer the question in part (a). Clearly state and justify your

Answer 1

Aulad we are resting to ; P = 0.002 HA ; pe 0.002 at a so a Let the number of battery stop opesutiang pesies to ur hous is ruubond variable x. Since the sample s the large, the random Variable x and proportions follow normal distribution: The rucne of test steutistic under null hypothesis given as a z = b - b Ipli- b) p = p = x 3 0.003 -0.002 0002X(-0.002) som : 15 5000 = 0.003 z 1.583 The critical value at level of signiticance level (a=0.01) is 2.33 which is greates than value of test stalistic (1.583) at Therefore, we can't reject will mykothesis & we conclude that the claim is to not true. (b). The vam of appes of bound of for bix Calculated as tollows, b + I POD) - 0.003 + 0.33 5000 0.93 1 - 0 503) - 0.00u8 Since, the value of uppes bound of confidence interval is greates than 0.002 erreich to mean that c.I contein the value of null wybothesis. Hence, the voll hy bothesis don't get rejected.

If you have a doubt about the above answer ask in the comment box otherwise please press like button.