Question

The advertised claim for batteries for cell phones is set at 48 operating hours, with proper charging procedures. In order to test the claim that less than 0.2% of the company's batteries will fail during the advertised time period, a study of 5000 batteries is carried out and 15 stop operating prior to 48 hours. Suppose p=the proportion of batteries operated less than 48 hours, the hypotheses of interest are, ?₀: ? = 0.002 and ?_?: ? < 0.002. What is the value of the test statistic for this hypothesis test?

Group of answer choices

z=0.002−0.0030.003(0.997)5000z=0.002−0.0030.003(0.997)5000

z=0.003−0.0020.002(0.998)5000z=0.003−0.0020.002(0.998)5000

z=0.002−0.0030.002(0.998)5000z=0.002−0.0030.002(0.998)5000

Answer 1

Solution:

?₀: ? = 0.002 and ?_?: ? < 0.002

n = 5000

x = 15

Let   be the sample proportion.

= x/n = 15/5000 = 0.003

The test statistic z is

z =

=  (0.003 - 0.002)/[0.002*(1 - 0.002)/5000]

=  (0.003 - 0.002)/[0.002*0.998/5000]

z =  (0.003 - 0.002)/[0.002*0.998/5000]