Question

A electronics manufacturer has developed a new type of remote control button that is designed to operate longer before failing to work consistently. A random sample of 19 of the new buttons is selected and each is tested in continuous operation until it fails to work consistently. The resulting lifetimes are found to have a sample mean of x¯x¯ = 1254.2 hours and a sample standard deviation of s = 104.3. Independent tests reveal that the mean lifetime of the best remote control button on the market is 1215 hours. Conduct a hypothesis test to determine if the new button's mean lifetime exceeds 1215 hours. Round all calculated answers to four decimal places.

1. The null hypothesis is H0:μ=1215H0:μ=1215 . What is the alternate hypothesis?

A. HA:μ=1254.2HA:μ=1254.2
B. HA:μ<1215HA:μ<1215
C. HA:μ≠1215HA:μ≠1215
D. HA:μ>1215HA:μ>1215
E. HA:μ>1254.2HA:μ>1254.2

2. Which of the following conditions must be met to perform this hypothesis test? Select all the correct answers.

A. The observations must be independent.
B. We must be able to expect that at least 5 buttons will fail to work consistently.
C. The lifetime of remote control buttons must be normally distributed.
D. The sample must be large enough so that at least 10 buttons fail and 10 succeed.
E. The number of remote control buttons tested must be normally distributed.

3. Calculate the test statistic ? =

4. Calculate the p-value

5. Calculate the effect size, Cohen's d, for this test. d^d^ =

6. The results of this test indicate we have a...
A. small
B. large
C. small to moderate
D. moderate to large
effect size, and...
A. strong evidence
B. extremely strong evidence
C. little evidence
D. very strong evidence
E. some evidence
that the null model is not compatible with our observed result.

Answer 1

Answer:

a)

Given,

Ha : u > 1215

b)

Here to perform the hypothesis test, the conditions are as follows

A. The observations must be independent.

C. The lifetime of remote control buttons must be normally distributed.

c)

test statistic t = (x-u)/(s/sqrt(n))

substitute values

= (1254.2 - 1215)/(104.3/sqrt(19))

= 1.64

d)

Degree of freedom = n - 1

= 19 - 1

= 18

P value P(t > 1.64) = 0.059182

e)

Effect size = (x-u)/s

= (1254.2 - 1215)/(104.3)

= 0.3758

f)

C. small to moderate

g)

C. little evidence