Respuesta :
Answer:
We conclude that the true average lifetime is smaller than what is advertised.
Step-by-step explanation:
The complete question is: Light bulbs of a certain type are advertised as having an average lifetime of 800 hours. The price of these bulbs is very favorable, so a potential customer has decided to go ahead with a purchase arrangement unless it can be conclusively demonstrated that the true average lifetime is smaller than what is advertised. A random sample of 21 bulbs was selected, the lifetime of each bulb determined, and the appropriate hypotheses were tested using SPSS, resulting in the accompanying output.
Sample Mean Sample Standard Deviation
Lifetime 738.44 38.30
Test the appropriate hypothesis for this study (i.e., state the hypotheses, compute the test statistics, compute the p-value, and make a conclusion). What conclusion would be appropriate for a significance level of 0.05?
We are given that Light bulbs of a certain type are advertised as having an average lifetime of 800 hours.
A potential customer has decided to go ahead with a purchase arrangement unless it can be conclusively demonstrated that the true average lifetime is smaller than what is advertised.
Let [tex]\mu[/tex] = true average lifetime of the light bulbs.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \geq[/tex] 800 hours {means that the true average lifetime is greater than or equal to what is advertised}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu[/tex] < 800 hours {means that the true average lifetime is smaller than what is advertised}
The test statistics that will be used here is One-sample t-test statistics because we don't know about the population standard deviation;
T.S. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample mean lifetime = 738.44 hours
s = sample standard deviation = 38.30 hours
n = sample of bulbs = 21
So, the test statistics = [tex]\frac{738.44-800}{\frac{38.30}{\sqrt{21} } }[/tex] ~ [tex]t_2_0[/tex]
= -7.36
The value of t-test statistics is -7.36.
Also, the P-value of test-statistics is given by;
P-value = P([tex]t_2_0[/tex] < -7.36) = Less than 0.05% {from t-table}
Since the P-value of our test statistics is less than the level of significance of 0.05, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.
Therefore, we conclude that the true average lifetime is smaller than what is advertised.
