How do I do this? Can someone give me steps please?
2 answers:
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Answer:
(a) Customer will not purchase the light bulbs at significance level of 0.05
(b) Customer will purchase the light bulbs at significance level of 0.01 .
Step-by-step explanation:
We are given that Light bulbs of a certain type are advertised as having an average lifetime of 750 hours. A random sample of 50 bulbs was selected, and the following information obtained:
Average lifetime = 738.44 hours and a standard deviation of lifetimes = 38.2 hours.
Let Null hypothesis, :
= 750 {means that the true average lifetime is same as what is advertised}
Alternate Hypothesis, :
< 750 {means that the true average lifetime is smaller than what is advertised}
Now, the test statistics is given by;
T.S. = ~
where, X bar = sample mean = 738.44 hours
s = sample standard deviation = 38.2 hours
n = sample size = 50
So, test statistics = ~
= -2.14
(a) Now, at 5% significance level, t table gives critical value of -1.6768 at 49 degree of freedom. Since our test statistics is less than the critical value of t, so which means our test statistics will lie in the rejection region and we have sufficient evidence to reject null hypothesis.
Therefore, we conclude that the true average lifetime is smaller than what is advertised and so consumer will not purchase the light bulbs.
(b) Now, at 1% significance level, t table gives critical value of -2.405 at 49 degree of freedom. Since our test statistics is higher than the critical value of t, so which means our test statistics will not lie in the rejection region and we have insufficient evidence to reject null hypothesis.
Therefore, we conclude that the true average lifetime is same as what it has been advertised and so consumer will purchase the light bulbs.
Answer:
Step-by-step explanation:
The fraction is the equivalent of ...
and the distributive property applies.