Lightbulbs of a certain type are advertised as having an average lifetime of 750 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 50 bulbs was selected, the lifetime of each bulb determined, and the appropriate hypothesis were tested usingMINITAB, resulting in the accompanying output.

Variable N Mean St Dev SEMean Z P -Value
lifetime 50 738.44 38.20 5.40 -2.14 0.016

a. What conclusion would be appropriate for a significance level of0.05 ?
b. A significance level of .01 what significance level wouldyou recommend ?

(a) The true average lifetime is smaller than what is advertised.

(b) The true average lifetime is not smaller than what is advertised.

Step-by-step explanation:

The hypothesis of this test can be defined as:

H₀: The true average lifetime is not smaller than what is advertised, i.e. μ > 750 hours.

Hₐ: The true average lifetime is smaller than what is advertised, i.e. μ < 750 hours.

The test is a left tailed test.

Decision rule:

If the p-value of the test is less than significance level (α) then the null hypothesis is rejected and vice-versa.

The p-value of the test is computed to be,

p-value = 0.016

(a)

The significance level is α = 0.05.

The p-value = 0.016 < α = 0.05.

As the p-value of the test is less than the significance level the null hypothesis was rejected.

Thus, the conclusion of the test at 5% level of significance is that the true average lifetime is smaller than what is advertised.

(b)

The significance level is α = 0.01.

The p-value = 0.016 > α = 0.01.

As the p-value of the test is more than the significance level the null hypothesis was failed to be rejected.

Thus, the conclusion of the test at 5% level of significance is that the true average lifetime is not smaller than what is advertised.

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step-by-step explanation: