Question

A government company claims that an average light bulb lasts 270 days. A researcher randomly selects 18 bulbs for testing. The sampled bulbs last an average of 260 days, with a standard deviation of 90 days. If the ceo's claim were true, what is the probability that 18 randomly selected bulbs would have an average life of no more than 260 days

Answer 1

Given:

A government company claims that an average light bulb lasts 270 days.

A researcher randomly selects 18 bulbs for testing.

The sampled bulbs last an average of 260 days, with a standard deviation of 90 days.

To find:

If the ceo's claim were true, what is the probability that 18 randomly selected bulbs would have an average life of no more than 260 days

Solution:

t - statistics for the data is given as follows:

$t=\dfrac{x-\mu}{\frac{s}{\sqrt n}}$

x = mean of the sample of bulbs =  260

μ = population mean = 270

s = standard deviation of the sample = 90

n = number of items in the sample = 18

$t=\dfrac{260-270}{\frac{90}{\sqrt 18}}$

$t = \dfrac{-10}{\frac{90}{3\sqrt 2}}$

$t = \dfrac{-10}{\frac{30}{\sqrt 2}}$

$t = \dfrac{-1 \times \sqrt 2}{3}$

t = - 0.471

For probability calculations, the number of degrees of freedom is n - 1, so here you need the t-distribution with 17 degrees of freedom.

The probability that t < - 0.471 with 17 degrees of freedom assuming the population mean is true, the t-value is less than the t-value obtained With 17 degrees of freedom and a t score of - 0.471, the probability of the bulbs lasting less than 260 days on average of 0.3218 assuming the mean life of the bulbs is 300 days.

Answer 2

Step-by-step explanation:

t - statistics for the data is given as follows:

t=\dfrac{x-\mu}{\frac{s}{\sqrt n}}t=

n

s

x−μ

x = mean of the sample of bulbs = 260

μ = population mean = 270

s = standard deviation of the sample = 90

n = number of items in the sample = 18

t=\dfrac{260-270}{\frac{90}{\sqrt 18}}t=

1

8

90

260−270

t = \dfrac{-10}{\frac{90}{3\sqrt 2}}t=

3

2

90

−10

t = \dfrac{-10}{\frac{30}{\sqrt 2}}t=

2

30

−10

t = \dfrac{-1 \times \sqrt 2}{3}t=

3

−1×

2

t = - 0.471

For probability calculations, the number of degrees of freedom is n - 1, so here you need the t-distribution with 17 degrees of freedom.

The probability that t < - 0.471 with 17 degrees of freedom assuming the population mean is true, the t-value is less than the t-value obtained With 17 degrees of freedom and a t score of - 0.471, the probability of the bulbs lasting less than 260 days on average of 0.3218 assuming the mean life of the bulbs is 300 days.