Question

Bottles of purified water are assumed to contain 250 milliliters of water. There is some variation from bottle to bottle because the filling machine is not perfectly precise. Usually, the distribution of the contents is approximately Normal. An inspector measures the contents of eight randomly selected bottles from one day of production. The results are 249.3, 250.2, 251.0, 248.4, 249.7, 247.3, 249.4, and 251.5 milliliters. Do these data provide convincing evidence at α = 0.05 that the mean amount of water in all the bottles filled that day differs from the target value of 250 milliliters?

Answer 1

Answer:

We accept H₀ we don´t have evidence of differences between the information from the sample and the population mean

Step-by-step explanation:

From data and excel (or any statistics calculator) we get:

X = 249,6 ml         and       s  1,26 ml

Sample mean and sample standard deviation respectively.

Population mean  μ₀  = 250 ml

We have a normal distribution but we dont know the standard deviation of the population. Furthermore we have a two tails test since we are finding if the sample give us evidence of differences ( in both senses ) when we compare them with the amount of water spec ( 250 ml )

Our test hypothesis are: null hypothesis       H₀    X = μ₀

Alternative Hypothesis                                 Hₐ     X ≠ μ₀

We also know that sample size is 8  therefore df  =  8 - 1   df = 7 , with this value and the fact that we are required to test at α = 0,05 ( two tails test)

t = 2,365

Then we evaluate our interval:

X ± t* (s/√n)   ⇒   249,6 ±  2,365 * ( 1,26/√8 )

 249,6 ±  2,365 * (1,26/2,83)  ⇒ 249,6 ±  2,365 *0,45

249,6 ± 1,052

P [ 250,652 ; 248,548]

Then the population mean 250 is inside the interval, therefore we must accept that the bottles have being  fill withing the spec. We accept H₀