Question

A manufacturer of salad dressings uses machines to dispense liquid ingredients into bottles that move along a filling line. The machine that dispenses dressings is working properly when the mean amount dispensed is 330 mL. The population standard deviation of the amount dispensed is 4 mL. A sample of 50 bottles is selected periodically, and the filling line is stopped if there is evidence that the mean amount dispensed is different from 330 mL.

Answer 1

ANSWER: Conclusion: Alternate Hypothesis H1: μ ≠ μ0 (Alternate Hypothesis) is true with at least 95% confidence. Yes; the sample shows a different average value than 330mL

Why??
SINGLE SAMPLE TEST, TWO-TAILED, 6 - Step Procedure for t Distributions, "two-tailed test"

Step 1: State the hypothesis to be tested.
Null Hypothesis H0: μ = μ0
Alternate Hypothesis H1: μ ≠ μ0

Step 2: Determine a planning value for α [level of significance] = 0.05

Step 3: From the sample data determine x-bar, s and n; then compute
Standardized Test Statistic: t = ( x-bar - μ0 )/( s/ SQRT(n) )

x-bar: Est. of the Pop. Mean (statistical mean of the sample) = 329.7
n: number of individuals in the sample = 50
s: sample standard deviation = 0.566
μ0: Population Mean = 330

Standardized Test Statistic t = ( 329.7 - 330 )/( 0.566 / SQRT( 50 )) = 3.748

Step 4: Using Students t distribution, 'lookup' the area outside of t = TDIST( 3.748 , 49 , 2 )

Step 5: Area in Step 4 is equal to P value = 0 based on n -1 = 49 df (degrees of freedom).

Table look-up value shows area under the 49 df curve outside of t = +/- 3.748 is (approx.)
P value = 0 [2 * 0.0002] by addition of both 'tails' of t distribution.

Step 6: For P ≥ α, fail to reject H0; and for P < α, reject H0 with 95% confidence in the conclusion.

Conclusion: Alternate Hypothesis H1: μ ≠ μ0 (Alternate Hypothesis) is true with at least 95% confidence.

Answer 2

Ohh my god... thankyou, soo sooo soo much

Answer 3

welcome

Answer 4

Wait can you please simplify
that some more?