Answer to Question #348793 in Statistics and Probability for Hakdog

Question #348793

A machine which has been regulated dispenses an average of 330 ml fruit concentrate per bottle. A random sample of 49 bottles filled by the machine has a mean content of 320 ml and a standard deviation of 50 ml.”?


1
Expert's answer
2022-06-08T13:24:03-0400

The parameter is average content of fruit concentrate per bottle.

The following null and alternative hypotheses need to be tested:

"H_0:\\mu=330"

"H_1:\\mu\\not=330"

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is "\\alpha = 0.05," "df=n-1=48" and the critical value for a two-tailed test is "t_c =2.010635."

The rejection region for this two-tailed test is "R = \\{t:|t|>2.010635\\}."

The t-statistic is computed as follows:


"t=\\dfrac{\\bar{x}-\\mu}{s\/\\sqrt{n}}=\\dfrac{320-330}{50\/\\sqrt{49}}=-1.4"


Since it is observed that "|t|=1.4<2.010635=t_c," it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed, "df=48" degrees of freedom, "t=-1.4" is "p=0.167944," and since "p=0.167944>0.05=\\alpha," it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean "\\mu"

is different than 330, at the "\\alpha = 0.05" significance level.


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