MATH SOLVE

5 months ago

Q:
# A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 420 gram setting. It is believed that the machine is underfilling the bags. A 24 bag sample had a mean of 414 grams with a standard deviation of 15. Assume the population is normally distributed. A level of significance of 0.1 will be used. Find the P-value of the test statistic. You may write the P-value as a range using interval notation, or as a decimal value rounded to four decimal places.

Accepted Solution

A:

Answer:We fail to reject the null hypothesis that the bag filling machine works correctly at the 420 gram setting at the level of significance of 0.1. The p-value of the test statistic is 0.0250Step-by-step explanation:We have the following null and alternative hypothesis[tex]H_{0}: \mu = 420[/tex] vs [tex]H_{1}: \mu < 420[/tex] lower-tail alternative.For n = 24, [tex]\bar{x} = 414[/tex] and [tex]s = 15[/tex].[tex]\bar{X}[/tex] is normally distributed with a mean [tex]\mu[/tex] and a standard deviation of [tex]\frac{15}{\sqrt{24}}[/tex] (approx). Therefore, we can use as test statistic[tex]Z = \frac{\bar{X}-420}{15/\sqrt{24}}[/tex] and the observed value is[tex]z = \frac{414-420}{15/\sqrt{24}} = -1.9596[/tex]p-value = P(Z < -1.9596) = 0.0250We can use a table from a book or a programming language to find this probability P(Z < -1.9596).You can use the instruction pnorm(-1.9596) in the R statistical programming language.Because the p-value is greater than 0.1 (0.0250 > 0.1) we fail to reject the null hypothesis at the level of significance of 0.1.