Question

A manufacturer of banana chips would like to know whether its bag filling machine works correctly at the 427427 gram setting. It is believed that the machine is underfilling the bags. A 1515 bag sample had a mean of 421421 grams with a standard deviation of 2828. A level of significance of 0.050.05 will be used. Assume the population distribution is approximately normal. Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places.

Answer 1

Solution :

Given that,

Population mean = = 427

Sample mean = = 421

Sample standard deviation = s = 28

Sample size = n = 15

Level of significance = = 0.05

This is a two tailed test.

The null and alternative hypothesis is,

Ho: 427

Ha: 427

The test statistics,

t = ( - )/ (s/)

= ( 421 - 427 ) / ( 28 / 15 )

= -0.830

Critical value of  the significance level is α = 0.05, and the critical value for a two-tailed test is

= 2.145

Rejection region : |t| > 2.145

Since it is observed that |t| = 0.830 < = 2.145 , it is then concluded that the null hypothesis is fail to rejected.