In: Statistics and Probability
1A manufacturer of chocolate chips would like to know whether its bag filling machine works correctly at the 408408 gram setting. It is believed that the machine is underfilling the bags. A 1616 bag sample had a mean of 401401 grams with a variance of 256256. Assume the population is normally distributed. A level of significance of 0.010.01 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.
2. A sample of 900900 computer chips revealed that 59%59% of the chips do not fail in the first 10001000 hours of their use. The company's promotional literature states that 61%61% of the chips do not fail in the first 10001000 hours of their use. The quality control manager wants to test the claim that the actual percentage that do not fail is different from the stated percentage. Determine the decision rule for rejecting the null hypothesis, H0H0, at the 0.100.10 level.
1)
Hypothesis : VS
The test statistic is ,
The p-value is ,
p-value=
The Excel function is , =TDIST(1.75,15,2)
Decision : Here , p-value=0.1005 > 0.010
Therefore , fail to reject Ho.
Conclusion : Hence , there is sufficient evidence to support the claim that the bag filling machine works correctly at the 408 gram setting.
i.e. It is not believed that the machine is underfilling the bags.
2)
Hypothesis: VS
The test statistic is ,
The critical values are ,
; From standard normal distribution table
Decision : Here , the value of the test statistic does not lies in the rejection region.
Therefore , fail to reject Ho.
Conclusion : Hence , there not sufficient evidence to support the claim that the the actual percentage that do not fail is different from the stated percentage.