Don\'t solve it with Minitab or Excel. Answer all questions CORRECTLY and CLEARL
ID: 3250340 • Letter: D
Question
Don't solve it with Minitab or Excel. Answer all questions CORRECTLY and CLEARLY for a thumbs-up (ex. label answers as #1, 2, etc. instead of as a garble of text).
Doris and Sakar, the co-directors of manufacturing at a soft drink bottling factory, need to determine whether a new filling machine is injecting the appropriate amount of cola into 16-ounce bottles. They select a random sample of 35 bottles and calculate the mean amount of cola fill to be 15.9 ounces with a standard deviation of 0.40 ounces. Let u true average amount of cola fii. Use the appropriate attached Minitab printout and the p-value approach, to perform the appropriate test of hypothesis. (Questions 1-5). 1. What is the appropriate null hypothesis? 2. What is the appropriate alternative hypothesis? 3. What is the p-value for this problem? 4. What is the appropriate rejection regional (Us a 0.05.) 5. What is the appropriate conclusion? 6. Use the appropriate attached Minitab printout to find a 90% confidence interval for the true average amount of cola fill. 7. Interpret the interval in Exercise 6 8. Based in the interval in Exercise 6, can one reject a claim that the true mean amount of cola fill is 16 ounces?Explanation / Answer
1)
H0: mu = 16
2)
Ha: mu ot equal to 16
3) n = 35 , x= 15.9 , s = 0.40
t = ( x - mean) / ( s /sqrt(n))
= ( 15.9 - 16) / ( 0.40 / sqrt(35))
= -1.479
p avlue is calculated using t = -1.479 , df = 34
p value = .148346.
4)
With alpha = .05, the critical values of z are -1.96 and +1.96. We reject if z < -1.96 or z > +1.96.
5)
we fail to reject the null hypothesis.
6)
t value at 90% CI = 1.303
CI = mean +/- t *(s/sqrt(n))
= 16 + /- 1.303 * (0.40 / sqrt(35))
= (15.9119 , 16.0880)
8)
we cannot reject the claim thatvthe true mean amount of cola fill 16 ounces
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