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The inspection division of the Lee County Weights μ. and Measures Department is interested in estimatin actual amount of soft drink that is placed in 2- to at the local bottling plant of a large nationally tles known soft-drink company. The bottling plant has ision that the standard deviation for 2-liter bottles is 0.05 liter. A random sanm ple of 100 2-liter bottles obtained from this botting plant indicates a sample average of 1.99 liters Section 8 (a) Set up a 95% confidence interval estimate of the (b) Does the population of soft-drink fill have to be (c) Explain why an observed value of 2.02 liters is not true average amount of soft drink in each bottle. normally distributed here? Explain. 8. unusual, even though it is outside the confidence interval you calculated. uppose that the sample average had been 1.97 liters. What would be your answer to (a)?

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batolisis

Answer:

hello your question is poorly written attached below is the complete question

answer

a)  ( 1.9802 , 1.9998 )

b) The population of soft drink does not have to be normally distributed in order to calculate the confidence interval for a sample mean.

c)  A single value of 2.02 is not unusual

d)   (1.9602, 1.9798)

Step-by-step explanation:

A) set up 95% confidence interval estimate

std = 0.05 liter

average ( mean ) X = 1.99 liters

n = 100

95% confidence interval = X ± [tex]Z_{95%}[/tex] ( std / √n )

                                         = 1.99 ± 1.96 ( 0.05 / √100 )

                                         = ( 1.9802 , 1.9998 )

B) The population of soft drink does not have to be normally distributed in order to calculate the confidence interval for a sample mean. as seen in this scenerio the value of n > 30 and sampling distribution of X  is required instead

C) A single value of 2.02 is not unusual because the confidence interval we calculated above is for 100 samples of the soft drink ( using true population mean ) but when we calculate for a one sample we will observe that the value ( 2.02 ) will not be unusual e.g. ( 2 ± 2*0.05 ) = ( 1.9, 2.1 )

D) when the sample average ( X )  = 1.97

confidence interval =  X ± [tex]Z_{95%}[/tex] ( std / √n )

                                = 1.97 ± 1.96 ( 0.05 / √100 )

                                =  (1.9602, 1.9798)

${teks-lihat-gambar} batolisis

There are four sub-parts in the question, whose answers are mentioned below:

Given information:

Mean of sample = [tex]\overline{x}[/tex] =1.99

Standard deviation of population= [tex]\sigma[/tex] = 0.05

Size of sample = n = 100

Part a: Setting 95% confidence interval

[tex]CI = \overline{x} \: \pm Z_{0.95}\dfrac{\sigma}{\sqrt{n}}\\CI = 1.99 \pm 1.96 \times \dfrac{0.05}{10}\\CI = (1.9802, 1.9998)\\[/tex]

Part b: Yes, the population has to be normally distributed here, since the CI here cannot be calculated without it being normally distributed.

Part c: It is because we took n = 100 and not for single value of observation. The single value of observation has 2 liter or value is 2, and 2.02 is near 2 thus not an unusual value.

Part d: If sample average be 1.97, then CI will be as follows:

[tex]CI = \overline{x} \: \pm Z_{0.95}\dfrac{\sigma}{\sqrt{n}}\\CI = 1.97 \pm 1.96 \times \dfrac{0.05}{10}\\CI = (1.9602, 1.9798)\\[/tex]

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