(1 point) The Get It at the Grove Company sells its orange juice in 1.89 liter c

Question

(1 point) The Get It at the Grove Company sells its orange juice in 1.89 liter cartons. A random sample of 12 cartons gives the following results (in liters) 1.73 2.03 2.62 1.83 1.69 1.92 2.17 1.53 1.95 1.79 (a) What is the population in this sample? A. The size of orange juice cartions. OB. The amount of juice in cartons of Get it in the Groves Orange Juice C. The size of cartons made by Get it in the Grove OD. The amount of juice in cartons of orange juice (b) What assumption is required to make inferences about the population mean from this data? A. that it is discrete B. that it is continuous C. that it is normal D. that it is independent (c) Will the data follow a normal distribution? Yes Report the P-value of the normality test: (d) Construct a 95% confidence interval estimate of the true mean volume in Get It at the Grove orange juice cartons. (Input answers to four decimal places) (e) Does your interval suggest the carton label of 1.89 L is inaccurate?

Explanation / Answer

Part a

Answer:

C. The size of cartons made by Get it in the Grove

(We are related to the population of sizes of cartons which are made by Get it in the Grove.)

Part b

Answer:

C. that it is normal

We know that the assumption of normality of the population data is required for making inferences or using testing of hypothesis.

Part c

Yes, data follow a normal distribution.

P-value of normality test is 0.907

Explanation:

Here, we have to check whether the population data follows a normal distribution or not. For checking this normality we have to use one sample Kolmogorov Smirnov test.

Null hypothesis: H0: Population data follows normal or approximate normal distribution.

Alternative hypothesis: Ha: Population data does not follow a normal or approximate normal distribution.

We assume 5% significance level for this test.

The results table for this test by using excel is given as below:

Amount

N

12

Normal Parameters

Mean

1.8967

Std. Deviation

0.30269

Most Extreme Differences

Absolute

0.163

Positive

0.163

Negative

-0.095

Kolmogorov-Smirnov Z

0.565

Asymp. Sig. (2-tailed)

0.907

From the above table it is observed that the p-value = 0.907 is greater than = 0.05, so we do not reject the null hypothesis that Population data follows normal or approximate normal distribution.

Part d

Here, we have to find 95% confidence interval of the true mean volume.

Confidence interval = Xbar -/+ t*S/sqrt(n)

From the given data, we have

Sample mean = Xbar = 1.896666667

Sample standard deviation = S = 0.302694966

Sample size = n = 12

Degrees of freedom = n – 1 = 12 – 1 = 11

Confidence level = c = 95% = 0.95

= 1 – c = 1 – 0.95 = 0.05

/2 = 0.05/2 = 0.025

Critical t value = 2.2010

(By using t-table)

Confidence interval = 1.896666667 -/+ 2.2010*0.302694966/sqrt(12)

Confidence interval = 1.896666667 -/+ 0.1923

Lower limit = 1.896666667 - 0.1923 = 1.7043

Upper limit = 1.896666667 + 0.1923 = 2.0890

Confidence interval = (1.7043, 2.0890)

1.7043 µ 2.0890

Part e

No, given interval does not suggest that carton label of 1.89 L is inaccurate.

(Because the value 1.89 L is lies within given confidence interval.)

Amount

N

12

Normal Parameters

Mean

1.8967

Std. Deviation

0.30269

Most Extreme Differences

Absolute

0.163

Positive

0.163

Negative

-0.095

Kolmogorov-Smirnov Z

0.565

Asymp. Sig. (2-tailed)

0.907

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