## A product comes in cans labeled “38 oz”. A random sample of 10 cans had the following weights: {34.6, 39.65, 34.75, 40, 39.5, 38

Question

A product comes in cans labeled “38 oz”. A random sample of 10 cans had the following weights:

{34.6, 39.65, 34.75, 40, 39.5, 38.9, 34.25, 36.8, 39, 37.2} with the following summary statistics:

n equals 10 space space top enclose X equals 37.465 space space s equals 2.27

Estimate the true mean weight of the cans of product with 95% confidence.

a. 34, 41 ounces
b. 35.9, 39.1 ounces
c. 36.1, 38.9 ounces
d. 37.465 ounces

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2 weeks 2021-09-11T01:47:03+00:00 1 Answer 0

b. 35.9, 39.1 ounces

Step-by-step explanation:

Previous concepts

A confidence interval is “a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval”.

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a “probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean”.

barX=37.465 represent the sample mean for the sample population mean (variable of interest)

s=2.27 represent the sample standard deviation

n=10 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

barX +- t*[(s)/(sqrt{n))]   (1)

In order to calculate the critical value t we need to find first the degrees of freedom, given by:

df=n-1=10-1=9

Since the Confidence is 0.95 or 95%, the value of alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: “=-T.INV(0.025,9)”.And we see that t_(alpha/2)=2.26

Now we have everything in order to replace into formula (1):

37.465-2.26[(2.27)/(sqrt{10))]=35.9

37.465+2.26[(2.27)/(sqrt{10))]=39.1

So on this case the 95% confidence interval would be given by (35.9;39.1)