Jack is a college athlete who has been weighing himself weekly on the same scale in the athletic center for the past few years. Jack's mean weight over the last nine weeks is 200 lbs. Over a period of a few months, Jack's weight measurements are approximately normal, with a standard deviation of about 3 lbs. Assume that 3 lbs is the true standard deviation in Jack's weight. Suppose that Jack asks you to compute a 90% confidence interval for his true mean weight.

a. Identify the positive critical value, z, for a 90% level of confidence using software or a table of critical values.
b. Derive the standard error of the mean, SE.
c. Compute the margin of error, m, for the 90% confidence interval for Jack's mean weight.

a) Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=\pm 1.64[/tex]

b) [tex] \bar X \sim N(\mu ,\frac{\sigma}{\sqrt{n}})[/tex]

And the standard error is given by:

[tex] SE = \frac{\sigma}{\sqrt{n}}=\frac{3}{\sqrt{9}}=1[/tex]

c) [tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)

And the margin of error is:

[tex] ME= z_{\alpha/2} \frac{\sigma}{\sqrt{n}} = 1.64 [/tex]

And then the confidence interval is be given by:

[tex] 200-1.64 = 198.36[/tex]

[tex] 200+1.64 = 201.64[/tex]

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".

[tex]\bar X= 200[/tex] represent the sample mean

[tex]\mu[/tex] population mean (variable of interest)

[tex]\sigma=3[/tex] represent the population standard deviation

n=9 represent the sample size

Part a

Since the Confidence is 0.90 or 90%, the value of [tex]\alpha=0.1[/tex] and [tex]\alpha/2 =0.05[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that [tex]z_{\alpha/2}=\pm 1.64[/tex]

Part b

The distribution for the sample mean is given by:

[tex] \bar X \sim N(\mu ,\frac{\sigma}{\sqrt{n}})[/tex]

And the standard error is given by:

[tex] SE = \frac{\sigma}{\sqrt{n}}=\frac{3}{\sqrt{9}}=1[/tex]

Part c

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

[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)

And the margin of error is:

[tex] ME= z_{\alpha/2} \frac{\sigma}{\sqrt{n}} = 1.64 [/tex]

And then the confidence interval is be given by:

[tex] 200-1.64 = 198.36[/tex]

[tex] 200+1.64 = 201.64[/tex]

raphealnwobi raphealnwobi · Answer 2 · 2022-04-12T12:21:46+02:00

The margin of error for the 90% confidence interval for Jack's mean weight is ± 4.935 lbs

How to calculate margin of error

The z score of 90% confidence interval is 1.645

The margin of error (E) is:

[tex]E = Z_\frac{\alpha }{2} *\frac{standard\ deviation}{\sqrt{sample\ size} } =1.645*\frac{3}{\sqrt{1} } =4.935[/tex]

The margin of error for the 90% confidence interval for Jack's mean weight is ± 4.935 lbs

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