Question

The diameter of ball bearings are known to be normally distributed with unknown mean and variance. A random sample of size 25 gave a mean of 2.5 cm. The 95% confidence interval had length 4 cm. Then

(a) The sample variance is 4.86.
(b) The sample variance is 26.03.
(c) The population variance is 4.84.
(d) The population variance is 23.47. (e) The sample variance is 23.47.

Answer 1

Given a 95% confidence interval length of 4 cm and a sample mean of 2.5 cm for ball bearings, we can calculate an approximate sample variance using the margin of error and sample size. However, without additional data on z-scores or t-scores, we cannot confirm the listed options for sample or population variance.

Step-by-step explanation:

The student's question relates to finding either the sample variance or the population variance using the information from a confidence interval. Given the fact that the 95% confidence interval for the mean diameter of ball bearings has a length of 4 cm, and knowing that the sample size is 25, which provides a mean of 2.5 cm, we can infer some properties of the variance.

The length of the confidence interval is the distance between the upper and lower bounds of the interval. Since the confidence interval is centered around the sample mean, which is 2.5 cm, half the length of the interval corresponds to the margin of error. Given the 95% confidence interval length of 4 cm, the margin of error (ME) is 2 cm (which is half of 4 cm).

Using the formula ME = z * (σ/√n), where z is the z-value corresponding to the 95% confidence level (approximately 1.96), σ is the standard deviation, and n is the sample size, we can solve for the standard deviation of the sample. Because we do not know the population standard deviation, we would use the sample standard deviation (s) instead, and the t-score would replace the z-value in the formula. However, given that neither t-score nor z-score are provided, and since the specific distribution of t-scores depends on degrees of freedom which correspond to sample size, precise calculations here are not possible without assumptions.

Nonetheless, we can demonstrate how sample variance would be calculated assuming the z-score could be appropriate. If ME is known to be 2 cm, then 2 = 1.96 * (s/√25), which leads to s being approximately 0.4 cm. Therefore, the sample variance (s^2) is approximately 0.4^2 cm^2, or 0.16 cm^2.

Unfortunately, the given information is insufficient to directly confirm any of the listed options (a) through (e) without additional context or data such as the precise critical value used for the confidence interval calculation.

Answer 2

`ME= (4)/(2)= 2`

And the margin of error is given by:

`ME= t_(\alpha/2)(s)/(√(n))`

`s= (ME √(n))/(t_(\alpha/2))= (2* √(25))/(2.0639)= 4.845`

And the sample variance is

`s^2 = 4.845^2 \approx 23.47`

So then the correct option would be:

(e) The sample variance is 23.47.

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".

`\bar X=2.5` represent the sample mean

`\mu` population mean (variable of interest)

s represent the sample standard deviation

n=25 represent the sample size

Solution to the problem

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

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=25-1=24`

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(0.025,24)".And we see that
`t_(\alpha/2)=2.0639`

The lenght of the interval is given by:

`2ME = 4`

`ME= (4)/(2)= 2`

And the margin of error is given by:

`ME= t_(\alpha/2)(s)/(√(n))`

And solving for the deviation we got:

`s= (ME √(n))/(t_(\alpha/2))= (2* √(25))/(2.0639)= 4.845`

And the sample variance is

`s^2 = 4.845^2 \approx 23.47`

So then the correct option would be:

(e) The sample variance is 23.47.