Question

A sample of 87 glass sheets has a mean thickness of 4.20 mm with a standard deviation of 0.10 mm. (a) Find a 98% confidence interval for the population mean thickness. (b) What is the level of the confidence interval (4.185, 4.215)? (c) How many glass sheets must be sampled so that a 98% confidence interval will specify the mean to within ±0.015? (d) Find a 90% confidence upper bound for the mean thickness.

Answer 1

To calculate confidence intervals for glass sheet thickness, we employ formulas using the z-score appropriate for the desired confidence level, the provided sample standard deviation, and the sample size. Adjustments to sample size can refine the precision of this estimate within a desired margin of error.

Step-by-step explanation:

Calculating Confidence Intervals for Glass Sheet Thickness

To answer the student's question: a) To find the 98% confidence interval for the population mean thickness of glass sheets, we can use the formula mean ± (z*standard deviation/√n), where z is the z-score corresponding to a 98% confidence level (approximately 2.326), standard deviation is given as 0.10 mm, and n is the sample size of 87. b) The level of confidence for the interval (4.185, 4.215) can be found using the z-score table and the standard deviation, and then checking which confidence level corresponds to the calculated z-score. c) To determine the number of glass sheets needed to obtain a 98% confidence interval within ±0.015 mm, we can use the formula n = (z*σ/E)^2, where σ is the standard deviation, E is the margin of error (0.015), and z is the z-score for a 98% confidence level. d) A 90% confidence upper bound can be found using a one-tailed z-test, where z is the z-score corresponding to 90% confidence (approximately 1.645).

The confidence interval estimates the true population mean (µ) of a statistic and is influenced by the confidence level, sample size, and sample standard deviation.

Answer 2

Step-by-step explanation:

From the information given,

Mean, μ = 4.2 mm

Standard deviation, σ = 0.1 mm

number of sample, n = 87

1) For a confidence level of 98%, the corresponding z value is 2.33.

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

4.2 ± 2.33 × 0.1/√87

= 4.2 ± 2.33 × 0.0107

= 4.2 ± 0.025

b) The lower end of the confidence interval is 4.2 - 0.025 =4.18

The upper end of the confidence interval is 4.2 + 0.025 =4.23

c) 0.015 = 2.33 × 0.1/√n

0.015/2.33 = 0.1/√n

0.00644 = 0.1/√n

√n = 0.1/0.00644 = 16

n = 16² = 256

d) For a confidence level of 90%, the corresponding z value is 1.645.

It becomes

4.2 ± 1.645 × 0.1/√87

= 4.2 ± 1.645 × 0.0107

= 4.2 ± 0.0176

The upper bound for the mean thickness is

4.2 + 0.0176 = 4.2176mm