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15.04.2020 •
Mathematics
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.
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Ответ:
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
Ответ:
We are 95% confident that the proportion of all Money magazine subscribers that made money in the previous year from their investments is between 0.6904 and 0.8218
Step-by-step explanation:
The confidence interval gives a range of values for which a population proportion will fall based on a given level of confidence, in this case it is 95%. This interval is calculated using the sample proportion. The sample proportion in this case is based on a number of samples examined. The interval we obtain in this case (0.6904; 0.8218) is the estimated range which we expect the population proportion to fall ; with a 95% confidence.