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# Finding Sample Amount for Population Mean

In a study, the carapace lengths (in mm) of Thenus lobster caught near Singapore measured: 78,66,65,63,60,60,58,56,52 and 50 ( the population data is approximately normal).
Find:
(round to the nearest tenth)

a. The sample mean
b. The 80% confidence intervals for the population mean=
c. The 50% confidence intervals for the population mean=
d. The 98% confidence intervals for the population mean=
e. How many would they need to sample to be 95% sure to be within 0.2 of the population mean?
f. How many more (than the original 10) would they need to sample to be 86% sure to be withing .3 of the population mean?

#### Solution Preview

Means and confidence intervals

a. The sample mean = 60.8
Hint:

X X - X̅ (X - X̅)^2
78 17.2 295.84
66 5.2 27.04
65 4.2 17.64
63 2.2 4.84
60 -0.8 0.64
60 -0.8 0.64
58 -2.8 7.84
56 -4.8 23.04
52 -8.8 77.44
50 -10.8 116.64
608 571.61

Sample mean, x̅ = sum of X/n = 608/10 = 60.8

Sample standard deviation = √ 1/n-1 sum of (X_i -x̅)^2 = √571.6/10-1 = 7.969385868

b. The 80% confidence intervals for the population mean = (57.3, 64.3)

Hint:
The 95% confidence interval is given by (x̅ - t_a/2, n-1 S /√n, x̅ + t_a/2, n-1 S/√N), where x̅ = 60.8, s = 7.969385868, n = 10, t_a/2 = 1.383028739
That is, (60.8 - 1.383028739 * 7.969385868/ √10, 60.8 + 1.383028739 * 7.969385868/√10)
= (57.3, 64.3)
Thus with 80% confidence we can claim that population mean is within (57.3, 64.3).
Details
Confidence Interval Estimate for the Mean

Data
Sample Standard Deviation 7.969385868
Sample Mean 60.8
Sample Size 10
Confidence Level 80%

Intermediate Calculations
Standard Error of the Mean 2.52014109
Degrees of ...

#### Solution Summary

The expert finds the sample amount for population means.

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