A simple random sample of 60 items resulted in a sample mean of 96. The population standard deviation is 16.
a. Compute the 95% confidence interval for the population mean (to 1 decimal).
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b. Assume that the same sample mean was obtained from a sample of 120 items. Provide a 95% confidence interval for the population mean (to 2 decimals).
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c. What is the effect of a larger sample size on the margin of error, does it increase, decrease,stays the same, or it cannot be determined?
I'll get you started.
a) CI95 = mean ± 1.96 (sd/√n)
mean = 96
sd = 16
n = 60
b) Use the same formula, except n = 120
c) Compare the margin of error in a) and b) to answer this question.
To solve these questions, we will use the formula for the confidence interval for the population mean:
Confidence Interval = sample mean +/- margin of error
The margin of error is calculated using the formula:
Margin of error = (z-value) * (standard deviation / square root of sample size)
Where the z-value is taken from the standard normal distribution for the desired confidence level. For a 95% confidence level, the z-value is approximately 1.96.
a. To compute the 95% confidence interval for the population mean:
Sample mean = 96
Standard deviation = 16
Sample size = 60
Using the formula, the margin of error is calculated as:
Margin of error = 1.96 * (16 / sqrt(60))
Now, we can calculate the confidence interval:
Lower bound = 96 - margin of error
Upper bound = 96 + margin of error
b. Assume the same sample mean, but now with a sample size of 120:
Sample mean = 96
Sample size = 120
Again, we will calculate the margin of error using the same formula as before:
Margin of error = 1.96 * (16 / sqrt(120))
Finally, compute the confidence interval:
Lower bound = 96 - margin of error
Upper bound = 96 + margin of error
c. The effect of a larger sample size on the margin of error is that it decreases. As the sample size increases, the margin of error decreases because there is more information available from the larger sample, resulting in a more precise estimate of the population mean.
To solve this problem, we can use the formula for the confidence interval:
Confidence interval = sample mean ± (critical value * standard deviation / square root of sample size)
a. For a sample size of 60:
We will use the t-distribution with n-1 degrees of freedom since the population standard deviation is not known.
The critical value for a 95% confidence level with 59 degrees of freedom is 2.000 (obtained from the t-distribution table).
The confidence interval is:
96 ± (2.000 * 16 / √60) = (96 ± 4.15)
Therefore, the 95% confidence interval for the population mean is (91.85, 100.15).
b. For a sample size of 120:
Again, we will use the t-distribution with n-1 degrees of freedom.
The critical value for a 95% confidence level with 119 degrees of freedom is 1.980 (obtained from the t-distribution table).
The confidence interval is:
96 ± (1.980 * 16 / √120) = (96 ± 2.91)
Therefore, the 95% confidence interval for the population mean is (93.09, 98.91).
c. The effect of a larger sample size on the margin of error is that it decreases. As the sample size increases, the standard error decreases, resulting in a narrower confidence interval. This means that the estimated population mean is likely to be more precise with a larger sample size.