A simple random sample of 60 items resulted

Grecia Cordova Ramirez

Grecia Cordova Ramirez

Answered question

2022-09-25

A simple random sample of 60 items resulted in a sample mean of 90. The population standard deviation is 13.

a. Compute the 95% confidence interval for the population mean (to 1 decimal).

( , ) I got 93.2 and 86.7, but I keep getting it wrong 

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).

( , )

c. What is the effect of a larger sample size on the margin of error?
 

Answer & Explanation

Jazz Frenia

Jazz Frenia

Skilled2023-05-29Added 106 answers

a. To compute the 95% confidence interval for the population mean, we can use the formula:
(x¯zσn,x¯+zσn)
where:
x¯ is the sample mean,
σ is the population standard deviation,
n is the sample size,
z is the critical value corresponding to the desired confidence level.
Given:
x¯=90,
σ=13,
n=60.
To find the critical value, we can use a standard normal distribution table or a statistical software. For a 95% confidence level, the critical value is approximately 1.96.
Plugging in the values, we get:
(901.961360,90+1.961360)
Simplifying the expression, we find:
(86.72,93.28)
Therefore, the 95% confidence interval for the population mean is (86.72,93.28).
b. Assuming the same sample mean of 90 and a larger sample size of 120, we can compute the 95% confidence interval using the same formula:
(x¯zσn,x¯+zσn)
Given:
x¯=90,
σ=13,
n=120.
Using the critical value of 1.96 for a 95% confidence level, we have:
(901.9613120,90+1.9613120)
Simplifying the expression, we find:
(88.16,91.84)
Therefore, the 95% confidence interval for the population mean with a sample size of 120 is (88.16,91.84).
c. The effect of a larger sample size on the margin of error is that it reduces the margin of error. As the sample size increases, the standard error decreases, resulting in a narrower confidence interval. A larger sample size provides more information about the population, leading to increased precision in estimating the population mean.

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