A simple random sample of 60 items resulted in a sample mean of 80. The population standard deviation is \sigma=15 Compute the 95\% confidence interval for the population mean.

UkusakazaL

UkusakazaL

Answered question

2021-08-03

A simple random sample of 60 items resulted in a sample mean of 80. The population standard deviation is σ=15
a) Compute the 95% confidence interval for the population mean. Round your answers to one decimal place.
b) Assume that the same sample mean was obtained from a sample of 120 items. Provide a 95% confidence interval for the population mean. Round your answers to two decimal places.
c) What is the effect of a larger sample size on the interval estimate?
Larger sample provides a-Select your answer-largersmallerItem 5 margin of error.

Answer & Explanation

Ian Adams

Ian Adams

Skilled2021-08-17Added 163 answers

a) The population standard deviation is 15, the sample size is 60, and the sample mean is 80.
Computation of critical value:
(1α)=0.95
α=0.05
α2=0.025
1α2=0.975
The critical value of z-distribution can be obtained using the excel formula =NORM.S.INV(0.975). The critical value is 1.96.
The margin of error is,
E=zα2(σn)
=1.96(1560)
=3.80
The margin of error is 3.80.
The 95% confidence interval for the population mean is,
C.I.=x±E
=80±3.80
=(803.8, 80+3.8)
=(76.2, 83.8)
Thus, the 95% confidence interval for the population mean is (76.2, 83.8).
b) The margin of error is,
E=zα2(σn)
=1.96(15120)
=2.68
The margin of error is 2.68.
The 95% confidence interval for the population mean is,
C.I.=x±E
=80±2.68
=(802.68, 80+2.68)
=(77.32, 82.68)
Thus, the 95% confidence interval for the population mean is (77.32, 82.68).
c) The margin of error for samples of sizes 60 and 120 is 3.80 and 2.68, respectively. It is evident that raising the sample size results in a lower margin of error value.
Therefore, the larger sample offers a smaller error margin.

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