Question

What sample size is needed if a ±1.5 lbs. margin of error is needed for computing the 95% confidence interval of the mean weight of U.S. adult males? Assume a population standard deviation of 12 lbs.

9. Find the 99% confidence interval for the mean number of times people blink per hour (bph), given that a sample of 60 people had a mean rate of 1230 bph. Assume a population standard deviation of 200 bph.

Answer 1

Answer 1

The following information has been provided: (1)Margin of Error E = 1.5 (2)Level of significance α=0.05 (3)Population Standard deviation σ =12 The critical value for the significance level α=0.05 is Zc =1.96. This can be found by using either Excel or by using the normal probability table. The following formula is used to compute the minimum sample size required to estimate the population mean μ within the required margin of error: Therefore, the sample size needed to satisfy the condition is n ≥ 245.8534, and it must be an integer number, we conclude that the minimum required sample size is

Answer 9

One-Sample Z test Confidence Interval

We need to construct the 99% confidence interval for the population mean μ. We have been provided with the following information: The sample mean is Xˉ=1230, the population standard deviation is σ=200, the sample size is n=60, and the significance level is α=0.01 Critical Value Based on the information provided, the significance level is α=0.01, therefore the critical value is Zc​=2.5758. This can be found by either using excel or the Z distribution table. Margin of Error The confidence interval: Therefore, based on the data provided, the 99% confidence interval for the population mean is 1163.4924<p<1296.5076, which indicates that we are 99% confident that the true population mean μ is contained by the interval

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