No porportion is given how large a sample is needed to estimate the true porport

Question

No porportion is given how large a sample is needed to estimate the true porportion within 2.5% points with 99% confidence & 50 white mice with average weight of 4.2 ounces standard deviation .7 estimate 90% confidence interval for the mean weight No porportion is given how large a sample is needed to estimate the true porportion within 2.5% points with 99% confidence & 50 white mice with average weight of 4.2 ounces standard deviation .7 estimate 90% confidence interval for the mean weight & 50 white mice with average weight of 4.2 ounces standard deviation .7 estimate 90% confidence interval for the mean weight

Explanation / Answer

1) for no estimate p=0.5

for margin of error E=0.025

for 99% CI, z=2.5758

hence sample size n=p(1-p)*(Z/E)2 =~2654

2) for std error =std deviation/(n)1/2 =0.099

and for 49 degree of freedom and 90% CI, t=1.6766

hence 90% CI =sample mean -/+t*std error =4.034 ; 4.366

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