A simple random sample was taken of 44 water bottles from a bottling plant’s war

Question

A simple random sample was taken of 44 water bottles from a bottling plant’s warehouse. The dissolved oxygen content (in mg/L) was measured for each bottle, with these results: 11.53, 8.35, 11.66, 11.54, 9.83, 5.92, 7.14, 8.41, 8.99, 13.81, 10.53, 7.4, 6.7, 8.42, 8.4, 8.18, 9.5, 7.22, 9.87, 6.52, 8.55, 9.75, 9.27, 10.61, 8.89, 10.01, 11.17, 7.62, 6.43, 9.09, 8.53, 7.91, 8.13, 7.7, 10.45, 11.3, 10.98, 8.14, 11.48, 8.44, 12.52, 10.12, 8.09, 7.34 Here the sample mean is 9.14 mg/L and the sample standard deviation is 1.78 mg/L. The population standard deviation of the dissolved oxygen content for the warehouse is known from long experience to be about = 2 mg/L.

(a) Find a 98% confidence interval for the unknown population mean dissolved oxygen content.

(b) Interpret your interval.

Explanation / Answer

a) here sample mean=9.14

for sample size n=44

therefore std error of mean=std deviation/(n)1/2 =2/(44)1/2 =0.3015

for 98% confidence level ; z score =2.326

hence 98% confidence interval =sample mean -/+ z*std error =8.4386 ; 9.8414

b) above interval gives 98% confidence to contain true population mean of the dissolved oxygen content for the warehouse

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