A simple random sample was taken of 44 water bottles from a bottling plant's warehouse. The dissolved oxygen content (in mg/L) w
as measured for each bottle, with the results below:
1.53, 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
a. Create a histogram and QQ plot of the data using R. Is it plausible that the data was drawn from a population that follows a normal distribution? Explain your answer.
b. Find a 98% condence interval for the population mean dissolved oxygen content.
c. Interpret the interval you created in (b)
Show all codes used in R.
1.53, 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
a. Create a histogram and QQ plot of the data using R. Is it plausible that the data was drawn from a population that follows a normal distribution? Explain your answer.
b. Find a 98% condence interval for the population mean dissolved oxygen content.
c. Interpret the interval you created in (b)
Show all codes used in R.
1 answer:
Answer:
a. Yes. p-value(0.05469)> 0.05
b,c- see details below
Step-by-step explanation:
a. #Use shapiro test to test for normality
Yes. The population has a p-value greater than 0.05:
0.05469>0.05
b.From our output, the 98% intervals are:
[8.19016,9.648021]
c. The new interval, [8.19016,9.648021] is wider, but we are more confident that it contains the true mean ( 8.919091)
#Attached histogram and snippet of the applicable code.
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