Please round answers as needed: A random sample of 22 lunch orders at Noodles an

Question

Please round answers as needed:

A random sample of 22 lunch orders at Noodles and Company showed a mean bill of $10.26 with a standard deviation of $5.21 Find the 99 percent confidence interval for the mean bill of all lunch orders (Round your answers to 4 decimal places.) A random sample of 16 pharmacy customers showed the waiting times below (in minutes). Find a 90 percent confidence interval for assuming that the sample is from a normal population (Round the value of t to 3 decimal places. Round your final answers to 3 decimal places.) A random sample of 10 shipments of stick-on labels showed the following order sizes. 15,953 27,967 28,890 50,574 17,767 24,216 19,119 48,642 19,219 32,360 Construct a 95 percent confidence interval for the true mean order size. (Round your answers to the nearest whole number.)

Explanation / Answer

a)[10.26-2.5758*5.21/sqrt(22),10.26+2.5758*5.21/sqrt(22)]=[7.3988,13.1211]

b)mean=17.1875,stddev=3.557013

90% conf int=[17.1875-1.64485*3.557013/sqrt(16),17.1875+1.64485*3.557013/sqrt(16)]=[15.725,18.650]

c)mean=28.4707 ,stddev=11.71924

95% conf int=[28.4707-1.96*11.71924/sqrt(10),28.4707+1.96*11.71924/sqrt(10)]=[21,36]

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