5. An engineering firm is evaluating their back charges. They claim that their average back charge is $1800. They randomly select 42 customers, and calculate the sample mean back charge to be $2010. If the standard deviation of back charges is $650, and alpha = 0.05, test the claim of the engineering firm. Perform an appropriate hypothesis test, showing the necessary calculations and/or explaining the process used to obtain the results.
Ho: µ = 1800
Ha: µ ≠ 1800
t-statistic = (2010-1800)/(650/sqrt(42)) = 2.0938
degree of freedom = 41
critical region: t<-2.0195 or t>2.0195
Null hypothesis is rejected because test statistic lies in critical region.
Average back charge is different from 1800.
6. A marketing firm wants to estimate the average amount spent by patients at the hospital pharmacy. For a sample of 65 randomly selected patients, the mean amount spent was $93.50 and the standard deviation was $13.25.
(a) Find a 95% confidence interval for the mean amount spent by patients at the pharmacy. Show your calculations and/or explain the process used to obtain the interval.
For 95% confidence level, z = 1.96
Maximum error of estimate = 1.96*13.25/sqrt(65) = 3.22
So, confidence interval is 93.50-3.22 to 93.50+3.22, that is, (90.28, 96.72)
(b) Interpret this confidence interval and write a sentence that explains it.
This means if mean amount is calculated using different samples of same sample size, 95% of times the sample mean would lie between 90.28 and 96.72.
7. A drug manufacturer wants to estimate the percentage of patients who experience cotton mouth when taking their heart medication. A survey of 450 patients who took their drug found that 55 experienced cotton mouth.
(a) Find a 98% confidence interval for the true proportion of patients who experienced cotton mouth. Show your calculations and/or explain the process used to obtain the interval.
For 98% confidence, z = 2.32635
Observed proportion = 55/450 = 0.12222
Maximum error of estimate = 2.32635*sqrt(0.12222*(1-0.12222)/450) = 0.03592
Lower limit = 0.12222 - 0.03592 = 0.0863
Upper limit = 0.12222 + 0.03592 = 0.15814
Confidence interval is (0.0863, 0.15814)
(b) Interpret this confidence interval and write a sentence that explains it.
This means if the survey is repeated multiple times using the same sample size, 98% of times observed proportion would be between 0.0863 and 0.15814
8. The heights of 10 sixth graders are listed in inches: {51, 61, 54, 60, 58, 62, 55, 58, 59, 56}.
(a) Find the mean, median, mode, sample variance, and range.
mean = 57.4
median = 58
mode = 58
sample variance = 11.6
range = 11
(b) Do you think that this sample might have come from a normal population? Why or why not?
The sample might have come from a normal population because mean, median and mode are quite close to each other.