Question

per vo-Woneer 1. Fifteen randomly selected dentists were timed while brushing their own teeth for a recent study The mean amount of time spent on brushing their teeth was 2.13 minutes with a standard deviation of 0.45 minutes. Calculate the Margin of error for the mean amount of time all dentists spend brushing their teeth. Use a 99% level of confidence. Assume that the population is normally distributed. What is the confidence interval? n = 15 d.f. = 14 7= 2.13 $x = 0.45 c=0.99 From Table: t«/2 = 2.575 E = (t. 2. (Sx/n) E = (round to two decimal places) t - Interval = (round to two decimal places) 2. A random sample of 253 working mothers was surveyed regarding the cost of health care. The mean amount of money spent on health care by these working mothers each year was $1300. Suppose that the population standard deviation is known to be $289 per year. Calculate the margin of error for the population mean using a 98% level of confidence. What is the confidence interval? n = 253 * = 1300 o=289 e=0.98 E = (2x/2) lo/n) (round to two decimal places) z-Interval = (round to two decimal places)

Answer 1

1Mean $(\bar{x})$ = 2.13
Sample size (n) = 15
Standard deviation (s) = 0.45
Confidence interval(in %) = 99

Required confidence interval = (2.13-0.35, 2.13+0.35)
Required confidence interval = (1.78, 2.48)

ta/2.n-1 = 2.97684273 Since we know that Margin of error = ta/2n-1 To Margin of error = 2.976842734 Margin of error = 0.35 And Confidence interval = £ta/2,n-17 Required confidence interval = (2.13–2.976842734 076020720.45121620020720.45 = 2.13+2.97684273 S 112.13

2
Mean $(\bar{x})$ = 1300
Sample size (n) = 253
Standard deviation (s) = 289
Confidence interval(in %) = 98
z @ 98% = 2.33

Since we know that
$\\Margin\; of\; error(E) = z\frac{s}{\sqrt{n}} \\Margin\; of\; error = 2.33\frac{289.0}{\sqrt{253}} \\Margin\;of\;error = 42.33\\And\\Confidence\; interval = \bar{x} \pm z\frac{s}{\sqrt{n}} \\Required\; confidence\; interval = (1300.0-2.33\frac{289.0}{\sqrt{253}}, 1300.0+2.33\frac{289.0}{\sqrt{253}})$
Required confidence interval = (1300.0-42.33, 1300.0+42.33)
Required confidence interval = (1257.67, 1342.33)

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