A company manufactures tennis balls. When its tennis balls are dropped from a he

Question

A company manufactures tennis balls. When its tennis balls are dropped from a height of 100 inches, the company wants the mean height the balls bounce upward to be 55.5 inches. This average is maintained by periodically testing a random sample of 25 tennis balls. A sample of 25 is randomly selected and tested. The mean bounce height of the sample is 56.0 inches and the standard deviation is 0.25 inch. Assume the bounce heights are approximately normally distributed. Create a 99% confidence interval for the mean height a tennis ball bounces. Using this interval, is the company making acceptable tennis balls? Explain your reasoning. 1. An employer wants to estimate the average heart rate of its employees. In a random sample of 22 of their employees, the average heart rate was found to be 76 with a standard deviation of 6.2. If heart rates in a random sample of people are normally distributed, find a 90% confidence interval for the average heart rate of the employees. 2. Using the information from number 2, how many employees would need to be sampled to ensure the average heart rate was within 3 bpm of the actual average? 3.

Explanation / Answer

Q1.
TRADITIONAL METHOD
given that,
sample mean, x =56
standard deviation, s =0.25
sample size, n =25
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 0.25/ sqrt ( 25) )
= 0.05
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 24 d.f is 2.797
margin of error = 2.797 * 0.05
= 0.14
III.
CI = x ± margin of error
confidence interval = [ 56 ± 0.14 ]
= [ 55.86 , 56.14 ]
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DIRECT METHOD
given that,
sample mean, x =56
standard deviation, s =0.25
sample size, n =25
level of significance, = 0.01
from standard normal table, two tailed value of |t /2| with n-1 = 24 d.f is 2.797
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 56 ± Z a/2 ( 0.25/ Sqrt ( 25) ]
= [ 56-(2.797 * 0.05) , 56+(2.797 * 0.05) ]
= [ 55.86 , 56.14 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 55.86 , 56.14 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean

the interval is above 55.5 inches, it is true

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