SOLUTION: A new piece of equipment is suspected of having faulty temperature control the desired temperature is supposed to be 68 dergree and a random sample of 45 readings are collected a

Question 1083817: A new piece of equipment is suspected of having faulty temperature control the desired temperature is supposed to be 68 dergree and a random sample of 45 readings are collected and the mean of these is 69.178 with a S.E of 0.483. at the level significance 0.05 . what would your concluison have been around this hypothisis
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
desired temperature is 68 degrees.

sample size is 45
mean of samples is 69.178
standard error of sample is .483

z-score of sample = (69.178 - 68) / .483 = 2.44

since the faulty readings could be high or low, a two tailed distribution is assumed.

a significance level of .05 would be divided by 2 to get an alpha of .025 on each end.

an alpha of .025 on each end gives you a critical z-score of plus or minus 1.96.

the z-score of 2.44 is greater than the critical z-score, therefore the results of the test are statistically significant and you can conclude that the temperature readings are not within the prescribed limits.

notice that you were given the standard error, and not the standard deviation of the sample.

standard error is equal to standard deviation divided by square root of sample size.

using this formula, the standard deviation of your sample would have been based on this formula to get:

se = sd / sqrt(ss)

se is standard error
sd is standard deviation
ss is sample size

.483 = sd / sqrt(45)

solve for sd to get sd = .483 * sqrt(45) = 3.24.

in order for the readings not to be considered out of limits, a test with a sample of 45 should have given you readings between what is shown below:

$$$

you could calculate this using the z-score formula.

that formula is z = (x-m)/se

at 95% two tailed confidence interval, the critical z-score is 1.96.

to find the high side limits for 95% confidence interval, formula becomes:

1.96 = (x-68)/.483

solve for x to get x = 1.96 * .483 + 68 = 68.95

similar calculations using a z-score of -1.96 will get you the low side raw score limit.

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