Given that,
Sample size= 83
Mean number= 39.04
Standard deviation= 11.51
We know the critical t-value for 95% confidence interval which is equal to 1.989.
We also know the formula for confidence interval,
CI=( mean number - critical t-value*standard deviation/(sample size)^(1/2), mean number + critical t-value*standard deviation/(sample size)^(1/2))
So, we have
CI= (39.04 - 1.989*11.51/83^(1/2), 39.04 + 1.989*11.51/83^(1/2)
CI= (39.04 - 2.513,39.04 + 2.513)
CI= (36.527,41.553)
Therefore, 95% confidence interval for these data is (36.527,41.553), and this result interpret that the true value for this survey sample lie in the interval (36.527,41.553).
I can’t see your math lab problem, it’s not possible unless you could have shown me the footage based on the problem
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glide reflection
Step-by-step explanation:
got it right on the test!
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Step-by-step explanation:
