Here are summary statistics for randomly selected weights of newborn girls: n equals 245, x over bar equals 30.4 hg, s equals 6.8 hg. Construct a confidence interval estimate of the mean. Use a 98% confidence level. Are these results very different from the confidence interval 28.2 hg less than mu less than 31.6 hg with only 15 sample values, x over bar equals 29.9 hg, and s equals 2.5 hg?
Answers
Answer:
Use A 95% Confidence Level. Are These Results Very Different From The Confidence Interval 28.9 Hg< μ < 31.9 Hg With Only 12 Sample Values, X=30.4 Hg, And ... Here are summary statistics for randomly selected weights of newborn girls: n=235, x=30.5 hg, s=6.7 hg. Construct a confidence interval estimate of the mean.
Explanation:
Answer:
The confidence intervals for the two samples are different but not extremely so, given the sample sizes are also very different.
Explanation:
From the above question,
They have given :
n equals 245, x over bar equals 30.4 hg, s equals 6.8 hg.
Construct a confidence interval estimate of the mean. Use a 98% confidence level.
Are these results very different from the confidence interval 28.2 hg less than mu less than 31.6 hg with only 15 sample values, x over bar equals 29.9 hg, and s equals 2.5 hg
We can use the formula for a 98% confidence interval for the mean of a normally distributed population:
CI = x̄ ± (tα/2 * s / √n)
where tα/2 is the critical value from a t-distribution table with n-1 degrees of freedom and α/2 is the level of significance (in this case, 0.01).
For the first sample with 245 observations, the critical value at 98% confidence is 2.450. So,
CI = 30.4 ± (2.450 * )
= (29.54, 31.26)
For the second sample with 15 observations, the critical value at 98% confidence is 3.751. So,
CI = 29.9 ± (3.751 * )
= (27.45, 32.35)
The confidence intervals for the two samples are different but not extremely so, given the sample sizes are also very different.
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