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olasank [31]
3 years ago
8

A No solution

Mathematics
2 answers:
malfutka [58]3 years ago
8 0
One solution I beleve
Anvisha [2.4K]3 years ago
6 0
I think Itsss B
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Here are summary statistics for randomly selected weights of newborn​ girls: nequals202​, x overbarequals28.3 ​hg, sequals6.1 hg
Lorico [155]

Answer:

The confidence interval is 27.5 hg less than mu less than 29.1 hg

(A) Yes, because the confidence interval limits are not similar.

Step-by-step explanation:

Confidence interval is given as mean +/- margin of error (E)

mean = 28.3 hg

sd = 6.1 hg

n = 202

degree of freedom = n-1 = 202-1 = 201

confidence level (C) = 95% = 0.95

significance level = 1 - C = 1 - 0.95 = 0.05 = 5%

critical value corresponding to 201 degrees of freedom and 5% significance level is 1.97196

E = t×sd/√n = 1.97196×6.1/√202 = 0.8 hg

Lower limit = mean - E = 28.3 0.8 = 27.5 hg

Upper limit = mean + E = 28.3 + 0.8 = 29.1 hg

95% confidence interval is (27.5, 29.1)

When mean is 28.3, sd = 6.1 and n = 202, the confidence limits are 27.5 and 29.1 which is different from 27.8 and 29.6 which are the confidence limits when mean is 28.7, sd = 1.8 and n = 17

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4 years ago
Please help and tell me how to do this <br> 20 POINTS
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The answers are :
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